Learn this powerful method for unlocking devilishly complicated problems about sets. 

Practice questions

First, try these challenging 700-level practice questions.

1) At Veridux Corporation, there are 250 employees. Of these, 90 are female, and the rest are males.  There are a total of 40 managers, and the rest of the employees are associates.  If there are a total of 135 male associates, how many female managers are there?

A. 15
B. 20
C. 25
D. 30
E. 35

2) In the county of Veenapaniville, there are a total of 50 high schools, of three kinds: 25 public schools, 16 parochial schools, and 9 private independent schools.  These 50 schools are divided between three districts: A, B, and C. District A has 18 high schools total. District B has 17 high schools total, and only two of those are private independent schools.  If District C has an equal number of each of the three kinds of schools, how many private independent schools are there in District A?

A. 2
B. 3
C. 4
D. 5
E. 6

This kind of question

It would not be surprising if you found one or both of those questions totally bewildering.  Without the right method, this type of question can be a nightmare.

First of all, let’s be clear about what we mean by “this type of question.”  These are set questions in which each element, each member, is classified according to two different variables, and each one of those variables has two or more categories.  For example, in question #1, the two variables are:

a) gender (male or female)
b) employee type (manager or associate)

Every one of those 250 employees is categorized both in terms of gender and in terms of employee type.  That’s key: every member is categorized in two different ways.  This type of question is screaming for a method of solution called the Double Matrix Method.

The Double Matrix Method: example #1

I will demonstrate the Double Matrix Method first by solving question #1, then question #2.  In the Double Matrix Method, you make a sectioned box, a rectangle with subdivisions.  The horizontal subdivisions will represent the categories of one variable, and the vertical subdivisions will represent the categories of the other variable.  Here’s the set-up for question #1: I temporarily put letters in the boxes, to facilitate talking about those boxes.

A is number of male managers, B is the number of female managers, and A + B = C is the total number of managers.  Similarly, D is the number male associates, and A + D = G is the total number of males.  In any row, you can add across the first two boxes and the sum will be the box on the right.  In any column, you can add the top two boxes, and the sum will be the bottom box.  The final number, I, is the “grand total”, the total number of employees in the whole problem, which obvious has to equal the sum of males + females, as well as the sum of managers + associates.

Here’s that matrix again with the numbers from the problem filled in — just the numbers stated in black-and-white in the problem.

Well, first let’s take care of the “totals”.  The numbers in the “totals” row must add up.  If 90 are females, the other 250 – 90 = 160 must be males.  Similarly, the numbers in the “totals” column must add up.  If 40 are managers, then the other 250 – 40 = 210 must be associates.

Now, in the “associate” row, 135 + E = 210, which means E = 75 — the other 75 associates must be female.

Now, to find B, which is what the question is asking, we need only look at the sum in the “female” column: B + 75 = 90, which means B = 15.  There are fifteen female managers in this company.  Thus, the answer = (A).

Instead of this route, once we found all the “totals” numbers, we could have used addition in the “male” column to find A = 25, then use addition across the “managers” to find B = 15 —- in fact, if you have a second route for finding the number, you should perform that too, to verify that you get the same number both ways.

The Double Matrix Method: example #2

Question #2 is very difficult: it’s debatable whether a set question this difficult even would appear on the GMAT.  Nevertheless, the Double Matrix Method unlocks it easily.  First of all, our matrix now has to have three categories for each variable.

As before, in any row, the sum across the first three entries will equal the right-most entry in the “totals” column; and similarly, the sum down the top three entries in any column will equal the bottom entry in the “total” row; and, of course, all the totals, in either the “totals” row or the “totals” column, have to add up to the grand total in the lower right-hand box.  Let’s start filling this in, following the information given in the question.  “In the county of Veenapaniville, there are a total of 50 high schools, of three kinds: 25 public schools, 16 parochial schools, and 9 private independent schools.”

“District A has 18 high schools total.”

“District B has 17 high schools total, and only two of those are private independent schools.”

“District C has an equal number of each of the three kinds of schools.”  Hmm.  That’s not a simple number.  Let’s put that on hold for a moment.  So far, from the matrix it looks like we could figure out the missing number in the “totals” row.  That number must be 15, because 18 + 17 + 15 = 50.

Now, go back to that last fact: “District C has an equal number of each of the three kinds of schools.”  Now that makes sense.  District C has 15 schools, so it must have 5 of each kind:

The question is asking: “how many private independent schools are there in District A?”  Well, now that’s just a matter of adding across the “private independent” row.  We know 2 + 2 + 5 = 9, so the missing number must be 2.

There’s still a good deal of information we can’t figure out from the information given in the question, but we are able to answer the specific question asked here.  There are 2 private independent schools in District A, so the answer = (A).

Notice, incidentally, if the question were framed differently, as a DS question, the Double Matrix method would allow us to see what questions we could and couldn’t answer based on the information given.  In a DS question, one would have to write out one Double Matrix for Statement #1, and a whole new Double Matrix for Statement #2.  Trying to save time by cramming all the information from both statements into a single matrix is an excellent way to make the classic mistake of conflating the information given in the two statements, one of the most common mistakes on Data Sufficiency.

Practice question

Here’s another very challenging question that you should be able to unlock with the help of the Double Matrix Method.

3) http://gmat.magoosh.com/questions/11

A good deal of success in math is simply about careful detail management.  If you are, by nature, someone already comfortable with math, highly organized and detail oriented, probably none of these mistakes will plague your work.  This is a post for folks who may be a little rusty in math, and need to be a little more careful with the basic work of detail management.

Dropping the negative sign

Suppose you are solving the equation

5 – 2x = 13

We want to isolate x.  One tactic would be to begin by subtracting 5 from both sides.  On the right, 13 – 5 = 8.  On the left, the 5′s cancel, but with what are we left?  It would be a mistake to subtract 5 and wind up with:

I have that in red, with an unequal sign, to emphasize that it is wrong.  Of course, the mistake is: when we subtract the 5 and get rid of it, the 2x term does not magically change from negative to positive.  It still has a negative sign in front of it.  Therefore, the next steps are:

-2x = 8

x = -4

Actually, if you notice any tendency toward making this mistake, I highly recommend: make your first step to add any subtracted variable to other side, to make it positive.  If your first step, automatically, is to make the variable positive, then  you will be considerably less likely to make this mistake.

Dividing by the numerator

Suppose you have this equation to solve:

Both sides are clearly divisible by 5, so one possible first step would be to divide by sides by 5.  On the right, .  On the left, the 5′s cancel, but the question is: what is on the left side after dividing by 5?  Just x?  No!  That’s a very tempting mistake to make!  In the equation above, x is in the denominator, and “being in the denominator” is not a condition that goes away just because a number in the numerator is cancelled.  If we divide by sides by 5, the proper result is

You can multiply by x, and divide by 4 to solve —- or, you simply could take the reciprocal of both sides (always a completely legitimate move when you have fraction = number or fraction = fraction).  Either way, the answer is .

Distributing a square

These next three mistakes are part of a broad category.  First of all, one of most fundamental laws underlying all arithmetic and algebra is a law called the Distributive Law.  Symbolically, it states

When read left-to-right, it is called distributing: we distribute A over (B + C).   When this same equation is read right-to-left, it is called “factoring out.”  See this post for a more extended panegyric to the Distributive Law, in a more advanced context.  That equation is 100% true, 100% of the time.  In words, we can say: multiplication distributes over addition and subtraction.  It’s one of the most fundamental laws in all of mathematics.

That pattern is very important, and has a wide variety of applications in elementary and advanced mathematics.  For some reason, though, this pregnant pattern is ripe for vast over-generalization.  The mind seems almost magnetically drawn to distributing all kind of things other than multiplication over addition and subtraction.

One example is: an exponent.  Suppose we are asked to expand algebraically the expression:

Be careful here, because unless you are a pro at math, your mind is going to be magnetically attracted to the wrong thing to do.  Here’s the wrong thing to do:

If you notice, this mistake involves following the Distributive Law pattern, but with an exponent rather than with multiplication.  That’s illegal.  What’s the correct procedure?  Well, squaring anything means multiplying it by itself, so the first step would be:

From there, you would FOIL out the expression.  That’s the step-by-step way to get to the answer.  It can be a very handy shortcut to have the following two patterns memorized.

The Square of a Sum:

The Square of a Difference:

Those formulas take into account the proper FOILing.  Memorizing these can be a time-saving shortcut and also might help you to remember to avoid Mistake #3 here.

Distributing a fraction

This is another mistake of the “over-extend the Distributive Law to regions that are not valid” variety.  Here is the succinct way to express this mistake.

In other words, you can neither combine nor separate fractions by additions in the denominator.  This one has far-reaching ramifications.  For example, in the following fraction …

… what can you cancel?  NOTHING!  If the 12 were over just the 3x, or if the 12 were just over the 8, then you would be able to cancel, but because you can’t separate the fraction, you can do absolutely no canceling.  BTW, in the fraction ….

… even though some cancellation is possible, you can’t do any while the fraction is still like this.  You have to separate it, by the addition in the numerator (a 100% legitimate move) and then you can cancel:

Another related mistake.  Suppose we have to solve the following equation.

While it’s true in general that you can take the reciprocal of both sides, unfortunately, you can only take the reciprocal of a single number or a single fraction, NOT a sum or difference of fractions.

The reciprocal of a sum is not the sum of the reciprocals.  How do you find the reciprocal of a sum?  You would have to add the two fractions, using a common denominator, combining them into a single fraction.  Here, by far the easiest solution would be to begin by subtracting 1/48 from both sides, and performing the fraction subtraction on the left side, so that you have a single fraction equals 1/x.  Then, you would legitimately be allowed to take the reciprocal of both sides to solve.

Distributing a root

The final mistake, yet another example of illegitimately over-extending the pattern of the Distributive Law, is distributing root signs.  Succinctly, this mistake says:

You cannot separate a square-root by addition or subtraction.  You can separate a root by multiplication or division: see this post for more on that.  You can see more about roots in general here.

If you have the equation…

… it is illegal to try to simplify that by taking a square-root of each term:

Think about it.  Mr. Pythagoras was a very intelligent individual.  If it were possible to simplify to a + b = c, then that’s how he would have stated the famous theorem.  The fact is: he had to state it as , because it is absolutely illegal to simplify that to a + b = c.

In fact, whereas the former is true for the three side of every right triangle, the latter is not true for the three sides of any triangle.  In fact, it constitutes a blatant violation of the Triangle Inequality, a law that is true for every possible triangle.

Summary

These are very common mistakes, and GMAT questions are often designed to elicit falling into one of these mistakes.  If you can simply learn and avoid these five mistakes, then you will avoid the common traps to which so many GMAT takers will readily succumb.

Few topics perplex students like circles on the GMAT. “What was the formula for area of a circle, again? I know pi is in there somewhere!” With a few simple insights, though, this entire topic simplifies.  First, let’s look at a problem. How do you find the do calculations for something like this?

Given that O is the center of the circle, and given the info in the diagram, how do we find (1) the length of arc AB? , and (2) the area of sector AOB (i.e. the shaded region)?  BTW, the word sector is the name for a “slice of pie” piece of a circle, and arclength is the curved length along the “crust” of the pie.   You can find a little more about arclength at this post.

Proportions

One of the greatest math “tricks” is proportional reasoning, and this is precisely what unlocks all the questions posed above.  Basically, to find either the length of an arc or the area of sector, we set up a part-to-whole ratio, and set this equal to a ratio for the angle.   In the diagram, the angle at the center of the circle is 80°.   A whole circle, all the way around, is always 360°, so this 80° angle takes up 80/360 = 8/36 = 2/9 of a circle.   Therefore, both the arclength and the area of the sector are 2/9 of their respective “wholes”.

Arclength

If arclength is the “part”, then the “whole” is the length all the way around the circle — we call that length circumference,  and its formula is

The part-over-whole of the angles has to equal the part-over-whole of these lengths:

For this diagram, this proportion becomes:

Area of a sector

If area of the sector is the “part”, then the area of the whole circle is the whole.  Of course, for the area of a circle, we use Archimedes‘ remarkable formula:

As with arclength, the part-over-whole of the angles has to equal the part-over-whole of these areas:

For this diagram, this proportion becomes:

BTW, a mental math tip — how did I just know, in my head, that 144/9 = 16, or in other words, that 9*16 = 144?  Well, think about it.

That logic makes it very easy to find factors of perfect squares.

Summary

If you remember these simple proportional reasoning tricks, you will be able to figure out anything the GMAT might ask about a part of a circle.   Here’s a free practice question on which to apply these skills.

1) http://gmat.magoosh.com/questions/87

What about GMAT circles slices you the wrong way? Let us know below.

On the GMAT Quantitative section, the Problem Solving questions are just the familiar five-choice multiple choice math problems you have seen on every standardized test since well before puberty.  Here, you have discovered a veritable treasure chest of Problem Solving sample questions.   Below is a link to thirty-two different articles on this blog, each with at least two Problem Solving questions.  The sample GMAT Problem Solving questions are often at the top of the article, although sometimes they are further down in the text.  The total number of sample Problem Solving problems available from this page is far more than 37, the total number of math questions you will see on a full Quantitative section of the GMAT.  In each blog, the solutions & explanations to the sample questions are at the ends of the articles.  (If the topic is less than crystal clear for you, you may find the article itself enlightening.)

1. Problems with Averages

http://magoosh.com/gmat/2012/gmat-averages-and-sums-formulas/

2. Distance, Rate, Time

http://magoosh.com/gmat/2012/gmat-distance-and-work-rate-formula/

3. Permutations & Combinations

http://magoosh.com/gmat/2012/gmat-permutations-and-combinations/

4. Factors & Prime Factorizations (five practice PS questions at the bottom of the article)

http://magoosh.com/gmat/2012/gmat-math-factors/

5. Advanced Geometric Solids

http://magoosh.com/gmat/2012/gmat-math-advanced-geometric-solids/

6. Estimation questions

http://magoosh.com/gmat/2012/the-power-of-estimation-for-gmat-quant/

7. Difficult Dice Questions

http://magoosh.com/gmat/2012/gmat-probability-difficult-dice-questions/

8. Difference of Two Squares

http://magoosh.com/gmat/2012/gmat-quant-difference-of-two-squares/

9. Sequences (five PS practice questions scattered through article)

http://magoosh.com/gmat/2012/sequences-on-the-gmat/

10. Remainders

http://magoosh.com/gmat/2012/gmat-quant-thoughts-on-remainders/

11. Work & Work Rate

http://magoosh.com/gmat/2012/gmat-work-rate-questions/

12. Circle & Line Diagrams

http://magoosh.com/gmat/2012/circle-and-line-diagrams-on-the-gmat/

13. Polygons

http://magoosh.com/gmat/2012/polygons-and-regular-polygons-on-the-gmat/

14. Set Problems, with Double Matrix Method

http://magoosh.com/gmat/2012/gmat-sets-double-matrix-method/

15. Set Problems, with Venn Diagrams

http://magoosh.com/gmat/2012/gmat-sets-venn-diagrams/

16. Scale Factor & Percent Change

http://magoosh.com/gmat/2012/scale-factors-on-the-gmat-percent-increases-and-decreases/

17. Standard Deviation

http://magoosh.com/gmat/2012/standard-deviation-on-the-gmat/

18. Radicals

http://magoosh.com/gmat/2012/simplifying-radical-expressions-on-the-gmat/

19. Function Notation

http://magoosh.com/gmat/2012/function-notation-on-the-gmat/

20. Algebraic Factoring

http://magoosh.com/gmat/2012/algebra-on-the-gmat-how-to-factor/

21. Hard Factorial Problems

http://magoosh.com/gmat/2012/gmat-factorials/

22. Backsolving from the answers

http://magoosh.com/gmat/2012/gmat-plugging-in-strategy-always-start-with-answer-choice-c/

23. Distance in the x-y plane

http://magoosh.com/gmat/2012/gmat-coordinate-geometry-distance-between-two-points/

24. Pythagoras!

http://magoosh.com/gmat/2012/pythagorean-triplets-to-memorize-for-the-gmat/

25. Lines in the x-y plane

http://magoosh.com/gmat/2012/gmat-math-lines-slope-in-the-x-y-plane/

26. Tricks for Calculating Combinations

http://magoosh.com/gmat/2012/gmat-math-calculating-combinations/

27. Parallel & Perpendicular Lines and Midpoints in the x-y plane

http://magoosh.com/gmat/2012/gmat-math-midpoints-and-parallel-vs-perpendicular-lines/

28. Probability: AND & OR Rules

http://magoosh.com/gmat/2012/gmat-math-probability-rules/

29. Probability: “at least” statements

http://magoosh.com/gmat/2012/gmat-math-the-probability-at-least-question/

30. Probability: counting problems

http://magoosh.com/gmat/2013/gmat-probability-and-counting-techniques/

31. Hard counting problems

http://magoosh.com/gmat/2013/gmat-counting-with-restrictions/

32. Probability: geometric probability

http://magoosh.com/gmat/2013/geometric-probability-on-the-gmat/

Also check out these GMAT Probability questions.

First, try these challenging GMAT practice problems. Give yourself a strict time-limit of six minutes for the set.

1) If A is a number, which of the following must be true for any A?

2) If F and G are integers, with F < G, which of the following must be true?

3) If J is an integer, which of the following must be true?

4) If p and q are two different odd prime numbers, such that p < q, then which of the following must be true?

Attacking “must be true” problems

For many math problems, one has to focus on finding the one right answer. That approach, with these problems, is highly problematic. If you are super-talented with math, fluent in Algebra, then perhaps you can scan the list of choices and immediately spot the correct answer. If you are talented enough to do that, then good for you!

For most folks, that approach is simply not feasible. The much more efficient solution involves elimination. You see, if any statement “must be true”, then it will be true for any value you pick. It takes incredible skill (or fantastic luck) to pick a single value that eliminates four answer choices and leaves only one. If that happens by chance, great, but don’t focus on that. Focus on speed and efficiency. Pick one value, eliminate what you can, then pick another, eliminate more, etc. Whittle down the choices until you are left with one. That’s the most efficient approach for these problems.

You have to pick values to plug in. At the start of picking, pick very easy values — with a few easy choices, you should be able to eliminate two or three answer choices, making your overall job much easier.

What values to pick?

If you picked the same values for all four questions above, that’s a problem. Those questions were specifically written so that the allowable numbers would be different in different questions.
When you hear “A is a number”, of what do you think? Remember that the general word “number”, like the general term “human being”, includes all types. Just as the individuals under the term “human being” constitute a bewildering variety, so do the citizens of the realm of “numbers.” Numbers include positive & negative, wholes & fractions & decimals, square-roots, pi, etc. etc. It’s a realm of perfect democracy — for example, -(pi)/5 is just as much a number as is 8. If, when a GMAT problem says “number”, your mind defaults to {1, 2, 3, 4, ..}, the “counting numbers”, then with all due respect, that’s really a third-grade way of thinking about the word “numbers”, and the GMAT will viciously punish that kind of thinking. You always must be aware of all possibilities for any category.

numbers = everything, positive/negative, zero, fractions, decimals, everything on the continuous infinity of the number line.

integers = positive and negative whole numbers = { … -3, -2, -1, 0, 1, 2, 3, …}

positive integers = the counting numbers, 1, 2, 3, 4, ..}; this list does NOT include zero, because zero is neither positive nor negative.

Remember that “odd integers” includes both positive and negatives, and “even integers” includes positives & zero & negatives. Prime numbers are, by definition, a subset of positive integers. There are no negative primes, and neither zero nor one is a prime number. It’s a very good idea to have the first few prime numbers memorized:

Primes = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, …)

Notice that 2 is the only even prime number (all other even numbers are divisible by 2, and hence are not prime). After 2, the primes are an irregularly spaced set of odd numbers that continues to infinity. In fact, let’s talk a moment about their pattern

The pattern of prime numbers

In this section, I am going to dip into very advanced math, far far beyond the GMAT realm. In modern mathematics, most work around the “pattern of prime numbers” revolves around the Riemann Hypothesis, the single hardest un-answered question in all of mathematics. The great mathematician Bernhard Riemann proposed this grand question as a conjecture in an 1859 paper, and it has utterly baffled the most brilliant mathematical minds on the planet every since. Forget about answering the question — you need close to a Ph.D. in mathematics just to understand what the question is asking and what it has to do with prime numbers! (A brief graph of the central player of the Riemann Hypothesis, the Riemann Zeta Function, is shown below.)

All of which is a long-winded way of saying: there is no easy algebraic formula that always produces prime numbers. A simple pattern for prime numbers absolutely does not exist. Thus, in a “must be true” question, any answer of the form “[simple algebra expression] is a prime number” cannot possibly be true all the time. That absolutely has to be a wrong answer.

Summary

If you had any big “aha!” while reading this article, you may want to give those problems another glance before reading the solutions below. If you have any further questions or observations, please let us know in the comments section below.

Practice problem explanations

1) First of all, notice that A is a “number” — could be positive or negative or zero, whole or fraction or negative. Let’s first try a very easy choice: A = 0. Then, for the five answer choices, we get:
(A) undefined
(B) false
(C) true
(D) true
(E) undefined

Remember, if some choice leads to an “undefined” value, a divide-by-zero error, then it is not “true” for that value. Error is just as good as false for eliminating answers. We are down to (C) & (D). Try another relatively simple choice, A = -2.
(C) 1 = 1, true
(D) +2 ≠ -2, false

Remember that if A is negative, a negative squared is a positive, and the square-root sign always has a positive output. Therefore, the left side of the (D) equation is always positive, and can’t be true if A is negative.
Answer = (C).

2) For this problem, both F & G are integers — they could be positive or negative or zero, but they must be whole numbers. Here’s a suggestion — one good choice would be to pick a positive number with smaller absolute value and a negative number with a larger absolute value: say, F = -4 and G = 1. Let’s true those in the answers:
(A) 16 < 1 false
(B) -64 < 1 true
(C) 4 < 1 false
(D) 4 < 1 false
(E) 1-16 = -15 > 0 false

With one magical choice, we were able to eliminate four answer choices. Keep in mind, the combination small positive/big negative is a good one, and so is pairing a negative number with 0 (e.g. F = -4, G = 0), because zero is greater than any negative number.
Answer = (B)

3) Here, J is an integer, so it can be positive or negative or zero, but not a fraction or a decimal. Zero is always a special case among integers, so let’s start there. J = 0: then,
(A) undefined error
(B) 0 > 0 false
(C) 0 > 0 false
(D) 1 > 0 true
(E) -1 > 0 false
Here, the special case was enough to eliminate four answers. Again, remember, if some choice leads to an “undefined” value, a divide-by-zero error, then it is not “true” for that value. Error is just as good as false for eliminating answers.
Answer = (D)

4) Here, p and q are odd prime numbers. First of all, eliminate (A), because there’s no simple algebraic rule that generates prime numbers. The easiest choices for numbers are p = 3 and q = 5. Try this:
(B) 3 + 5 = 8 is divisible by 4 — true
(C) 5 – 3 = 2 is not divisible by 4 — false
(D) 5 + 3 + 1 = 9 = (5^2) – (4^2) — true
(E) (3^2) + (5^2) = 34

This last number requires comment. Every odd number is the difference between two perfect squares, because adjacent squares (n + 1)2 – n2 = 2n + 1. When we subtract two even squares or two odd squares, the number is always divisible by 4. Thus, 34 cannot be the difference of any pair of squares. Thus, on the basis of this choice, we can eliminate (E). We are left with (B) & (D).

Try p = 3 and q = 7.
(B) 3 + 7 = 10 is not divisible by 4 — false
(D) 3 + 7 + 1 = (6^2) – (5^2) — true

In fact, since p & q are both odd, (p + q + 1) must be odd, and as stated above, any odd number is the difference of two perfect squares. That’s why (D) must be true.
Answer = (D)

First, a few practice GMAT math questions.

1) Last year, the price of a vacation package was P.  At the beginning of this year, the price went up 40%.  Lucas used a travel voucher to purchase the vacation package at 30% off this year’s price.   In terms of P, how much did Lucas pay?

(A) P + 10

(B) 1.1*P

(C) 1.12*P

(D) 0.9*P

(E) 0.98*P

2) Given that xy > 0, which of the following must also be greater than zero?

3) The numbers m, n, and K are all positive integers.  Given that m is a factor of K, that n is also a factor of K, and m < n, which of the following must also be a positive integer factor of K?

4) Marcia took a trip consisting of three segments at three different speeds: she drove a distance of (5D) at a speed of (2v), then a distance of (4D) at a speed of (3v), then a distance of D at a speed of (6v).  In terms of D & v, what was the total time of Marcia’s trip?

Answers and explanations will follow this article.

Picking numbers

It’s absolutely true that, when the GMAT gives a problem with variables in the answer choices, one possible approach is to pick numbers for the variables.  Students sometimes ask, “How do I pick numbers? Can you give me a general strategy for picking good numbers in all cases?”  Well, let’s start with the second question, which has a crystal clear answer: “No!”  You see, there are several different kinds of questions in which picking numbers could be an appropriate strategy, and the best numbers to pick in any case depend much more on the specifics of the question.  Some numbers that would be excellent choice in one question would be the worst possible choice in another question.  For many questions, the best numbers to pick are inextricably tied up with the mathematics in the question itself.  Some of this I will explain in the course of this article, and other things I will explain in the solution.

Numbers

First, let’s start with how you think about the word “number”.  What pops into your head when I say that word?  If what pops into your head is the set {1, 2, 3, 4, 5, …}, then with all due respect, you are operating with a third-grade level definition of the word “number”, and the GMAT Quant section will mercilessly punish you for such naïveté.  The word “number” is a broad category, encompassing everything on the number line: zero is a number; 4/7 is a number; negative four is a number; negative one fifth is a number; pi is number; the square-root of 17 is a number; etc. etc. All of these possibilities must leap to mind when a problem mentions a “number”.  It’s good to appreciate the categories of numbers, as well as their different properties.

Number sense is the ability to see patterns among numbers and sense which ones are best for a problem.  The person who asks the question at the top, how to pick numbers — what that person really needs is number sense!!  In other words, the person who asks for a general rule for picking numbers is really, at a very profound level, asking the wrong question.  Although that person might not even be familiar with the terms, what that person really needs to develop is number sense and mathematical thinking in general. These two posts say more about these larger issues, but I will say a few general things about picking numbers here.

Percent problems

One of the most braindead strategies is picking 100 for the starting value in every percent problem.  Everyone has heard this strategy.  Everyone who picks numbers does precisely this, as if it were a reflex.  That, in a nutshell, is the problem.  You see, if the test writers know that everyone makes the same choice, then they design answer choices designed to trip up people who make that all-too-obvious choice.   Think a little out of the box here.  Instead of 100, pick 500 or 1000 — you should be able to figure out round number percents with those relatively quickly.

Extending that strategy even a little further: if anything is a super-obvious choice to pick, then rather than pick that super-obvious thing, pick something a shade away from super-obvious.

If the question is one of a very typical type, a percent increase or decrease followed by another percent increase or decrease, then I would strongly advise against picking numbers.  Instead, I would recommend using multipliers to handle percent changes.

Dimensional analysis

If different variable represent quantities of different units, then make sure you understand what combination of them will have the correct units for the answer choice.   In Physics, this thinking about the units of different quantities is called dimensional analysis.   For example if J is in units of miles per gallon, and M is in units of miles, then to get units of gallons, I would need M/J — if the question is asking for “how many gallons”, then the correct answer would have to be a number times M/J.   Any other combinations (J/M, M*J, etc.) are automatically incorrect.  This helps you eliminate answers choices, which makes the process of plugging in numbers easier.

Know the traps

Many different math questions have very predictable traps: the mistake that a large number of test takers routinely make, time and time again.  GMAT problems always have answer choices that reflect these traps: in some cases, it’s as if the GMAT simply sets up a trap answer like a big butterfly net, and they watch hordes of test takers run headlong into it.  For every such trap you learn to recognize and avoid, you will be that much ahead of so many other test takers.

One big trap involves a sequence of percent increases & decreases.  There are a few common ones involving fractions.  Still more involve exponent properties and powers.  Some very common ones involve incorrect variations on three basic algebra formulas.  Here’s a post on the most common math mistakes.

Avoid one

Avoid picking 1 as a value for a variable.  The reason is: one has so many special properties that no other number has.   For example, all powers of 1 equal one: one is the only number that has this particular property.  Think if you had to distinguish choices involving different powers of that variable: picking one for the value would make all the power equal the same thing, making it impossible to distinguish which answers should be eliminated.  One is often the worst possible choice.

The only exception to this is — sometimes, it’s worthwhile to pick very easy numbers, like 1 and 2, just to do a very quick elimination of a couple of answers.  If that’s your goal, 1 can be a perfectly acceptable choice on a first pass.  It will never be a choice that eliminate everything, but if your goal is to begin by eliminating low-hanging fruit, then by all means pick it.

To pick or not to pick

Finally, let’s step back.  Why do folks pick numbers for problems with variable?  One reason is that sometimes, quite rarely, one can either eliminate many answers or even get to a single right answer very quickly by picking numbers.  Most often, though, this is not the case.  In fact, often it’s much quicker and much easier to do the algebra directly.  After all, this is precisely why mathematicians invented algebra in the first place: it’s a powerful efficient problem solving system.

Most of the time, most habitual number-pickers are choosing this strategy, not because it’s necessarily faster or easier or more efficient or better in any particular way, but simply because these folks are absolutely petrified of algebra.  For these folks, algebra is a fire-breathing dragon, and folks will do anything, no matter how awkward or cumbersome, simply to avoid facing that dragon.   If this describes you, then I have some hard medicine for you: do algebra.  From this point forward, in any problem with variable, you are forbidden to pick numbers, and you must practice the algebra.  You see, for the GMAT, you will have to know some algebra, and the only way you get better at algebra is by practicing it at every opportunity you can get.

Summary

If you had any flashes of insight while reading this article and exploring the links, you may want to give the questions at the top another look.   Here’s another problem of another sort, from inside Magoosh:

5) http://gmat.magoosh.com/questions/123

Finally, if you have any questions, please let us know in the comments section at the bottom.

Practice problem explanations

1) First of all, let’s talk about the trap.  Even if you don’t learn how to solve this problem, learn to recognize this trap, possibly the single most enticing trap on the entire GMAT.  When a percent increase or decrease is followed by another percent increase or decrease, you cannot simply add or subtract the two percents.  If you increase something by 40% and then decrease it by 30%, that is absolutely NOT a 10% increase.   That is the trap answer to which hordes of GMAT test takers are drawn like bugs to a bug zapper.

Problems are this sort are not ones on which I would recommend picking numbers at all.  Instead, I would recommend using multipliers for the percent changes.  The multiplier for a 40% increase would be (1 + 0.40) = 1.40.  The multiplier for a 30% decrease would be (1 – 0.30) = 0.70.  The net effect of both would be

Final price = (1.40)*(0.70)*P = 0.98*P

Answer = (E)

Even though it’s not as efficient, here’s a picking-numbers approach, just so you can see how the numbers play out. Let p = $1000.   This increases by 40%, to $1400.  Now, this is going to decrease by 30% — but notice!  It decreases not by 30% of the original, $1000; rather, it decrease by 30% of the new price, $1400.  After the 40% increase, the price of $1000 doesn’t even exist anymore: the only price that exists at that point is the $1400 price; that’s what decreases by 30%.  Well, 10% of $1400 is $140, and so, 30% = 3*10% = 3*$140 = $420.  The price decrease to $1400 – $420 = $980, which is 98% of P or 0.98*P.

2) This is a “must be true” problem.  First of all, remember that if the product of two numbers is positive, either both are positive or both are negative.  Since we want something that “must be greater than zero”, i.e. must be positive, I am going to pick negative numbers for my values of x & y.  Probably all five answer choices would be positive if I picked to positive numbers.  I am going to pick negatives, because any answer choice that winds up negative will be something I can eliminate, and on “must be true” problem, my goal is to eliminate answers.

I’ll start with very simple choices, x = –1 and y = –2.

(A) (–1) + (–2) = –3 — negative, so we can eliminate this.

(B) (+1) + (–2) = –1 — negative, so we can eliminate this.

(C) (-3)^2 + 1 = 9 + 1 = 10 — can’t eliminate yet

(D) 1 – 4 + 100 = +97 — can’t eliminate yet

(E) (-1)^2 = 1 — can’t eliminate yet

So, with one easy choice, we have eliminated two answer choices, (A) & (B).  I notice that x-squared minus y-squared, in choice (D) — that expression could be negative.  If it’s a small negative, then adding a hundred would make it positive, so to eliminate that, we need the subtracted term to be big.  Use x = –1 and y = –30.

(D) (+1) – 900 + 100 = –799 — negative, so we can eliminate this.

Now, we are down to just (C) & (E).  Notice, both of them have something in parentheses squared.  Of course, if we square a positive, we get a positive, and if we square a negative, we get a positive, so it would seem that square anything gives us a positive, right?  Almost.  The one exception: when you square zero, you get zero, and zero is not positive.  Zero is neither positive or negative.  Therefore, if either of these answers could have an output of zero, then it is not always positive.  Notice, if we pick x = –2 and y = –2, then the sum in the parentheses of (E) is, ((–2) – (–2)) = 0, and the sum in the parentheses of (C) is, ((–2) + (–2)) = –4

(C) +16 — can’t eliminate

(E) 0 — because zero is not positive, we can eliminate this.

In fact, if we add two negative numbers, or any two positive numbers, then we never can get zero.  Therefore, the inside of the parentheses of (C) will never equal zero: it will always be a non-zero number; and therefore, when we square, it must be positive.  That’s why (C) is the only possible answer.

3) First, pick a K with a lot of factors: say K = 24.  Now, for m and n, pick two factors of 24 that are not factors of each other — for example, pick the two biggest factors of 24 less than 24 itself: n = 12 and m = 8

(A) 8 + 12 = 20 — not a factor of 24

(B) 8*12 = 96 — not a factor of 24

(C) (12*12)/8 = (3*12)/2 = (3*6) = 18 — not a factor of 24

(D) 24/8 = 3 — that’s a factor of 24

(E) 24/96 — not even an integer, so certainly not a factor

The only one that works is (D), the correct answer.

Notice that, in (C), to find twelve squared divided by eight, the absolute worst, most unstrategic thing one could do would be to square 12, get 144, and then have to divide that by 8 without a calculator.  Notice, instead, I left the two 12′s unmultiplied while I canceled factors from the denominator, easily reducing the entire process to single-digit multiplication.  Cancel before you multiply is one of the biggest non-calculator tricks on GMAT math.

4) First of all, let’s use a little dimensional analysis, discussed above. If D is a distance, and v is a speed, we know v = D*T, or T = D/v.  Anything with a D in the numerator and a v in the denominator will be a time, but if those two are reversed, it will not have the right units.  We can eliminate (B) & (D) right away on these grounds.

On this one, I recommend an algebraic approach.

(distance #1) = (rate #1)*(time #1)

5D = (2v)*(time #1)

time #1 = (5D)/(2v)

(distance #2) = (rate #2)*(time #2)

4D = (3v)*(time #2)

time #2 = (4D)/(3v)

(distance #3) = (rate #3)*(time #3)

D = (6v)*(time #1)

time #3 = (D)/(6v)

TOTAL TIME = (time #1) + (time #2) + (time #3)

= (5D)/(2v) + (4D)/(3v) + (D)/(6v)

= (D/v)[5/2 + 4/3 + 1/6]

= (D/v)[15/6 + 8/6 + 1/6]

= (D/v)[24/6]

= (D/v)[4]

= (4D)/v

Answer = (A)

While it’s somewhat less efficient, here’s a picking number approach.  We want D to be a number with lots of factors, including a factor of 6, because it’s going to be divided by 2 and 3 and 6 times something.  Let’s pick D = 60 and v = 10

First leg: distance = 5D = 300 mi, at a speed of 2v = 20 mph, that’s T = D/V = 300/20 = 15 hours.

Second leg: distance = 4D = 240 mi, at a speed of 3v = 30 mph, that’s T = D/V = 240/30 = 8 hours.

Third leg: distance of D = 60 mi, at a speed of 6v = 60 mph —that will take exactly one hour.

TOTAL TIME = 15 hr + 8 hr + 1 hr = 24 hours

Well, D/V = 60 mi/10 mph = 6 hours, and clearly, we need to multiply this by 4 to get 24 hours, not by 10/11 or 11/10, so (A) is the answer.

First, a bank of eight practice problems

1) A certain zoo has mammal and reptiles and birds, and no other animals.  The ratio of mammals to reptiles to birds is 11: 8:5.  How many birds are in the zoo?

Statement (1): there are twelve more mammals in the zoo than there are reptiles

Statement (2): if the zoo acquired 16 more mammals, the ratio of mammals to birds would be 3:1

2) In a certain large company, the ratio of college graduates with a graduate degree to non-college graduates is 1:8, and ratio of college graduates without a graduate degree to non-college graduates is 2:3.  If one picks a random college graduate at this large company, what is the probability this college graduate has a graduate degree?

3) Dan’s car gets 32 miles per gallon.  If gas costs $4/gallon, then how many miles can Dan’s car go on $50 of gas?

1. 61.25 miles
2. 256 miles
3. 400 miles
4. 1600 miles
5. 6400 miles

4) For a certain concert, the price of balcony tickets was exactly half the price of orchestra tickets.  The ratio of balcony to orchestra tickets sold was 3:2.  What was the price of one orchestra ticket?

Statement (1): the total revenue taken in from tickets of both kinds was $4200

Statement (2): the difference between the number of balcony tickets sold and the number of orchestra tickets sold was 40

5) At a certain high school, there are three sports: baseball, basketball, and football.  Some athletes at this school play two of these three, but no athlete plays in all three.  At this school, the ratio of (all baseball players) to (all basketball players) to (all football players) is 15:12:8.  How many athletes at this school play baseball?

Statement (1): 40 athletes play both baseball and football, and 75 play football only and no other sport

Statement (2): 60 athletes play only baseball and no other sport

6)  In a certain class, the ratio of girls to boys is 5:4.  How many girls are there?

Statement (1): If four new boys joined the class, the number of boys would increase by 20%.

Statement (2): If the number of girls increases by 50%, then after such an increase, the probability that a randomly chosen student would be a boy would be 8/23

7) If 28 passes to a show cost $420, then at the same rate, how much will 42 passes cost?

1. $500
2. $560
3. $630
4. $700
5. $840

8) Metropolitan Concert Hall was half full on Tuesday night.  How many seats are in the Hall?

Statement (1): If the number of people in the Hall increased by 20% from Tuesday night to Wednesday night, then the Hall would be 60% full on Wednesday night.

Statement (2): If 20 more people showed up on Tuesday night, that would have increased the number of people in the Hall by 4%.

Answers will come at the end of the article.

Ratios and proportions: a review

Probably ratios and proportions have been on your radar since some time in grade school or middle school, when they are introduced.  You remember there are a lot of “mathy” facts related to these things, but it’s a bit blurry.  Here are a few quick facts as reminders.

1. What is a ratio?  Fundamentally, a ratio is a fraction, and is subject to all the laws of fractions.   “Ratios” and “fractions” are mathematically identical.

2. What is a proportion?  Is it the same as a ratio?  No, a proportion is NOT the same as a ratio.  Whereas a ratio is single fraction by itself (e.g. 1/3), a proportion is an equation that sets two ratios equal to each other (e.g. 1/3 = 4/12).   See the fraction post for a refresher on what you can and can’t cancel in a proportion.

3.  When geometric figures are similar, the sides are proportional.  Geometric similarity is a topic rife with ratios and proportions.  One helpful idea discussed in that post is the idea of a scale factor, which, it turns out, is helpful in many many proportional situations well beyond anything having to do with geometry.

4.  Percents and probabilities are specialized cases of ratios, and either gives you very much the same kind of information.

Those are four simple ideas, and there’s one more, but it takes a little setting up to express.   Let’s divide the mathematical information that can appear in a problem into two categories.  The first category is “ratio  information”, and this includes any statements about percents or probability information.   The second I’ll call, for lack of a better term, “count information” — not a percent or ratio, but an actual count — i.e. how many people or animals or whatever; it could be how many in any particular group, the sum or difference of multiple groups, or how many are in the whole population.   This leads us the final important simple idea:

5.  To get count information as an output, you need some count information as an input.  If you all you have is ratio information as an input, it is impossible to get count information as an output.

Ratios and portioning

Suppose, in some population, there are three kinds of things, A & B & C, and they are in a proportion of 3:8:4.  That’s ratio information.  Suppose we want to know either the percent that A makes of the whole, or the ratio of A to the whole.  That’s also ratio information, so we should be able to calculate it from that given ratio.

It can be very helpful to understand ratios in terms of “parts”.  For every 3 parts of A, there are 8 parts of B and 4 parts of C.  To get the whole in the same ratio, we simply add up the parts — 3 + 8 + 4 = 15 parts.  Keep in mind, we have zero information about the actual size of the population — we have no count information.  This simply indicates the size of the population in the same ratios, so we could say A to the whole is 3/15 = 1/5, and B to the whole is 8/15, and C to the whole is 4/15.  This means that A is 1/5 of the whole, or 20%.

Similarly, suppose girls are 4/7 of a class.  Girls are four parts, and the whole is seven parts, so boys must be the other three parts.  (Here, for mathematical simplicity, I will conform to very conventional gender assumptions, with sincere apologies to all transgender and transsexual readers.)  Proceeding, we see that boys must be 3/7 of the class, the ratio of boys to girls must be 3:4.  Thinking about “parts” and portioning can be a powerful way to expand the information you get from any given ratio.

Summary

If you had any realizations while reading this blog, you may want to go back and give the practice problems another glance, before proceeding to the solutions.  Here’s another practice question, involving two less-than-lovable baseball teams:

9) http://gmat.magoosh.com/questions/60

If you have any questions about this article, please let us know in the comment section at the bottom!

Practice problem explanations

1) A short way to do this problem.  The prompt gives us ratio information.  Each statement gives use some kind of count information, so each must be sufficient on its own.   From that alone, we can conclude: answer = D.  This is all we have to do for Data Sufficiency.

Here are the details, if you would like to see them.

Statement (1): there are twelve more mammals in the zoo than there are reptiles

From the ratio in the prompt, we know mammals are 11 “parts” and reptiles are 8 “parts”, so mammals have three more “parts” than do reptiles.  If this difference of three “parts” consists of 12 mammals, that must mean there are four animals in each “part.”  We have five bird “parts”, and if each counts as four animals, that’s 5*4 = 20 birds.  This statement, alone and by itself, is sufficient.

Statement (2): if the zoo acquired 16 more mammals, the ratio of mammals to birds would be 3:1

Let’s say there are x animals in a “part” — this means there are currently 11x mammals and 5x birds.  Suppose we add 16 mammals.  Then the ratio of (11x + 16) mammals to 5x birds is 3:1. —-

(11x + 16)/(5x) = 3/1 = 3

11x + 16 = 3*(5x) = 15x

16 = 15x  – 11x

16 = 4x

4 = x

So there are four animals in a “part”.  The birds have five parts, 5x, so that’s 20 birds.   This statement, alone and by itself, is sufficient.

Both statements sufficient.  Answer = D.

2) We are given ratio information, and we are asked for ratio information: a probability.  That’s fine.  For simplicity, let’s use the abbreviations:

P = college graduates with a graduate degree

Q = college graduates without a graduate degree

R = non-college graduates

The two ratios we are given is

P:R = 1:8

Q:R = 2:3

We have combine the ratios, by making the common term the same.   The non-college graduates, R, are the common member, accounting for 8 parts in the first ratio, and 3 parts in the second, so we have to multiply the first ratio by 3/3, and the second by 8/8.

P:R = 1:8 = 3:24

Q:R = 2:3 = 16:24

P:Q:R = 3:16:24

Now, the probability about which the question asks  is about only college graduates, so ignore the non-college graduates, and just focus on the ratio among college graduates:

P:Q = 3:16

There are 3 + 16  = 19 parts in total, and of those, 3 are the folks in P, so that’s a probability of 3/19.  Answer = (D).

3) To get from dollars to gallons, we have to start with the $50, and divide by ($4/gallon) — that cancels the unit of dollars, and leaves us with gallons —– 50/4 gallons.  (For the moment, I’ll leave it as that un-simplified fraction).

Now, we need to get from gallons to distance.  We multiply by (32 miles/gallon), to cancel the units of gallons, and leave only miles —-

(50/4) gallons*(32 miles/gallon) = (32*50)/4 miles = (8*50) miles = 400 miles

Notice, we didn’t do anything with the four in the denominator until we got a 32 in the numerator with which we could cancel.

Answer = (C)

4) We know the price of a balcony ticket, B = (1/2)*C, or 2B = C, where C = the price of orchestra ticket.  The 3:2 ratio tells us that, for some n, the concert sold 3n balcony tickets and 2n orchestra tickets.

Statement (1): the total revenue taken in from tickets of both kinds was $4200

We know total revenue would be (3n) balcony ticket plus (2n) orchestra tickets:

(3n)*B + (2n)*C = (3n)*B + (2n)* 2B = 3nB + 4nB = 7nB = $4200,  or nB = 600.  The problem with this — we have two variables, n and B.  In one extreme case, we could say n = 1 (sold 3 balcony ticket and 2 orchestra tickets), and a balcony ticket cost $600.  At another extreme, we could say n = 600 (sold 1800 balcony ticket and 1200 orchestra tickets) and a balcony ticket cost $1.  This information alone does allow us to determine a definitive answer to the prompt question.   This statement, alone and by itself, is insufficient.

Statement (2): the difference between the number of orchestra tickets sold and the number of balcony tickets sold was 40

Well, balcony tickets are 3 parts, and orchestra tickets are two parts, so there’s one part of difference between them.  The statement lets us know: one part = 40, so that allows us to figure out: we sold 3*40 = 120 balcony tickets, and 2*40 = 80 orchestra tickets.  That’s great, but unfortunately, with this statement, we get absolutely no financial information, so we can’t solve for a price.   This statement, alone and by itself, is insufficient.

Combined statements: now, put this altogether.  From the first statement, we got down to the equation nB = 600, and the second statement, in essence, tells us n = 40.  Therefore, B = 600/40 = 60/4 = $15.  That’s the price of a balcony ticket.  The price of an orchestra ticket is twice that, $30.  With both pieces of information, we were able to solve for this.   Together, the statements are sufficient.

Statement are sufficient together but not individually.  Answer = C

5) This is a tricky one.  The ratio 15:12:8 double-counts some students, and even triple-counts those students who play all three sports.   In terms of a triple Venn diagram:

So, in that diagram

a = folks who play baseball only

b = folks who play football only

c = folks who play basketball only

d = folks who play baseball and football only

e = folks who play baseball and basketball only

f = folks who play football and basketball only

g = folks who play all three sports

At the outset, that’s seven unknowns!!

Now, notice:

everyone who plays baseball = a + d + e + g

everyone who plays basketball = c + e + f + g

everyone who plays football = b + d + f + g

So, the ratio given in the problem is:

(a + d + e + g):(c + e + f + g):(b + d + f + g) = 15:12:8

Of course, these are all fractions, so we can’t simplify, for example, by subtracting g from each term —- we are not allows to simply a fraction by subtracting equal terms from the numerator and the denominator!   In fact, there is no way to simplify.

We would like to find the total number of baseball players, a + d + e + g.

Statement #1 tells us that d = 40 and b = 75.  Thus

total number of baseball players = a + 40 + e + g

We don’t have enough information to calculate this, and we don’t have enough ratio information to solve.   This statement, alone and by itself, is insufficient.

Statement #2 tells us a = 60.  Thus

total number of baseball players = 60 + d + e + g

We don’t have enough information to calculate this.  This statement, alone and by itself, is insufficient.

Combined statements: Now we know a = 60, b = 75, and d = 40.

total number of baseball players = 60 + 40 + e + g = 100 + e + g

total number of football players = 75 + 40 + f + g = 115 + f + g

We know the ratio of these two quantities, baseball to football, is 15:8.  Unfortunately, that would give us only one equation for three unknowns, e & f & g.  If we can’t solve for these, then we can solve for the total number of baseball players.   Even with both statements combined, we cannot determine the answer.

Both statements combined are insufficient.   Answer = E

6) Statement (1): If four new boys joined the class, the number of boys would increase by 20%.  This means, those four new boys count as 20% of the original boys, which means 2 new boys would be 10% of the original boys, which means there must have been 20 original boys.   If we know how many boys, we can use the ratio to calculate how many girls.   This statement, alone and by itself, is sufficient.

Statement (2): If the number of girls increases by 50%, then after such an increase, the probability that a randomly chosen student would be a boy would be 8/23.  The prompt gives us “ratio information”, and this statement also gives us more “ratio information”.  We have absolutely no “count information”, so we can’t figure out the count or number of anything.   This statement, alone and by itself, is insufficient.

First statement sufficient, second insufficient.  Answer = A

7) We can solve this question by setting up a proportion:

Divide the left fraction, numerator and denominator, by 7:

Now, cancel the factor of 4:

For this multiplication, use the doubling & halving trick — double 15 is 30, and half of 42 is 21:

x = (15)*(42)  = 30*21 = $630

Answer = C

8) The prompt gives us “ratio information”.

Statement #1 also gives us “ratio information”, so there is no way we can calculate a count, such as total number of seats in the hall.  This statement, alone and by itself, is insufficient.

Statement #2: 20 people would be 4% of the audience.  Divide by two — 10 people would have be 2%; now  multiply by five — 50 people would have been 10% of the audience; now multiply by ten — 500 people would be 100% of the audience on Tuesday night.  Since the concert hall was half full, there must be 1000 seats in total.  This piece of information allows us to solve for the answer to the prompt question.   This statement, alone and by itself, is sufficient.

First statement insufficient, second sufficient.  Answer = B

We recently issued a call for you, and other students, to submit questions that we could answer in a video series. There were a lot of great submissions, and the ones that we thought useful for a broad swath of students, we answered in our video series. These videos will be released over the next few weeks.

But some of the questions require a longer explanation. So in this post, I want to answer those other questions that came in.

Question 1

Of the 120 passengers on Flight 750, 50% are female. 10% of the passengers sit in first class, and the rest of the passengers sit in coach class. If 1/3 of the passengers in first class are male, how many females are there in coach class?

44

48

50

52

56

Here we have a math problem that you might see on the GMAT. This question would rank as an easy to medium level question on the test. Let’s look at how to solve this problem.

First thing I always do when I approach a problem is wade through the information and write down the facts. So, here is what I’m told:

• 120 people
• 50% female
• 10% of passengers sit in first class
• ⅓ of passengers in first class are male

Now, I will interpret this information and see what else it tells me:

• 50% female means there are 60 males and 60 females on the plane
• 10% of passengers sit in first class means that 12 people are seated in first class. If 12 people are in first class, there are 108 people in coach.
• ⅓ of passengers in first class are male, which means that 4 males and 8 females are seated in first class

Now our job is to find out how many females are seated in coach. This may seem like a tall order, but if we use a Double Matrix this will be fairly simple. First we need to make a table with what we know:

Alright now we can start filling in our boxes with the other information and we will be able to solve for how many females are sitting in coach.

Now with some simple subtraction (108 – 56 = 52 or 60 – 8 = 52), we have our answer. You could’ve used a Venn diagram to find the solution, but I find the Double Matrix more straightforward. If you don’t, spend some time learning about the Double Matrix because it is an important tool have in your math toolbox.

Question 2

What is a ratio?

This is a question that we have addressed in a few other posts (here, here, and here). And I am not sure if I can provide a better explanation than some of these other resources (Wikipedia or Khan Academy), but I will give it a shot.

Simply stated, a ratio expresses a relationship between two things. These two things are connected for some reason and exert some influence on each other. Whether it is the relationship between miles and time, the relationship between a distance on a map to an actual distance, or the relationship of teachers to students in a school, they all tell us something about the relationship.

For example, when we talk about the speed limit, we are talking about a relationship, one of the fundamental relationships in the universe—time and distance. So speed is really just a ratio that tells us how far we have traveled, or could travel, in a certain amount of time.

A ratio can be expressed two different ways—with a colon (:) or with a fraction bar (/). As we know with speed, we usually say “per” to indicate the fraction bar or the colon. So when your friend turns to you and says, “We’re going a 165 miles per hour,” you could write this as:

165 miles : 1 hour

165 miles / 1 hour

One crucial point to know is that a ratio doesn’t necessarily tell you totals. Let me show you what I mean. If I told you that a school has 1 teacher for every 32 students, I haven’t told you anything that helps you know how many students are at the school and how many teachers are at the school.

But with another piece of information, you’d be able to figure out all this information. So, if I told you that there are 12 teachers at the school, then we can determine how many students are at the school. If 1 teacher has 32 students, then 2 teachers have 64 students, and 3 teachers have 96 students, and…. Hopefully, you can see that we are just counting by 32 until we have counted 12 teachers. That is the same as a simple multiplication formula: 12 * 32 = 384.

So this is not the end all for your understanding of ratios—this is the beginning! Take the time to review some of the other resources to deepen your understanding of ratios.

First, a few practice problems.  Remember: no calculator!

1) If $5,000,000 is the initial amount placed in an account that collects 7% annual interest, which of the following compounding rates would produce the largest total amount after two years?

1. compounding annually
2. compounding quarterly
3. compounding monthly
4. compounding daily
5. All four of these would produce the same total

2) If A is the initial amount put into an account, R is the annual percentage of interest written as a decimal, and the interest compounds annually, then which of the following would be an expression, in terms of A and R, for the interest accrued in three years?

3) At the beginning of January 2003, Elizabeth invested money in an account that collected interest, compounding more frequently than a year. Assume the annual percentage rate of interest remained constant.  What is the total amount she has invested after seven years?

Statement #1: her initial investment was $20,000

Statement #2: the account accrued 7% annual interest

4) Sarah invested $38,700 in an account that paid 6.2% annual interest, compounding monthly.  She left the money in this account,  collecting interest for a full three-year period.  Approximately how much interest did she earn in the last month of this period?

1. $239.47
2. $714.73
3. $2793.80
4. $7,888.83
5. $15,529.61

Solutions to these will be given at the end of the article.

Simple Interest

In grade school, you learn about simple interest, largely because we want to teach little kids something about the idea of interest, and that’s the only kind of interest that children can understand.  No one anywhere in the real world actually uses simple interest: it’s a pure mathematical fiction.

Here’s how it works.  There’s an initial amount A, and an annual percentage P.  At the end of each year, you get interest in the amount of P percent of A — the same amount every year.  Suppose the initial amount is $1000, and the annual percentage is 5%.  Well, 5% of $1000 is $50, so each year, you would get the fixed value of $50 in interest.  If you plotted the value of the account (principle + interest) vs. time, you would get a straight line.

Again, this is a fiction we teach children, the mathematical equivalent of Santa Claus.  This never takes place in the real world.

Compound Interest, compounding annually

With compound interest, in each successive year or period, you collect more interest not merely on the principle but on all the interest you have accrued up to that point in time.   Interest on interest: that’s the big idea of compound interest.

Here’s how it plays out.  Again, there’s an initial amount A, and an annual percentage P, and we also have to know how frequently we are compounding.  For starters, let’s just say that we are compounding annually, once a year at the end of the year.   In fact, let’s say we have $1000, and the annual interest rate is 5%.  Well, in the first year, we would earn five percent on $1000, and gain $50 in interest.   The first year is exactly the same as the simple interest scenario.  After that first year, we now have $1050 in the account, so at the end of the second year, we gain 5% of $1050, or $52.50, for a new total of $1102.50.  Now, that’s our new total, so at the end of the third year, we gain 5% of $1102.50, or $55.12 (rounded down to the nearest penny), for a new total of $1157.62.  At the end of three years, the simple interest scenario would give us $1150, so the compound interest gains us an extra $7.62 —- not much, but then again, $1000 is not a lot to have invested.  You can see that, with millions or billions of dollars, this would be a significant difference.

Here, I was demonstrating everything step-by-step for clarity, but if we wanted to calculate the total amount after a large number of years, we would just use a formula.  We know that each year, the amount increased by 5%, and we know that 1.05 is the multiplier for a 5% increase.  After 20 years, the amount in the account would have experienced fifteen 5% increases, so the total amount would be

We don’t have to do it step-by-step: we can just jump to the answer we need, using multipliers.  Of course, for this exact value, we would need a calculator, and you don’t get a calculator on the GMAT Quantitative section.  Sometimes, though, the GMAT lists some answers in “formula form”, and you would just have to recognize this particular expression, , as the right formula for this amount.

If we graph compound interest against time, we get an upward curving graph (purple), which curves away from the simple interest straight line (green):

The curve of the graph, that is to say, the multiplying effect of the interest, gets more pronounces as time goes on.

BIG IDEA #1: as long as there is more than one compounding period, then compound interest always earns more than simple interest.  

Other compounding periods

A year is a long time to wait to get any interest.   Historically, some banks have compounded over shortened compounding period.  Here is a table of common compounding periods:

Technically, the fraction for “compounding daily” would be 1/365 in a non-leap year and 1/366 in a leap year; alternatively, one could use 1/365.25 for every year.

Now, how does this work?  Let’s say the bank gives 5% annual, compounding quarterly.  It would be splendid if the bank wanted to give you another 5% each and every quarter, but that’s not how it works.  The bank takes the percentage rate of interest and multiplies it by the corresponding fraction.  For 5% annual, compounding quarterly, we would multiply (5%)*(1/4) = 1.25%.  That’s the percentage increase we get each quarter.   The multiplier for a 1.25% increase is 1.0125.  Suppose we invest $1000 initially and keep the money in this account for seven years: that would be 7*4 = 28 compounding periods, so there are twenty-eight times in that period in which the account experiences a 1.25% increase.  Thus, the formula would be

For compounding quarterly, we divide the annual rate by four and compound four times each year.  For compounding monthly, we divide the annual rate by twelve and compound twelve times a year.   Similarly, for daily or any other conceivable compounding period.

How do the amounts of interest accrued compare for different compounding periods?  To compare this, let’s pick a larger initial value, $1,000,000, and collect over a longer period, 20 years.  Below are the total amounts, after twenty years, on an initial deposit of one million dollars compound at 5% annual:

• simple interest = $2,000,000
• compounding annually = $2,653,297.71
• compounding quarterly = $2,701484.94
• compounding monthly = $2,712,640.29
• compounding daily = $2,718,095.80
• compounding hourly = $2,718,274.07

As we go down that list, notice the values keep increasing as we decrease the size of the compounding period (and, hence, increase the total number of compound periods).  This leads to:

BIG IDEA #2: We always get more interest, and larger account value overall, when the compounding period decreases; the more compounding periods we have, the more interest we earn. 

Admittedly, the difference between “compounding daily” and “compounding hourly” only turn out to be a measly $178 on a million dollar investment over a twenty year period. A infinitesimally small difference, but technically, it is still an increase to move from “compounding daily” to “compounding hourly.”

Both ways

Notice that so far, we have been talking about investments and interest that you earn.   All of this, everything in this article, works equally well for debt and interest that you have to pay.  Just as the compounding effect, over time, magically multiplies an investment, so the same compounding effect will sink you deeper and deeper into debt.  This, in a nutshell, is the lose-lose proposition of credit card debt.

A little more than you need to know …

You may have noticed that, as the compounding periods get smaller and smaller, the amount increase in every diminishing steps.  In fact, as you may suspect, the total amount you possibly could earn from decreasing the compound period reaches a ceiling, a limit.  This limit is called “compounding continuously.”  The mathematics of this involves the special irrational number e, named for Leonard Euler (1707 – 1783), who is often considered the single greatest mathematician of all times.

e = 2.71828182845904523536028747135266249775724709369995 …

The formula for calculating continuously compound interest involves e, and is more complicated than anything you need to understand for the GMAT.   I will simply point out, in the example above, with one million dollars invested at 5% annual for 20 years, the limit of continuously compounding would be 1 million times e, which is $2,718,281.83.  You do not need to understand why that is or how this was calculated.

Most banks use monthly compounding interest, for accounts and for mortgages: this make sense for accounts with monthly statement or payments.  Credit cards tend to use continually compounding interest, because charges or payments could occur at any point, at any time of any day of the month.    In addition to understanding the mathematics of compound interest, it’s good to have a general idea of how it works in the real world: after all, the history or logic of compound interest would be a very apt topic for a Reading Comprehension passage or a Critical Reasoning prompt on the GMAT!

Summary

If you had any “aha’s” while reading this article, you may want to go back a take another look at the four practice problems above.  If you would like to express anything on these themes, or if you have a question about anything I said in this article, please let us know in the comments section.

Practice problem explanations

1) The smaller the compounding period is, the greater the number of times the interest will be compounded.  Of course, if we compound monthly instead of quarterly, then we are compounding by 1/12 of the annual rate each time, instead of 1/4.  The number of times we compound goes up, but the percentage by which we compound each time goes down.  Naively, you may think that those two would cancel out, but they don’t.  As discussed above, as the compounding period gets smaller, the total amount of interest earned goes up.  Therefore, we will get the most with the smallest compound period, daily.   Answer = (D)

2) Notice that, since R is the annual percent as a decimal, we can form a multiplier simply by adding one: (1 + R).  That’s very handy!  We will explore two different methods to get the answer.

Method One: Step-by-step

Starting amount = A

After one year, we multiply by the multiplier once

That’s the total amount at the end of the first year. The amount A is the original principle, and AR is the interest earned.

At the end of the second year, that entire amount is multiplied by the multiplier.  We need to FOIL.

That’s the total amount at the end of the second year. The amount A is the original principle, and the rest is the interest earned.

At the end of the third year, this entire amount is again multiplied by the multiplier.

That’s the total amount at the end of the third year. The amount A is the original principle, and the rest is the interest earned.

Answer = (C)

Method Two: some fancy algebra

Over the course of three years, the initial amount A is multiplied by the multiplier (1 + R) three times.  Thus, after three years,

Now, if you happen to know it offhand, we can use the cube of a sum formula:

Thus,

and

Answer = (C)

3) In order to determine the total amount at the end of an investment, we would need to know three things: (a) the initial deposit; (b) the annual percentage rate; and (c) the compounding period.

Statement #1 tells us the initial deposit but not the annual percentage rate. Insufficient.

Statement #2 tells us the annual percentage rate but not the initial deposit. Insufficient.

Together, we know both the initial deposit and the annual percentage rate, but we still don’t know the compounding period.  All we know is that it’s less than a year, but quarterly compounding vs. monthly compounding vs. daily compounding would produce different total amounts at the end.  Without knowing the exact compounding period, we cannot calculate a precise answer.  Even together, the statements are insufficient.

Answer = (E)

4) Without a calculator available, this is a problem screaming for estimation.  The problem even uses the magic word “approximately” to indicate that estimation is a good idea, and the answer choices are spread far apart, making it easier to estimate an individual answer.

Let’s round the deposit up to $40,000, and the percentage down to 6% annual.  Compounding monthly means each month, Sarah will accrue 6/12 = 0.5% in interest.  Well, 1% of $40,000  is $400.  Divide by 2: then 0.5% of $40,000 would be $200.  That would be the simple interest amount, as well as the interest in the first month.  We expect the amount in the last month to be a little more than this, but certain not even as large as double this amount.  The only possible answer is (A).

Success on the GMAT depends on many skills, and one skill that can help a student make enormous progress is the skill of asking excellent questions.  In order to discuss this skill, first let’s look at a relatively straightforward practice problem.

1. 14
2. 26
3. 28
4. 32
5. 38

I will show a full solution at the bottom of this article.  Right now, let’s just suppose that you got this question wrong and wanted to ask about it.  The correct answer is the lowest of the five answer choices, so for example, if you went for the trap answer of 32, you fell right into one of the mistakes that the question was specifically designed to elicit.  Questions writers regular incorporate trap answers like big butterfly nets, and simply let hordes of test-takers run into them.  The more you, as a student, can recognize these traps, the more you can set yourself apart from the  mass of ordinary test takers.  I will discuss this more in the solutions below.

Asking poor questions

Suppose you did this questions, got it wrong, and didn’t have access to an immediate explanation for the question.  Suppose you wanted to get an explanation from an expert, either on a blog or on a forum such as GMAT Club.  How would you ask your question?

Here’s what many people might say:

1) I don’t understand.  Please help.

2) I don’t get this.

3) Please explain.

I see questions of this ilk often.  These are abysmally poor questions.  If I were to assign a grade to such a question, I would give an F.  Asking a questions of this sort does not indicate any willingness, on the part of the student, to engage the material and put significant energy into the process of coming to deeper understand.

Notice, that questions like this require almost no effort to write.  One could write one of these questions and post it in under 15 seconds.  Right there, that’s an indication that it’s not necessarily going to help you.  You are not going learn the most and make the most progress unless you put a significant amount of effort into your own cause.

Another problem with these question is: think about it from the perspective of the expert who gets such a question.  I certainly understand how to solve this problem myself, but if  student tells me simply “I don’t understand. Please help,” then I don’t know where that student is stuck.  Does that student not know how to factor quadratics at all?  Did the student fall for the trap answer of (D)?  Did the student make a simple arithmetic mistake?  Does the student not understand algebra in the least?  Given that meager question, I have absolutely no idea what would best help the student.

The presumption of such a question is that the teacher, not the student is responsible for the process of education.  This is incorrect.  Education is not something a teacher or test prep company does to you.  Education is not a spectator sport.  Education is primarily something you do to yourself, by yourself, for yourself, and teachers or test prep companies simply provide the support and resources to enable you to educate yourself.

The art of asking excellent questions

The habits of excellence can set a student apart from others.  Excellence always takes more work, and many students avoid it for that very reason.  Many people don’t realize that, in choosing to exert a minimum of effort in many things related to their studies, they are implicitly choosing mediocrity.  The great Law of Mediocrity is: if you put in about the same effort as most people, then you probably will get about the same results as most people.  The corollary is: if you want results that stand out, you need to put in outstanding effort.  The habits of excellence require outstanding effort, and one of these habits is asking excellent questions.

Here are some examples of excellent questions:

4) When I looked at the problem, I saw right away that we could solve the first equation by subtracting 8 from both sides and solving x² = 9 to get x = 3.  I didn’t know how to solve that other equation, so I am glad I didn’t have to deal with it.  From there, I simply plugged 3 into 3x + 23 to get 9 + 23 = 32.  I chose (D), but that does not agree with the OA.  What did I do wrong?

5) I began by moving the 8 to the other side, and then I solved x² = 25 to get x = ±5.  I then factored the quadratic: x² – 2x – 15 = (x + 3)(x – 5) = 0, x = –3 or x = +5, so because of the combined constraints of both equations, we see x must equal +5, and therefore 3x + 23 = 38, which would be (E), but that’s not the OA given. 

Both of these are much much better questions.  Notice, these two students made vastly different mistakes.  The first needs to learn some basic algebra facts, while the second appears to have made a very superficial arithmetic error but really knows her algebra well.  Notice that an excellent question tells a mini-story of your entire experience of the problem.  “When I first saw the problem, I thought this or tried this. Then when I did this, I thought this ...”  An excellent question makes clear all the steps, the entire thought process.  First of all, this is tremendously helpful to the expert answering the question: when we read the question, we can fine-tune our answer to exactly what the student needs, and thus provide an answer that gives the student the most help.  More importantly, though, formulating this questions involves self-reflection on the part of the student.  Self-reflection is the very substance of education.  A student who is not actively reflecting on his process and his progress will not be learning nearly as fast as he could.  When a student puts in the effort to formulate an excellent question, this very process forces her to consider all the nuances of the question; she may thereby answer her own question, which is ideal, but if she doesn’t and asks her question, her mind will be primed to receive the focused response that an expert will give her.  Learning is not simply about getting all the information, because many students get all the information in front of them and simply can’t remember it all or apply it all.  Learning also has to involve preparing one’s mind to absorb the information one will receiving, and asking an excellent question, detailed and well thought-out, is one of the best ways to guarantee that you will retain and assimilate the answer you receive.

Summary

One of the habits of excellence is asking excellent questions.  Many students don’t do this, and are mystified about why they continue to struggle and cannot retain or apply the valuable information they have learned.  I urge you to embrace the habit of asking excellent questions and all the habits of excellence.  If you have experiences with your own practice of asking excellent questions, we would love to hear from you in the comments section.

Practice problem explanation

1) In the first equation, we need to subtract 8 from both sides, which gives us:

When we ourselves take a square root, we need to remember to include both the positive and the negative root.  See this GRE post for details.  We get two roots from this:

From this equation alone, we cannot determine unique value of x.  We have to factor the second equation.

The only root that is common to both equations is x = –3, so this must be the value of x.  This means

3x + 23 = 3(–3) + 23 = –9 + 23 = 23 – 9 = 14

Answer = (A) 

Hello!

This week, we’re tackling the GMAT math through a question that a student brought up. At first, the problem seems like a plug-and-chug kind of problem. But with the help of a crucial insight, this problem can be solved much quicker! Watch the video to find out what that insight is.

Here’s a still of the final board work:

Feel free to leave me any questions or comments you have below!

Here are twelve challenging problems related to the topic of exponents & roots.  Remember, no calculator.

(A) 17

(B) 19

(C) 21

(D) 23

(E) 27

4) Rank the following quantities in order, from smallest to biggest.

(A) I, II, III

(B) I, III, II

(C) II, I, III

(D) III, I, II

(E) III, II, I

5) Rank the following quantities in order, from smallest to biggest.

(A) I, II, III

(B) I, III, II

(C) II, I, III

(D) III, I, II

(E) III, II, I

(A) 96

(B) 120

(C) 144

(D) 192

(E) 288

(A) 4,000

(B) 8,000

(C) 16,000

(D) 25,000

(E) 125,000

8) Rank the following quantities in order, from smallest to biggest.

(A) I, II, III

(B) I, III, II

(C) II, I, III

(D) II, III, I

(E) III, II, I

(A) 1/5

(B) 2/13

(C) 2/15

(D) 5/3

(E) 15/2

Exponents and roots

For a review of some of the basics, see these blogs:

1) Exponent Properties on the GMAT

2) Adding and Subtracting Powers on the GMAT

3) Roots

4) Dividing by a Square Root

5) Mistake: Distributing Roots

6) Practice Problems on Powers and Roots

If reading any of those blogs gives you some insight, you might want to give the problems a second look before proceeding the solutions below.

Practice problems explanations

1) For this one, we can eliminate choices (A) & (B) because they make ghastly exponent mistakes.  For the other three, we would have to add and subtract the powers of each, to see which work, trying each by trial and error.

Instead, here’s an elegant solution.  Notice that the product . So, this is 80 times a power of 3.  Now, notice that 80 = 81 – 1.  In other words, we can easily express the factor 80 as the difference of two powers of 3.  Thus.

Answer = (E).

2) We have to begin with some clever noticing.  Notice, first of all, that 409 = 400 + 9.  This suggests a difference of two squares factoring pattern.  Notice that the numerator can be expressed as 159,919 = 160,000 – 81.  That’s a difference of two squares that can be factored!

The first parenthesis produces 409.  The second parenthesis contains another difference of two squares:

Therefore, if we divide by sides by (409)(17), we get that the fraction equals 23.

Answer = (D).

3) Well, we know that 15 = 3*5, and we have the exponent that gives a 5, but how do we get an exponent that gets a 3?  Well, of course [pmath]6^1 = 6[/pamth], and therefore

and therefore

Answer = (E).

4) First of all, the fact that 120 is a multiple of 30 draws our attention to that comparison.  The ratio 120:30 = 4:1, so we just would have to compare those powers of 2 and 17.

So, III is bigger than I.  Now, notice that the exponents of I and II are also in a convenient ratio — 120:72 = 5:3.  We can use those exponents on 2 and 3.

Therefore, II is smaller than I.  From smallest to biggest: II, I, III.

Answer = (C).

5) First of all, clearly

So, II is bigger than I.  Now, what about III?  When we take higher order roots, the values move closer to one.  If the number starts larger than one, then higher and higher roots make it smaller, closer to one.  If the number starts between 0 and 1, then higher and higher roots make it larger, closer to one.  Therefore, III is larger than II.  From smallest to biggest, I, II, III.

Answer = (A).

6) First of all, the mistake: we CANNOT add through

That incorrect thinking would lead to the trap answer of (B).  Instead, we have to simplify each square root on the left.

Answer = (D).

7) This one becomes clearer if we change the roots to fractional exponents.

Answer = (A).

8) The exponent of 3, which is 42, is close to 40.  If 2 and 3 had exponents of 60 and 40, respectively, those would be in a ratio easy to reduce — 60:40 = 3:2.  Clearly

Raise both sides of that inequality to the power of 20.

Therefore, III is bigger than I.

Now, let’s think about II.  The square root of 2 is 2 to the power of 1/2, so 2 times the square root of two, the contents of the parentheses, would be 2 to the power of 1.5.  Multiply the exponents: 1.5*35 = 35 + 17.5 = 52.5 — that would be the resultant exponent of 2.  Clearly, this is a lower power of 2 than given in statement I.  So, II is less than I.

From smallest to biggest is II, I, III.

Answer = (C).

9) An increase of 50% corresponds to a multiplier of 1.5, so the above information can be written as

To get all the K’s on one side, we will divide by the power on the left, K to the power of 5/4.  This will mean we have to subtract the exponents.  What do we get when we subtract 5/4 from 3/2?

Thus, after the division, we have:

Now, raise both sides to the fourth power to solve for K:

Answer = (E).

10) First, we have to express 0.15 as a fraction:

Also, 16 is 2 to the power of 4, so the equation with b tells us:

Bases are the same, so we can equate the exponents.

Answer = (E).

11) As is usual, all the answer choices have been rationalized, so we have to rationalize the denominator of the prompt fraction.   This means multiplying by the conjugate over itself: that would be 7 minus 3 times the square root of 5.

In the denominator, we use the difference of two squares pattern.   In the numerator, we simply FOIL.

Answer = (D).

BTW, you don’t need to know this for the GMAT, but this OA answer is the reciprocal of the Golden Ratio, and answer choice (E) equals the Golden Ratio.

12) We need to express each side as a power of 5.  We will use fractional exponents for the roots.

Equate the exponents.

Multiply both sides by 6 to clear the fractions.

Answer = (B).

1) Jennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price?

1. 100T/(NB)
2. TB/(100N)
3. 100TN/B
4. ((T/N) – B)/(100B)
5. 100(T – NB)/N

2) At the store, Sam bought a shirt and a toaster.  There was an 8% sales tax on each item, and with tax, Sam paid a total of K.  If the price of the toaster before tax was T, what, in terms of K and T, is the price of the shirt before tax?

1. 0.92(K – T)
2. 0.92K – T
3. 0.92(K – 1.08T)
4. (K – T) /1.08
5. (K/1.08) – T

3) Before January, the price of a dress was D and the price of a matching pair of shoes was H.  In January, the price of the dress increased by 40% and the price of the shoes increased by 50%.  In March, Roberta bought both items with a 30% discount.  If D = 5H, which of the following represents the amount that Roberta paid?

1. D + 40
2. D + H – 1
3. D + 2H
4. 5.95H
5. 1.21D

4) A marketer bought N crates of empty cardboard gift boxes.  Each crate held Q individual gift boxes, and the lot of N crates was purchases at a wholesale price of W dollars.  This marketer will sell collections of J cardboard gift boxes to retailers, at a price of P dollars for each collection.  (Note: J is a divisor of Q.)  The marketer knows that, when he has sold all the cardboard gift boxes this way, he wants to net a total profit of Z dollars on the entire transaction.  What price P must he charge, to net this profit? Express P in terms of N, Q, W, J, and Z.

5) In the rectangle above, the ratio of (length):(height) = A, which is a number greater than one.  Which of the following expresses the ratio of (diagonal):(length) in terms of A?

6) In the diagram above, BF is an altitude drawn to the base AC, and AC is the side of square ACDE.  If BF = h, and the area of triangle ABC equals Q, find the area of the square (shaded) in terms of h and Q.

Solutions are given below.

Variables in the Answer Choices

The GMAT loves these question, not least because so many real-world situations something of this feel: we have a few changing variables (number of items sold, or stock price, or commodity price, etc.) and we want to figure out something else that depends on these changing quantities.

There are two basic approaches: an algebraic approach vs. a numerical approach.  Each one has advantages and disadvantages.  Ultimately, you should practice both approaches on a number of these problems, to get a sense of what comes most naturally for you.

If you know how to do the algebra, it is efficient and it always leads to an unambiguous answer choice.  If you can do the algebra, that’s great, but the disadvantage is that it’s sometimes hard to know how to perform this approach.

Picking numbers is inefficient but easier.   It important to know that there is an art to picking numbers.   Of course, if I pick a set of numbers for the variables in the prompt, and an answer choice doesn’t produce the desired answer, I can eliminate it, but if these numbers do produce the desired answer, I can draw no conclusion from that.  Picking numbers is, de facto, a process of elimination approach.   We have to pick numbers until four answer choices have been clearly eliminated.  Obviously, the test loves to design questions so that the most obvious choices for numbers eliminate few or no answers.  Each round of picking numbers is simple and relatively quick, but when you begin, you have no idea how many rounds it will take before four answer choices have been eliminated.  That’s the inefficient part.

In the problems below, I will demonstrate both approaches for each problem, and in demonstrating the numerical approach, I will discuss in context some points of strategy for picking numbers.

Summary

If you have been working with these problems, and have your own thoughts about the algebraic approach vs. the picking-numbers approach, we would love to hear from you.  Let us know in the comment section below!

Practice problem explanations

As you read these explanations, pay attention to which solution seems more natural and more doable for you.

1) Algebraic Solution: If she makes a total profit of T for N watches, then that must be a profit of T/N for each watch.  That must be the markup above cost on each watch, the amount of the increase.  Well, percent increase = (amount of increase)/(starting amount) x 100% = (T/N)/B *100 = 100T/(NB)

Answer = (A)

That Algebraic solution was elegant if you saw it, but if not, here’s a full solution with picking numbers.

Numerical Solution: The price of each watch, I will pick B = 7, a prime number.   For the percent increase, I will make this easy.  Picking 100% is too easy, and too predictable, so I am going to pick 400% — 400% of 7 is 28, and the $7 watches are marked up $25, then the selling price is $35.  The profit per watch is $28, and so if she sells N = 11 watches (another prime number), that would be a profit of T = 28*11 = $308.  Leaving T in unmultiplied form will make it easier to cancel.

OK, now we will plug in T = 308 = 11*28, N = 11, and B = 7, and hope to get 400 as our answer.

(A) 100T/(NB) = 100*11*28/(11*4) = 100*28/7 = 110*4 = 400 = works!

(B) TB/(100N) = 11*28*7/(100*11) = 28*7/100 = doesn’t work!

(C) 100TN/B = 100*11*28*11/7 = 100*11*4*11 = doesn’t work!

(D) ((T/N) – B)/(100B) = [(11*28/11) – 7]/(7*100) = (28 – 7)/(7*100) = 21/(7*100) = 3/100 = doesn’t work

(E) 100(T – NB)/N = 100(11*28 – 11*7)/7 = 100*11*21/7 = 100*11*3 = doesn’t work!

We were lucky here.  With one choice of numbers, we were able to eliminate four answer choices, leaving (A) as the only possible answer.

2) Algebraic Solution: Call the price of the shirt S.  The price of the toaster is T.  The total before tax is (S + T).  Then an 8% tax was applied.  For an 8% increase, we use the multiplier 1.08.  Thus, K = 1.08*(S + T). Solve this for S.

S + T = K/1.08

S = K/1.08 – T

Answer = (E)

Numerical Solution: we make our lives much easier if we first recognize the percent trap.  The multiplier 1.08 is the correct multiplier for an 8% increase, and it makes sense that this would appear somehow in the correct answer.   BUT, 0.92 is the multiplier for an 8% decrease.  This is tricky: an 8% decrease is not the opposite of an 8% increase; it does not “undo” an 8% increase.  In other words, an 8% increase followed by an 8% decrease does NOT get us back to the original starting point.  That’s a major percent trap.  There is a 8% increase in the problem, but no 8% decrease: this means that any answer in which the multiplier for an 8% decrease, 0.92, appears, it is automatically wrong.   We can immediately eliminate (A) & (B) & (C) without plugging in anything.

OK, let’s say T = 50, shirt = 150, so total before tax is $200. (Again, avoiding the cliché price of $100 for a percent.)  Now, 8% of $200 is $16, for a total of K = 216.  Because dividing by a decimal isn’t fun, I am going to change the division by 1.08 into multiplication by the fraction 100/108.  We are looking for an answer that equals 150, the price of the shirt.

(D) (K – T) /1.08 = (216 – 50)/1.08 = 156*100/108 = 39*100/27 = 13*100/9

well, 13/9 doesn’t equal 1.5, so this doesn’t work.

(E) (K/1.08) – T = 216/1.08 – 50 = 216*(100/108) – 50 = 2*108*(100/108) – 50

= 2*100 – 50 = 200 – 50 = 150

That’s the price of the shirt!  Choice (E) is the answer!

3) Algebraic Solution: dress increase to 1.4D, and the shoes increase to 1.5H.  The multiplier for a 30% decrease is 0.7, so the total amount could be written as 0.7*(1.4D + 1.5H).  This does not appear as an answer.  Since H is smaller, write D in terms of H: D = 5H.  Then the total is 0.7*(1.4(5H) + 1.5H) = 0.7*(7H + 1.5H) = 4.9H + 1.05H = 5.95H.  That’s answer (D).

Numerical Solution: What do you think every single drooling GMAT test taker is going to pick for the price of the dress?  You guessed it: $100 dress and $20 shoes.  It’s a spectacularly bad idea to make that choice.  As will happen on the real GMAT, I was anticipating this choice, and composed trap answers that would work for this test.  If you plug this in, I guarantee you will not be able to eliminate four answers at once.

Think about making a choice that is still relatively easy, but not everyone would make it.

Make the dress D = 300, so H = 60.  D increased by 40% — 10% of 300 is 30, so 40% must be 4*30 = 120, and the new dress price is $420.  The shoes increase by 50%, to $90.  Together, after the increase, dress + shoes would cost 420 + 90 = $510.  Now, Roberta gets a 30% decrease on this price.  Well, 10% of 510 is 51, so 30% would be 3*51 = 153.  510 – 153 = 357.  The correct answer must equal 357.

(A) D + 40 = 340 = not correct

(B) D + H – 1 = 300 + 60 – 1 = 359 = close, but no cigar!

(C) D + 2H = 300 + 2*60 = 420 = not even close

(D) 5.95H = 5.95*60 — hmmm — I am going to think of 60 as 10*2*3.  First, 59.5*10 = 59.5. Then 59.5*1 = 119.  Then 119*3 = 357.  This works.

(E) 1.21D = 1.21*300 = (1 + 0.2 + 0.01)*300 = 300 + 60 + 3 = 363 = not correct

The only possible answer is (D).

4) Algebraic Solution: So, the marketer initially paid W: that was his outlay, his cost.   There are Q boxes in each crate, and J boxes make a collection, so there are Q/J collections in each crate, and NQ/J collections in total.  If he charges a price P, his revenue would be PNQ/J.  Now, profit equals revenue minus cost, so Z = PNQ/J – W .  Solve this for P

Numerical Solution: Here, we have several choices to make, and it would be a good idea to pick numbers that aren’t the same, or aren’t even divisible by each other if that’s not required. Picking some different prime numbers is good way eliminate more than one answer that comes out to the correct value.

I will say the price is 2, a nice round prime number.   Let’s say J = 3, so Q must be a multiple of that — say Q = 30.  Let’s N = 7.  This means he has 10 collections from every box, and 70 collections from the total lot.  If he charges $2 per collection, that’s a revenue of $140.  Let’s say his outlay was W = $95; then his profit is Z = $45.

OK, that’s an unusual enough set of number that we expect no more than one solution will work.  We will plug in N = 7, Q = 30, W = 95, J = 3, and Z = 45, and we hope to get an output of P = 2.

Thus, because we chose numbers well, the only possible answer is (B).

5) Algebraic Solution: Let length = L, height = H, and diagonal = D.  Clearly, these three are related by the Pythagorean Theorem

Divide both sides by L squared.

The right side is the square of the answer we seek.  The left side is the square of the reciprocal of A.

Take a square root of both sides; remember that we cannot distribute the square root across addition:

Answer = (E)

Numerical Solution: Let’s choose an easy 3-4-5 triangle for the height-length-diagonal.   Then A = 4/3, and (diagonal):(length) = 5/4.  We want to plug in A = 4/3 and get an answer of 5/4.

This choice of numbers eliminated four answers and leaves only one, so (E) must be the answer.

6) Algebraic Solution: Let s be the side of the square, which means it is also the base of the triangle.  We know that the area Q must equal one half times base s times height h.

Q = (1/2)sh

Solve this for s.

2Q = sh

Now, square this to get the area of the square:

Answer = (D)

Numerical Solution: Let’s say that h = 3, and that the side of the square is 8.  (Notice, I picked these so neither was a factor of the other, and I made one even so the area of the triangle Q would be a whole number.)  Then the area of the triangle is Q = (0.5)(3)(8) = 12, and the area of the square is 64.  Therefore, we should plug in h = 3 and Q = 12, and get an output of 64.

This choice of numbers eliminated four answers and leaves only one, so (D) must be the answer.

Here are fifteen challenging questions for GMAT math practice, with explanations below.  Can you keep the GMAT Quant pace, doing these in under 90 seconds each?   As always, no calculator!

1) Let abcd be a general four-digit number and all the digits are non-zero.  How many four-digits numbers abcd exist such that the four digits are all distinct and such that a + b + c = d?

(A) 6

(B) 7

(C) 24

(D) 36

(E) 42

2) Let abcd be a general four-digit number.  How many odd four-digits numbers abcd exist such that the four digits are all distinct, no digit is zero, and the product of a and b is the two digit number cd?

(A) 4

(B) 6

(C) 12

(D) 24

(E) 36

3) There are 500 cars on a sales lot, all of which have either two doors or four doors.  There are 165 two-door cars on the lot.  There are 120 four-door cars that have a back-up camera.   Eighteen percent of all the cars with back-up cameras have standard transmission.  If 40% of all the cars with both back-up cameras and standard transmission are two-door cars, how many four-door cars have both back-up cameras and standard transmission?

(A) 18

(B) 27

(C) 36

(D) 45

(E) 54

4) At Mnemosyne Middle School, there are 700 students: all the students are boys or girls in the 4^th or 5^th grade.   There are 320 students in the 4^th grade, and there are 210 girls in the 5^th grade.  Fifty percent of the 5^th graders and 40% of the 4^th graders take Mandarin Chinese.   Ninety 5^th grade boys do not take Mandarin Chinese.  The number of 4^th grade girls taking Mandarin Chinese is less than half of the number of 5^th grade girls taking Mandarin Chinese.  Which of the following could be the number of 4^th grade boys in Mandarin Chinese?

(A) 10

(B) 40

(C) 70

(D) 100

(E) 130

5)  A hundred identical cubic boxes are currently arranged in four cubes: a single cubic box, a 2 x 2 x 2 cube, a 3 x 3 x 3 cube, and a 4 x 4 x 4 cube.  These four are not touching each other.  All outward faces are painted and all inward faces are not painted.  These four cubes are going to be dismantled and reassembled as a flat 10 x 10 square.  The top and all the edges of this 10 x 10 square must be painted, but there is no requirement for paint on the bottom.  How many individual faces will have to be painted to accommodate the requirements of this new design?

(A) 0

(B) 5

(C) 9

(D) 16

(E) 27

6) Twelve points are spaced evenly around a circle, lettered from A to L.  Let N be the total number of isosceles triangles, including equilateral triangles, that can be constructed from three of these points. A different orientation of the same lengths counts as a different triangle, because a different combination of points form the vertices.   What is the value of N?

(A) 48

(B) 52

(C) 60

(D) 72

(E) 120

7)  Theresa is a basketball player practicing her free throws.  On her first free throw, she has a 60% chance of making the basket.  If she has just made a basket on her previous throw, she has a 80% of making the next basket.  If she has just failed to make a basket on her previous throw, she has a 40% of making the next basket.  What is the probability that, in five throws, she will make at least four baskets?

8) Suppose a “Secret Pair” number is a four-digit number in which two adjacent digits are equal and the other two digits are not equal to either one of that pair or each other.  For example, 2209 and 1600 are “Secret Pair” numbers, but 1333 or 2552 are not.  How many “Secret Pair” numbers are there?

(A) 720

(B) 1440

(C) 1800

(D) 1944

(E) 2160

9) In the coordinate plane, a circle with its center on the negative x-axis has a radius of 12 units, and passes through (0, 6) and (0, – 6).  What is the area of the part of this circle in the first quadrant?

10) In the coordinate plane, line L passes above the points (50, 70) and (100, 89) but below the point (80, 84).  Which of the following could be the slope of line L?

(A) 0

(B) 1/2

(C) 1/4

(D) 2/5

(E) 6/7

11) At the beginning of the year, an item had a price of A.  At the end of January, the price was increased by 60%.  At the end of February, the new price was decreased by 60%.  At the end of March, the new price was increased by 60%.  At the end of April, the new price was decreased by 60%.  On May 1^st, the final price was approximately what percent of A?

(A) 41%

(B) 64%

(C) 100%

(D) 136%

(E) 159%

12) Suppose that, at current exchange rates, $1 (US) is equivalent to Q euros, and 1 euro is equivalent to 7Q Chinese Yuan.   Suppose that K kilograms of Chinese steel, worth F Chinese Yuan per kilogram, sold to a German company that paid in euros, can be fashioned into N metal frames for chairs.  These then are sold to an American company, where plastic seats & backs will be affixed to these frames.  If the German company made a total net profit of P euros on this entire transaction, how much did the US company pay in dollars for each frame?

13) At the Zamenhof Language School, at least 70% of the students take English each year, at least 40% take German each year, and between 30% and 60% take Italian each year.  Every student must take at least one of these three languages, and no student is allowed to take more than two languages in the same year.  What is the possible percentage range for students taking both English and German in the same year?

(A) 0% to 70%

(B) 0% to 100%

(C) 10% to 70%

(D) 10% to 100%

(E) 40% to 70%

14) On any given day, the probability that Bob will have breakfast is more than 0.6.  The probability that Bob will have breakfast and will have a sandwich for lunch is less than 0.5.  The probability that Bob will have breakfast or will have a sandwich for lunch equals 0.7.  Let P = the probability that, on any given day, Bob will have a sandwich for lunch.  If all the statements are true, what possible range can be established for P?

(A) 0 < P < 0.6

(B) 0 ≤ P < 0.6

(C) 0 ≤ P ≤ 0.6

(D) 0 < P < 0.7

(E) 0 ≤ P < 0.7

(A) – 64

(B) – 7

(C) 38

(D) 88

(E) 128

Explanations for this problem are at the end of this article.

More Practice

Here are twenty-eight other articles on this blog with free GMAT Quant practice questions.  Some have easy questions, some have medium, and few have quite challenging questions.

1) GMAT Geometry: Is It a Square?

2) GMAT Shortcut: Adding to the Numerator and Denominator

3) GMAT Quant: Difficult Units Digits Questions

4) GMAT Quant: Coordinate Geometry Practice Questions

5) GMAT Data Sufficiency Practice Questions on Probability

6) GMAT Quant: Practice Problems with Percents

7) GMAT Quant: Arithmetic with Inequalities

8) Difficult GMAT Counting Problems

9) Difficult Numerical Reasoning Questions

10) Challenging Coordinate Geometry Practice Questions

11) GMAT Geometry Practice Problems

12) GMAT Practice Questions with Fractions and Decimals

13) Practice Problems on Powers and Roots

14) GMAT Practice Word Problems

15) GMAT Practice Problems: Sets

16) GMAT Practice Problems: Sequences

17) GMAT Practice Problems on Motion

18) Challenging GMAT Problems with Exponents and Roots

19) GMAT Practice Problems on Coordinate Geometry

20) GMAT Practice Problems: Similar Geometry Figures

20) GMAT Practice Problems: Variables in the Answer Choices

21) Counting Practice Problems for the GMAT

22) GMAT Math: Weighted Averages

23) GMAT Data Sufficiency: More Practice Questions

24) Intro to GMAT Word Problems, Part I

25) GMAT Data Sufficiency Geometry Practice Questions

26) GMAT Data Sufficiency Logic: Tautological Questions

27) GMAT Quant: Rates and Ratios

28) Absolute Value Inequalities

Summary

These are hard problems.  When you read the solutions, don’t merely read them passively.  Study the strategies used, and do what you can to retain them.  Learn from your mistakes!

Practice Problem Explanations

1) We need sets of three distinct integers {a, b, c} that have a sum of one-digit number d.  There are seven possibilities:

1. a) {1, 2, 3}, sum = 6
2. b) {1, 2, 4}, sum = 7
3. c) {1, 2, 5}, sum = 8
4. d) {1, 3, 4}, sum = 8
5. e) {1, 2, 6}, sum = 9
6. f) {1, 3, 5}, sum = 9
7. g) {2, 3, 4}, sum = 9

For each set, the sum-digit has to be in the one’s place, but the other three digits can be permutated in 3! = 6 ways in the other three digits.  Thus, for each item on that list, there are six different possible four-digit numbers.  The total number of possible four-digit numbers would be 7*6 = 42.  Answer = (E)

2)  The fact that abcd is odd means that cd must be an odd number and that a & b both must be odd.  That limits the choices significantly.  We know that neither a nor b can equal 1, because any single digit number times 1 is another single digit number, and we need a two-digit product—there are no zeros in abcd.  We also know that neither a nor b can equal 5, because any odd multiple of 5 ends in 5, and we would have a repeated digit: the requirement is that all four digits be distinct.

Therefore, for possible values for a & b, we are limited to three odd digits {3, 7, 9}.  We can take three different pairs, and in each pair, we can swap the order of a & b. Possibilities:

1. use {3, 7}, product = 21, abcd could be 3721 or 7321
2. use {3, 9}, product = 27, abcd could be 3927 or 9327
3. use {7, 9}, product = 63, abcd could be 7963 or 9763

Those six are the only possibilities for abcd.

Answer = (B)

3) Total number of cars = 500

2D cars total = 165, so

4D cars total = 335

120 4D cars have BUC

“Eighteen percent of all the cars with back-up cameras have standard transmission.”

18% = 18/100 = 9/50

This means that the number of cars with BUC must be a multiple of 50.

How many 2D cars can we add to 120 4D cars to get a multiple of 50?  We could add 30, or 80, or 130, but after that, we would run out of 2D cars.  These leaves three possibilities for the total number with BUC:

If a total of 150 have BUC, then 18% or 27 of them also have ST.

If a total of 200 have BUC, then 18% or 36 of them also have ST.

If a total of 250 have BUC, then 18% or 45 of them also have ST.

Then we are told: “40% of all the cars with both back-up cameras and standard transmission are two-door car.”

40% = 40/100 = 2/5

This means that number of cars with both back-up cameras and standard transmission must be divisible by 5.  Of the three possibilities we have, only the third words.

Total cars with BUC cams = 250 (120 with 4D and 130 with 2D)

18% or 45 of these also have ST.

40% of that is 18, the number of 2D cars with both BUC and ST.

Thus, the number of 4D cars with both BUC and ST would be

45 – 18 = 27

Answer = (B)

4) 700 student total

4G = total number of fourth graders

5G = total number of fifth graders

We are told 4G = 320, so 5G = 700 – 320 = 380

5GM, 5GF = fifth grade boys and girls, respectively

We are told 5GF = 210, so 5GM = 380 – 210 = 170

4GC, 5GC = total number of 4^th or 5^th graders, respectively taking Chinese

We are told

5GC = 0.5(5G) = 0.5(380) = 190

4GC = 0.4(4G) = 0.4(320) = 128

4GFM, 4GMC, 5GFC, 5GMC = 4^th/5^th grade boys & girls taking Chinese

We are told that, of the 170 fifth grade boys, 90 do not take Chinese, so 170 = 90 = 80 do.  Thus 5GMC = 80.

5GMC + 5GFC = 5GC

80 + 5GFC = 190

5GFC = 110

We are told:

4GFM < (0.5)(5GFC)

4GFM < (0.5)(100)

4GFM < 55

Thus, 4GFM could be as low as zero or as high as 54.

4GMC = 4GC – 4GFM

If 4GFM = 0, then 4GMC = 128 – 0 = 128

If 4GFM = 54, then 4GMC = 128 – 54 = 74

Thus, fourth grade boys taking Mandarin Chinese could take on any value N, such that 74 ≤ N ≤ 128.  Of the answer choices listed, the only one that works is 100.

Answer = (D)

5) The single cube has paint on all six sides.  Each of the eight boxes in the 2 x 2 x 2 cube has paint on three sides (8 corner pieces).  In the 3 x 3 x 3 cube, there are 8 corner pieces, 12 edge pieces (paint on two sides), 6 face pieces (paint on one side), and one interior piece (no paint).  In the 4 x 4 x 4 cube, there are 8 corner pieces, 24 edge pieces, 24 face pieces, and 8 interior pieces.  This chart summarizes what we have:

For the 10 x 10 flat square, we will need 4 corner pieces that have paint on three sides, 32 edge pieces that have paint on two sides (top & side), and 64 middle pieces that have paint on one side (the top).

We could use either the single total box or any of the 24 corner boxes for the four corners of the square.  That leaves 21 of these, and 35 edge boxes, more than enough to cover the 32 edges of the square.  The remaining ones, as well as all 30 face boxes, can be turned paint-side-up to fill in the center.  The only boxes that will need to be painted, one side each, are the 9 interior boxes.  Thus, we have 9 sides to paint.

Answer = (C)

6) Here’s a diagram.

First, let’s count the equilateral triangles.  They are {AEI, BFJ, CGK, DHL}.  There are only four of them.

Now, consider all possible isosceles triangles, excluding equilateral triangles, with point A as the vertex.  We could have BAL, CAK, DAJ, and FAH.  All four of those have a line of symmetry that is vertical (through A and G).  Thus, we could make those same four triangles with any other point as the vertex, and we would never repeat the same triangle in the same orientation.  That’s 4*12 = 48 of these triangles, plus the 4 equilaterals, is 52 total triangles.

Answer = (B)

7) There are five basic scenarios for this:

Case I: (make)(make)(make)(make)(any)

If she makes the first four, then it doesn’t matter if she makes or misses the fifth!

Case II: (miss)(make)(make)(make)(make)

Case III: (make)(miss)(make)(make)(make)

Case IV: (make)(make)(miss)(make)(make)

Case V: (make)(make)(make)(miss)(make)

Put in the probabilities:

Case I: (0.6)(0.8)(0.8)(0.8)

Case II: (0.4)(0.4)(0.8)(0.8)(0.8)

Case III: (0.6)(0.2)(0.4)(0.8)(0.8)

Case IV: (0.6)(0.8)(0.2)(0.4)(0.8)

Case V: (0.6)(0.8)(0.8)(0.2)(0.4)

Since all the answers are fractions, change all of those to fractions.  Multiply the first by (5/5) so it has the same denominator as the other products.

Case I: (3/5)(4/5)(4/5)(4/5)(5/5) = 960/5^5

Case II: (2/5)(2/5)(4/5)(4/5)(4/5) = 256/5^5

Case III: (3/5)(1/5)(2/5)(4/5)(4/5) = 96/5^5

Case IV: (3/5)(4/5)(1/5)(2/5)(4/5) = 96/5^5

Case V: (3/5)(4/5)(4/5)(1/5)(2/5) = 96/5^5

Add the numerators.  Since 96 = 100 – 4, 3*96 = 3(100 – 4) = 300 – 12 = 288.

288 + 256 + 960 = 1504

P = 1504/5^5

Answer = (E)

8) There are three cases: AABC, ABBC, and ABCC.

In case I, AABC, there are nine choices for A (because A can’t be zero), then 9 for B, then 8 for C.  9*9*8 = 81*8 = 648.

In case II, ABBC, there are 9 choices for A, 9 for B, and 8 for C.  Again, 648.

In case III, ABCC, there are 9 choices for A, 9 for B, and 8 for C.  Again, 648.

48*3 = (50 – 2)*3 = 150 – 6 = 144

3*648 = 3(600 + 48) = 1800 + 144 = 1948

Answer = (D)

9)

We know that the distance from A (0,6) to B (0, – 6) is 12, so triangle ABO is equilateral.  This means that angle AOB is 60°.   The entire circle has an area of

A 60° angle is 1/6 of the circle, so the area of sector AOB (the “slice of pizza” shape) is

The area of an equilateral triangle with side s is

Equilateral triangle AOB has s = 12, so the area is

If we subtract the equilateral triangle from the sector, we get everything to the right of the x-axis.

Again, that’s everything to the right of the x-axis, the parts of the circle that lie in Quadrants I & IV.  We just want the part in Quadrant I, which would be exactly half of this.

Answer = (C)

10) One point is (50, 70) and one is (100, 89): the line has to pass above both of those.  Well, round the second up to (100, 90)—if the line goes above (100, 90), then it definitely goes about (100, 89)!

What is the slope from (50, 70) to (100, 90)?  Well, the rise is 90 – 70 = 20, and the run is 100 – 50 = 50, so the slope is rise/run = 20/50 = 2/5.  A line with a slope of 2/5 could pass just above these points.

Now, what about the third point?  For the sake of argument, let’s say that the line has a slope of 2/5 and goes through the point (50, 71), so it will pass above both of the first two points.  Now, move over 5, up 2: it would go through (55, 73), then (60, 75), then (65, 77), then (70, 79), then (75, 81), then (80, 83).  This means it would pass under the third point, (80, 84).  A slope of 2/5 works for all three points.

We don’t have to do all the calculations, but none of the other slope values works.

Answer = (D)

11) The trap answer is 100%: a percent increase and percent decrease by the same percent do not cancel out.

Let’s say that the A = $100 at the beginning of the year.

End of January, 60% increase.  New price = $160

End of February, 60% decrease: that’s a decrease of 60% of $160, so that only 40% of $160 is left.

10% of $160 = $16

40% of $160 = 4(16) = $64

That’s the price at the end of February.

End of March, a 60% increase: that’s a increase of 60% of $64.

10% of $64 = $6.40

60% of $64 = 6(6 + .40) = 36 + 2.4 = $38.40

Add that to the starting amount, $64:

New price = $64 + $38.40 = $102.40

End of April, 60% decrease: that’s a decrease of 60% of $102.40, so that only 40% of $102.40 is left.

At this point, we are going to approximate a bit.  Approximate $102.40 as $100, so 40% of that would be $40.  The final price will be slightly more than $40.

Well, what is slightly more than $40, as a percent of the beginning of the year price of $100?  That would be slightly more than 40%.

Answer = (A)

12) The K kilograms, worth F Chinese Yuan per kilogram, are worth a total of KF Chinese Yuan.  The German company must pay this amount.

Since 1 euro = (7Q) Chinese Yuan, then (1/(7Q)) euro = 1 Chinese Yuan, and (KF/7Q) euros = KF Chinese Yuan.  That’s the amount that the Germans pay to the Chinese.

That is the German company’s outlay, in euros.  Now, they make N metal chairs, and sell them, making a gross profit of P euros.

That must be the total revenue of the German company, in euros.  This comes from the sale to the American company.  Since $1 = Q euros, $(1/Q) = 1 euro, so we change that entire revenue expression to euros to dollars, we divide all terms by Q.

That must be the total dollar amount that leaves the American company and goes to the German company.  This comes from the sale of N metal frames for chairs, so each one must have been 1/N of that amount.

Answer = (A)

13) First, we will focus on the least, the lowest value.  Suppose the minimum of 70% take English, and the minimum of 40% take German.  Even if all 30% of the people not taking English take German, that still leaves another 10% of people taking German who also have to be taking English.  Thus, 10% is the minimum of this region.

Now, the maximum.  Both the German and English percents are “at least” percents, so either could be cranked up to 100%.  The trouble is, though, that both can’t be 100%, because some folks have to take Italian, and nobody can take three languages at once.  The minimum taking Italian is 30%.  Let’s assume all 100% take German, and that everyone not taking Italian is taking English: that’s 70% taking English, all of whom also would be taking German.  Thus, 70% is the maximum of this region.

Answer = (C)

14) Let A = Bob eats breakfast, and B = Bob has a sandwich for lunch.  The problem tells us that:

P(A) > 0.6

P(A and B) < 0.5

P(A or B) = 0.7

First, let’s establish the minimum value.  If Bob never has a sandwich for lunch, P(B) = 0, then it could be that P(A and B) = 0, which is less than 0.5, and it could be that P(A) = 0.7, which is more than 0.6, so that P(A or B) = 0.7.  All the requirements can be satisfied if P(B) = 0, so it’s possible to equal that minimum value.

Now, the maximum value.  Since P(A or B) = 0.7, both P(A) and P(B) must be contained in this region.  See the conceptual diagram.

The top line, 1, is the entire probability space.  The second line, P(A or B) = 0.7, fixes the boundaries for A and B.  P(A) is the purple arrow, extending from the right.  P(B) is the green arrow extending from the left.  The bottom line, P(A and B) < 0.5, is the constraint on their possible overlap.

Let’s say that P(A) is just slightly more than 0.6.  That means the region outside of P(A), but inside of P(A or B) is slightly less than 1.  That’s the part of P(B) that doesn’t overlap with P(A).    Then, the overlap has to be less than 0.5.  If we add something less than 1 to something less than 5, we get something less than 6.  P(B) can’t equal 0.6, but it can any value arbitrarily close to 0.6.

Thus, 0 ≤ P(B) < 0.6.

Answer = (B)

15)

Answer = (E)

The post Challenging GMAT Math Practice Questions appeared first on Magoosh GMAT Blog.

First, try these challenging GMAT practice problems. Give yourself a strict time-limit of six minutes for the set.

1) If A is a number, which of the following must be true for any A?

2) If F and G are integers, with F < G, which of the following must be true?

3) If J is an integer, which of the following must be true?

4) If p and q are two different odd prime numbers, such that p < q, then which of the following must be true?

Attacking “must be true” problems

For many math problems, one has to focus on finding the one right answer. That approach, with these problems, is highly problematic. If you are super-talented with math, fluent in Algebra, then perhaps you can scan the list of choices and immediately spot the correct answer. If you are talented enough to do that, then good for you!

For most folks, that approach is simply not feasible. The much more efficient solution involves elimination. You see, if any statement “must be true”, then it will be true for any value you pick. It takes incredible skill (or fantastic luck) to pick a single value that eliminates four answer choices and leaves only one. If that happens by chance, great, but don’t focus on that. Focus on speed and efficiency. Pick one value, eliminate what you can, then pick another, eliminate more, etc. Whittle down the choices until you are left with one. That’s the most efficient approach for these problems.

You have to pick values to plug in. At the start of picking, pick very easy values — with a few easy choices, you should be able to eliminate two or three answer choices, making your overall job much easier.

What values to pick?

If you picked the same values for all four questions above, that’s a problem. Those questions were specifically written so that the allowable numbers would be different in different questions.
When you hear “A is a number”, of what do you think? Remember that the general word “number”, like the general term “human being”, includes all types. Just as the individuals under the term “human being” constitute a bewildering variety, so do the citizens of the realm of “numbers.” Numbers include positive & negative, wholes & fractions & decimals, square-roots, pi, etc. etc. It’s a realm of perfect democracy — for example, -(pi)/5 is just as much a number as is 8. If, when a GMAT problem says “number”, your mind defaults to {1, 2, 3, 4, ..}, the “counting numbers”, then with all due respect, that’s really a third-grade way of thinking about the word “numbers”, and the GMAT will viciously punish that kind of thinking. You always must be aware of all possibilities for any category.

numbers = everything, positive/negative, zero, fractions, decimals, everything on the continuous infinity of the number line.

integers = positive and negative whole numbers = { … -3, -2, -1, 0, 1, 2, 3, …}

positive integers = the counting numbers, 1, 2, 3, 4, ..}; this list does NOT include zero, because zero is neither positive nor negative.

Remember that “odd integers” includes both positive and negatives, and “even integers” includes positives & zero & negatives. Prime numbers are, by definition, a subset of positive integers. There are no negative primes, and neither zero nor one is a prime number. It’s a very good idea to have the first few prime numbers memorized:

Primes = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, …)

Notice that 2 is the only even prime number (all other even numbers are divisible by 2, and hence are not prime). After 2, the primes are an irregularly spaced set of odd numbers that continues to infinity. In fact, let’s talk a moment about their pattern

The pattern of prime numbers

In this section, I am going to dip into very advanced math, far far beyond the GMAT realm. In modern mathematics, most work around the “pattern of prime numbers” revolves around the Riemann Hypothesis, the single hardest un-answered question in all of mathematics. The great mathematician Bernhard Riemann proposed this grand question as a conjecture in an 1859 paper, and it has utterly baffled the most brilliant mathematical minds on the planet every since. Forget about answering the question — you need close to a Ph.D. in mathematics just to understand what the question is asking and what it has to do with prime numbers! (A brief graph of the central player of the Riemann Hypothesis, the Riemann Zeta Function, is shown below.)

All of which is a long-winded way of saying: there is no easy algebraic formula that always produces prime numbers. A simple pattern for prime numbers absolutely does not exist. Thus, in a “must be true” question, any answer of the form “[simple algebra expression] is a prime number” cannot possibly be true all the time. That absolutely has to be a wrong answer.

Summary

If you had any big “aha!” while reading this article, you may want to give those problems another glance before reading the solutions below. If you have any further questions or observations, please let us know in the comments section below.

Practice problem explanations

1) First of all, notice that A is a “number” — could be positive or negative or zero, whole or fraction or negative. Let’s first try a very easy choice: A = 0. Then, for the five answer choices, we get:
(A) undefined
(B) false
(C) true
(D) true
(E) undefined

Remember, if some choice leads to an “undefined” value, a divide-by-zero error, then it is not “true” for that value. Error is just as good as false for eliminating answers. We are down to (C) & (D). Try another relatively simple choice, A = -2.
(C) 1 = 1, true
(D) +2 ≠ -2, false

Remember that if A is negative, a negative squared is a positive, and the square-root sign always has a positive output. Therefore, the left side of the (D) equation is always positive, and can’t be true if A is negative.
Answer = (C).

2) For this problem, both F & G are integers — they could be positive or negative or zero, but they must be whole numbers. Here’s a suggestion — one good choice would be to pick a positive number with smaller absolute value and a negative number with a larger absolute value: say, F = -4 and G = 1. Let’s true those in the answers:
(A) 16 < 1 false
(B) -64 < 1 true
(C) 4 < 1 false
(D) 4 < 1 false
(E) 1-16 = -15 > 0 false

With one magical choice, we were able to eliminate four answer choices. Keep in mind, the combination small positive/big negative is a good one, and so is pairing a negative number with 0 (e.g. F = -4, G = 0), because zero is greater than any negative number.
Answer = (B)

3) Here, J is an integer, so it can be positive or negative or zero, but not a fraction or a decimal. Zero is always a special case among integers, so let’s start there. J = 0: then,
(A) undefined error
(B) 0 > 0 false
(C) 0 > 0 false
(D) 1 > 0 true
(E) -1 > 0 false
Here, the special case was enough to eliminate four answers. Again, remember, if some choice leads to an “undefined” value, a divide-by-zero error, then it is not “true” for that value. Error is just as good as false for eliminating answers.
Answer = (D)

4) Here, p and q are odd prime numbers. First of all, eliminate (A), because there’s no simple algebraic rule that generates prime numbers. The easiest choices for numbers are p = 3 and q = 5. Try this:
(B) 3 + 5 = 8 is divisible by 4 — true
(C) 5 – 3 = 2 is not divisible by 4 — false
(D) 5 + 3 + 1 = 9 = (5^2) – (4^2) — true
(E) (3^2) + (5^2) = 34

This last number requires comment. Every odd number is the difference between two perfect squares, because adjacent squares . When we subtract two even squares or two odd squares, the number is always divisible by 4. Thus, 34 cannot be the difference of any pair of squares. Thus, on the basis of this choice, we can eliminate (E). We are left with (B) & (D).

Try p = 3 and q = 7.
(B) 3 + 7 = 10 is not divisible by 4 — false
(D) 3 + 7 + 1 = (6^2) – (5^2) — true

In fact, since p & q are both odd, (p + q + 1) must be odd, and as stated above, any odd number is the difference of two perfect squares. That’s why (D) must be true.
Answer = (D)

The post GMAT Quant: Must Be True Problems appeared first on Magoosh GMAT Blog.

First, a few practice GMAT math questions.

1) Last year, the price of a vacation package was P.  At the beginning of this year, the price went up 40%.  Lucas used a travel voucher to purchase the vacation package at 30% off this year’s price.   In terms of P, how much did Lucas pay?

(A) P + 10

(B) 1.1*P

(C) 1.12*P

(D) 0.9*P

(E) 0.98*P

2) Given that xy > 0, which of the following must also be greater than zero?

3) The numbers m, n, and K are all positive integers.  Given that m is a factor of K, that n is also a factor of K, and m < n, which of the following must also be a positive integer factor of K?

4) Marcia took a trip consisting of three segments at three different speeds: she drove a distance of (5D) at a speed of (2v), then a distance of (4D) at a speed of (3v), then a distance of D at a speed of (6v).  In terms of D & v, what was the total time of Marcia’s trip?

Answers and explanations will follow this article.

Picking numbers

It’s absolutely true that, when the GMAT gives a problem with variables in the answer choices, one possible approach is to pick numbers for the variables.  Students sometimes ask, “How do I pick numbers? Can you give me a general strategy for picking good numbers in all cases?”  Well, let’s start with the second question, which has a crystal clear answer: “No!”  You see, there are several different kinds of questions in which picking numbers could be an appropriate strategy, and the best numbers to pick in any case depend much more on the specifics of the question.  Some numbers that would be excellent choice in one question would be the worst possible choice in another question.  For many questions, the best numbers to pick are inextricably tied up with the mathematics in the question itself.  Some of this I will explain in the course of this article, and other things I will explain in the solution.

Numbers

First, let’s start with how you think about the word “number”.  What pops into your head when I say that word?  If what pops into your head is the set {1, 2, 3, 4, 5, …}, then with all due respect, you are operating with a third-grade level definition of the word “number”, and the GMAT Quant section will mercilessly punish you for such naïveté.  The word “number” is a broad category, encompassing everything on the number line: zero is a number; 4/7 is a number; negative four is a number; negative one fifth is a number; pi is number; the square-root of 17 is a number; etc. etc. All of these possibilities must leap to mind when a problem mentions a “number”.  It’s good to appreciate the categories of numbers, as well as their different properties.

Number sense is the ability to see patterns among numbers and sense which ones are best for a problem.  The person who asks the question at the top, how to pick numbers — what that person really needs is number sense!!  In other words, the person who asks for a general rule for picking numbers is really, at a very profound level, asking the wrong question.  Although that person might not even be familiar with the terms, what that person really needs to develop is number sense and mathematical thinking in general. These two posts say more about these larger issues, but I will say a few general things about picking numbers here.

Percent problems

One of the most braindead strategies is picking 100 for the starting value in every percent problem.  Everyone has heard this strategy.  Everyone who picks numbers does precisely this, as if it were a reflex.  That, in a nutshell, is the problem.  You see, if the test writers know that everyone makes the same choice, then they design answer choices designed to trip up people who make that all-too-obvious choice.   Think a little out of the box here.  Instead of 100, pick 500 or 1000 — you should be able to figure out round number percents with those relatively quickly.

Extending that strategy even a little further: if anything is a super-obvious choice to pick, then rather than pick that super-obvious thing, pick something a shade away from super-obvious.

If the question is one of a very typical type, a percent increase or decrease followed by another percent increase or decrease, then I would strongly advise against picking numbers.  Instead, I would recommend using multipliers to handle percent changes.

Dimensional analysis

If different variable represent quantities of different units, then make sure you understand what combination of them will have the correct units for the answer choice.   In Physics, this thinking about the units of different quantities is called dimensional analysis.   For example if J is in units of miles per gallon, and M is in units of miles, then to get units of gallons, I would need M/J — if the question is asking for “how many gallons”, then the correct answer would have to be a number times M/J.   Any other combinations (J/M, M*J, etc.) are automatically incorrect.  This helps you eliminate answers choices, which makes the process of plugging in numbers easier.

Know the traps

Many different math questions have very predictable traps: the mistake that a large number of test takers routinely make, time and time again.  GMAT problems always have answer choices that reflect these traps: in some cases, it’s as if the GMAT simply sets up a trap answer like a big butterfly net, and they watch hordes of test takers run headlong into it.  For every such trap you learn to recognize and avoid, you will be that much ahead of so many other test takers.

One big trap involves a sequence of percent increases & decreases.  There are a few common ones involving fractions.  Still more involve exponent properties and powers.  Some very common ones involve incorrect variations on three basic algebra formulas.  Here’s a post on the most common math mistakes.

Avoid one

Avoid picking 1 as a value for a variable.  The reason is: one has so many special properties that no other number has.   For example, all powers of 1 equal one: one is the only number that has this particular property.  Think if you had to distinguish choices involving different powers of that variable: picking one for the value would make all the power equal the same thing, making it impossible to distinguish which answers should be eliminated.  One is often the worst possible choice.

The only exception to this is — sometimes, it’s worthwhile to pick very easy numbers, like 1 and 2, just to do a very quick elimination of a couple of answers.  If that’s your goal, 1 can be a perfectly acceptable choice on a first pass.  It will never be a choice that eliminate everything, but if your goal is to begin by eliminating low-hanging fruit, then by all means pick it.

To pick or not to pick

Finally, let’s step back.  Why do folks pick numbers for problems with variable?  One reason is that sometimes, quite rarely, one can either eliminate many answers or even get to a single right answer very quickly by picking numbers.  Most often, though, this is not the case.  In fact, often it’s much quicker and much easier to do the algebra directly.  After all, this is precisely why mathematicians invented algebra in the first place: it’s a powerful efficient problem solving system.

Most of the time, most habitual number-pickers are choosing this strategy, not because it’s necessarily faster or easier or more efficient or better in any particular way, but simply because these folks are absolutely petrified of algebra.  For these folks, algebra is a fire-breathing dragon, and folks will do anything, no matter how awkward or cumbersome, simply to avoid facing that dragon.   If this describes you, then I have some hard medicine for you: do algebra.  From this point forward, in any problem with variable, you are forbidden to pick numbers, and you must practice the algebra.  You see, for the GMAT, you will have to know some algebra, and the only way you get better at algebra is by practicing it at every opportunity you can get.

Summary

If you had any flashes of insight while reading this article and exploring the links, you may want to give the questions at the top another look.   Here’s another problem of another sort, from inside Magoosh:

5) http://gmat.magoosh.com/questions/123

Finally, if you have any questions, please let us know in the comments section at the bottom.

Practice problem explanations

1) First of all, let’s talk about the trap.  Even if you don’t learn how to solve this problem, learn to recognize this trap, possibly the single most enticing trap on the entire GMAT.  When a percent increase or decrease is followed by another percent increase or decrease, you cannot simply add or subtract the two percents.  If you increase something by 40% and then decrease it by 30%, that is absolutely NOT a 10% increase.   That is the trap answer to which hordes of GMAT test takers are drawn like bugs to a bug zapper.

Problems are this sort are not ones on which I would recommend picking numbers at all.  Instead, I would recommend using multipliers for the percent changes.  The multiplier for a 40% increase would be (1 + 0.40) = 1.40.  The multiplier for a 30% decrease would be (1 – 0.30) = 0.70.  The net effect of both would be

Final price = (1.40)*(0.70)*P = 0.98*P

Answer = (E)

Even though it’s not as efficient, here’s a picking-numbers approach, just so you can see how the numbers play out. Let p = $1000.   This increases by 40%, to $1400.  Now, this is going to decrease by 30% — but notice!  It decreases not by 30% of the original, $1000; rather, it decrease by 30% of the new price, $1400.  After the 40% increase, the price of $1000 doesn’t even exist anymore: the only price that exists at that point is the $1400 price; that’s what decreases by 30%.  Well, 10% of $1400 is $140, and so, 30% = 3*10% = 3*$140 = $420.  The price decrease to $1400 – $420 = $980, which is 98% of P or 0.98*P.

2) This is a “must be true” problem.  First of all, remember that if the product of two numbers is positive, either both are positive or both are negative.  Since we want something that “must be greater than zero”, i.e. must be positive, I am going to pick negative numbers for my values of x & y.  Probably all five answer choices would be positive if I picked to positive numbers.  I am going to pick negatives, because any answer choice that winds up negative will be something I can eliminate, and on “must be true” problem, my goal is to eliminate answers.

I’ll start with very simple choices, x = –1 and y = –2.

(A) (–1) + (–2) = –3 — negative, so we can eliminate this.

(B) (+1) + (–2) = –1 — negative, so we can eliminate this.

(C) (-3)^2 + 1 = 9 + 1 = 10 — can’t eliminate yet

(D) 1 – 4 + 100 = +97 — can’t eliminate yet

(E) (-1)^2 = 1 — can’t eliminate yet

So, with one easy choice, we have eliminated two answer choices, (A) & (B).  I notice that x-squared minus y-squared, in choice (D) — that expression could be negative.  If it’s a small negative, then adding a hundred would make it positive, so to eliminate that, we need the subtracted term to be big.  Use x = –1 and y = –30.

(D) (+1) – 900 + 100 = –799 — negative, so we can eliminate this.

Now, we are down to just (C) & (E).  Notice, both of them have something in parentheses squared.  Of course, if we square a positive, we get a positive, and if we square a negative, we get a positive, so it would seem that square anything gives us a positive, right?  Almost.  The one exception: when you square zero, you get zero, and zero is not positive.  Zero is neither positive or negative.  Therefore, if either of these answers could have an output of zero, then it is not always positive.  Notice, if we pick x = –2 and y = –2, then the sum in the parentheses of (E) is, ((–2) – (–2)) = 0, and the sum in the parentheses of (C) is, ((–2) + (–2)) = –4

(C) +16 — can’t eliminate

(E) 0 — because zero is not positive, we can eliminate this.

In fact, if we add two negative numbers, or any two positive numbers, then we never can get zero.  Therefore, the inside of the parentheses of (C) will never equal zero: it will always be a non-zero number; and therefore, when we square, it must be positive.  That’s why (C) is the only possible answer.

3) First, pick a K with a lot of factors: say K = 24.  Now, for m and n, pick two factors of 24 that are not factors of each other — for example, pick the two biggest factors of 24 less than 24 itself: n = 12 and m = 8

(A) 8 + 12 = 20 — not a factor of 24

(B) 8*12 = 96 — not a factor of 24

(C) (12*12)/8 = (3*12)/2 = (3*6) = 18 — not a factor of 24

(D) 24/8 = 3 — that’s a factor of 24

(E) 24/96 — not even an integer, so certainly not a factor

The only one that works is (D), the correct answer.

Notice that, in (C), to find twelve squared divided by eight, the absolute worst, most unstrategic thing one could do would be to square 12, get 144, and then have to divide that by 8 without a calculator.  Notice, instead, I left the two 12’s unmultiplied while I canceled factors from the denominator, easily reducing the entire process to single-digit multiplication.  Cancel before you multiply is one of the biggest non-calculator tricks on GMAT math.

4) First of all, let’s use a little dimensional analysis, discussed above. If D is a distance, and v is a speed, we know v = D*T, or T = D/v.  Anything with a D in the numerator and a v in the denominator will be a time, but if those two are reversed, it will not have the right units.  We can eliminate (B) & (D) right away on these grounds.

On this one, I recommend an algebraic approach.

(distance #1) = (rate #1)*(time #1)

5D = (2v)*(time #1)

time #1 = (5D)/(2v)

(distance #2) = (rate #2)*(time #2)

4D = (3v)*(time #2)

time #2 = (4D)/(3v)

(distance #3) = (rate #3)*(time #3)

D = (6v)*(time #1)

time #3 = (D)/(6v)

TOTAL TIME = (time #1) + (time #2) + (time #3)

= (5D)/(2v) + (4D)/(3v) + (D)/(6v)

= (D/v)[5/2 + 4/3 + 1/6]

= (D/v)[15/6 + 8/6 + 1/6]

= (D/v)[24/6]

= (D/v)[4]

= (4D)/v

Answer = (A)

While it’s somewhat less efficient, here’s a picking number approach.  We want D to be a number with lots of factors, including a factor of 6, because it’s going to be divided by 2 and 3 and 6 times something.  Let’s pick D = 60 and v = 10

First leg: distance = 5D = 300 mi, at a speed of 2v = 20 mph, that’s T = D/V = 300/20 = 15 hours.

Second leg: distance = 4D = 240 mi, at a speed of 3v = 30 mph, that’s T = D/V = 240/30 = 8 hours.

Third leg: distance of D = 60 mi, at a speed of 6v = 60 mph —that will take exactly one hour.

TOTAL TIME = 15 hr + 8 hr + 1 hr = 24 hours

Well, D/V = 60 mi/10 mph = 6 hours, and clearly, we need to multiply this by 4 to get 24 hours, not by 10/11 or 11/10, so (A) is the answer.

The post How to Plug in Numbers on GMAT Math Questions appeared first on Magoosh GMAT Blog.

First, a bank of eight practice problems

1) A certain zoo has mammal and reptiles and birds, and no other animals.  The ratio of mammals to reptiles to birds is 11: 8:5.  How many birds are in the zoo?

Statement (1): there are twelve more mammals in the zoo than there are reptiles

Statement (2): if the zoo acquired 16 more mammals, the ratio of mammals to birds would be 3:1

2) In a certain large company, the ratio of college graduates with a graduate degree to non-college graduates is 1:8, and ratio of college graduates without a graduate degree to non-college graduates is 2:3.  If one picks a random college graduate at this large company, what is the probability this college graduate has a graduate degree?

3) Dan’s car gets 32 miles per gallon.  If gas costs $4/gallon, then how many miles can Dan’s car go on $50 of gas?

(A) 61.25 miles
(B) 256 miles
(C) 400 miles
(D) 1600 miles
(E) 6400 miles

4) For a certain concert, the price of balcony tickets was exactly half the price of orchestra tickets.  The ratio of balcony to orchestra tickets sold was 3:2.  What was the price of one orchestra ticket?

Statement (1): the total revenue taken in from tickets of both kinds was $4200

Statement (2): the difference between the number of balcony tickets sold and the number of orchestra tickets sold was 40

5) At a certain high school, there are three sports: baseball, basketball, and football.  Some athletes at this school play two of these three, but no athlete plays in all three.  At this school, the ratio of (all baseball players) to (all basketball players) to (all football players) is 15:12:8.  How many athletes at this school play baseball?

Statement (1): 40 athletes play both baseball and football, and 75 play football only and no other sport

Statement (2): 60 athletes play only baseball and no other sport

6)  In a certain class, the ratio of girls to boys is 5:4.  How many girls are there?

Statement (1): If four new boys joined the class, the number of boys would increase by 20%.

Statement (2): If the number of girls increases by 50%, then after such an increase, the probability that a randomly chosen student would be a boy would be 8/23

7) If 28 passes to a show cost $420, then at the same rate, how much will 42 passes cost?

(A) $500
(B) $560
(C) $630
(D) $700
(E) $840

8) Metropolitan Concert Hall was half full on Tuesday night.  How many seats are in the Hall?

Statement (1): If the number of people in the Hall increased by 20% from Tuesday night to Wednesday night, then the Hall would be 60% full on Wednesday night.

Statement (2): If 20 more people showed up on Tuesday night, that would have increased the number of people in the Hall by 4%.

Answers will come at the end of the article.

Ratios and proportions: a review

Probably ratios and proportions have been on your radar since some time in grade school or middle school, when they are introduced.  You remember there are a lot of “mathy” facts related to these things, but it’s a bit blurry.  Here are a few quick facts as reminders.

1. What is a ratio?  Fundamentally, a ratio is a fraction, and is subject to all the laws of fractions.   “Ratios” and “fractions” are mathematically identical.

2. What is a proportion?  Is it the same as a ratio?  No, a proportion is NOT the same as a ratio.  Whereas a ratio is single fraction by itself (e.g. 1/3), a proportion is an equation that sets two ratios equal to each other (e.g. 1/3 = 4/12).   See the fraction post for a refresher on what you can and can’t cancel in a proportion.

3.  When geometric figures are similar, the sides are proportional.  Geometric similarity is a topic rife with ratios and proportions.  One helpful idea discussed in that post is the idea of a scale factor, which, it turns out, is helpful in many many proportional situations well beyond anything having to do with geometry.

4.  Percents and probabilities are specialized cases of ratios, and either gives you very much the same kind of information.

Those are four simple ideas, and there’s one more, but it takes a little setting up to express.   Let’s divide the mathematical information that can appear in a problem into two categories.  The first category is “ratio  information”, and this includes any statements about percents or probability information.   The second I’ll call, for lack of a better term, “count information” — not a percent or ratio, but an actual count — i.e. how many people or animals or whatever; it could be how many in any particular group, the sum or difference of multiple groups, or how many are in the whole population.   This leads us the final important simple idea:

5.  To get count information as an output, you need some count information as an input.  If you all you have is ratio information as an input, it is impossible to get count information as an output.

Ratios and portioning

Suppose, in some population, there are three kinds of things, A & B & C, and they are in a proportion of 3:8:4.  That’s ratio information.  Suppose we want to know either the percent that A makes of the whole, or the ratio of A to the whole.  That’s also ratio information, so we should be able to calculate it from that given ratio.

It can be very helpful to understand ratios in terms of “parts”.  For every 3 parts of A, there are 8 parts of B and 4 parts of C.  To get the whole in the same ratio, we simply add up the parts — 3 + 8 + 4 = 15 parts.  Keep in mind, we have zero information about the actual size of the population — we have no count information.  This simply indicates the size of the population in the same ratios, so we could say A to the whole is 3/15 = 1/5, and B to the whole is 8/15, and C to the whole is 4/15.  This means that A is 1/5 of the whole, or 20%.

Similarly, suppose girls are 4/7 of a class.  Girls are four parts, and the whole is seven parts, so boys must be the other three parts.  (Here, for mathematical simplicity, I will conform to very conventional gender assumptions, with sincere apologies to all transgender and transsexual readers.)  Proceeding, we see that boys must be 3/7 of the class, the ratio of boys to girls must be 3:4.  Thinking about “parts” and portioning can be a powerful way to expand the information you get from any given ratio.

Summary

If you had any realizations while reading this blog, you may want to go back and give the practice problems another glance, before proceeding to the solutions.  Here’s another practice question, involving two less-than-lovable baseball teams:

9) http://gmat.magoosh.com/questions/60

If you have any questions about this article, please let us know in the comment section at the bottom!

Practice problem explanations

1) A short way to do this problem.  The prompt gives us ratio information.  Each statement gives use some kind of count information, so each must be sufficient on its own.   From that alone, we can conclude: answer = D.  This is all we have to do for Data Sufficiency.

Here are the details, if you would like to see them.

Statement (1): there are twelve more mammals in the zoo than there are reptiles

From the ratio in the prompt, we know mammals are 11 “parts” and reptiles are 8 “parts”, so mammals have three more “parts” than do reptiles.  If this difference of three “parts” consists of 12 mammals, that must mean there are four animals in each “part.”  We have five bird “parts”, and if each counts as four animals, that’s 5*4 = 20 birds.  This statement, alone and by itself, is sufficient.

Statement (2): if the zoo acquired 16 more mammals, the ratio of mammals to birds would be 3:1

Let’s say there are x animals in a “part” — this means there are currently 11x mammals and 5x birds.  Suppose we add 16 mammals.  Then the ratio of (11x + 16) mammals to 5x birds is 3:1. —-

(11x + 16)/(5x) = 3/1 = 3

11x + 16 = 3*(5x) = 15x

16 = 15x  – 11x

16 = 4x

4 = x

So there are four animals in a “part”.  The birds have five parts, 5x, so that’s 20 birds.   This statement, alone and by itself, is sufficient.

Both statements sufficient.  Answer = D.

2) We are given ratio information, and we are asked for ratio information: a probability.  That’s fine.  For simplicity, let’s use the abbreviations:

P = college graduates with a graduate degree

Q = college graduates without a graduate degree

R = non-college graduates

The two ratios we are given is

P:R = 1:8

Q:R = 2:3

We have combine the ratios, by making the common term the same.   The non-college graduates, R, are the common member, accounting for 8 parts in the first ratio, and 3 parts in the second, so we have to multiply the first ratio by 3/3, and the second by 8/8.

P:R = 1:8 = 3:24

Q:R = 2:3 = 16:24

P:Q:R = 3:16:24

Now, the probability about which the question asks  is about only college graduates, so ignore the non-college graduates, and just focus on the ratio among college graduates:

P:Q = 3:16

There are 3 + 16  = 19 parts in total, and of those, 3 are the folks in P, so that’s a probability of 3/19.  Answer = (D).

3) To get from dollars to gallons, we have to start with the $50, and divide by ($4/gallon) — that cancels the unit of dollars, and leaves us with gallons —– 50/4 gallons.  (For the moment, I’ll leave it as that un-simplified fraction).

Now, we need to get from gallons to distance.  We multiply by (32 miles/gallon), to cancel the units of gallons, and leave only miles —-

(50/4) gallons*(32 miles/gallon) = (32*50)/4 miles = (8*50) miles = 400 miles

Notice, we didn’t do anything with the four in the denominator until we got a 32 in the numerator with which we could cancel.

Answer = (C)

4) We know the price of a balcony ticket, B = (1/2)*C, or 2B = C, where C = the price of orchestra ticket.  The 3:2 ratio tells us that, for some n, the concert sold 3n balcony tickets and 2n orchestra tickets.

Statement (1): the total revenue taken in from tickets of both kinds was $4200

We know total revenue would be (3n) balcony ticket plus (2n) orchestra tickets:

(3n)*B + (2n)*C = (3n)*B + (2n)* 2B = 3nB + 4nB = 7nB = $4200,  or nB = 600.  The problem with this — we have two variables, n and B.  In one extreme case, we could say n = 1 (sold 3 balcony ticket and 2 orchestra tickets), and a balcony ticket cost $600.  At another extreme, we could say n = 600 (sold 1800 balcony ticket and 1200 orchestra tickets) and a balcony ticket cost $1.  This information alone does allow us to determine a definitive answer to the prompt question.   This statement, alone and by itself, is insufficient.

Statement (2): the difference between the number of orchestra tickets sold and the number of balcony tickets sold was 40

Well, balcony tickets are 3 parts, and orchestra tickets are two parts, so there’s one part of difference between them.  The statement lets us know: one part = 40, so that allows us to figure out: we sold 3*40 = 120 balcony tickets, and 2*40 = 80 orchestra tickets.  That’s great, but unfortunately, with this statement, we get absolutely no financial information, so we can’t solve for a price.   This statement, alone and by itself, is insufficient.

Combined statements: now, put this altogether.  From the first statement, we got down to the equation nB = 600, and the second statement, in essence, tells us n = 40.  Therefore, B = 600/40 = 60/4 = $15.  That’s the price of a balcony ticket.  The price of an orchestra ticket is twice that, $30.  With both pieces of information, we were able to solve for this.   Together, the statements are sufficient.

Statement are sufficient together but not individually.  Answer = C

5) This is a tricky one.  The ratio 15:12:8 double-counts some students.   In terms of a triple Venn diagram:

So, in that diagram

a = folks who play baseball only

b = folks who play football only

c = folks who play basketball only

d = folks who play baseball and football only

e = folks who play baseball and basketball only

f = folks who play football and basketball only

g = folks who play all three sports

We know from the prompt that g = 0, but at the outset, that’s still six unknowns!!

Now, notice:

everyone who plays baseball = a + d + e

everyone who plays basketball = c + e + f

everyone who plays football = b + d + f

So, the ratio given in the problem is:

(a + d + e):(c + e + f):(b + d + f) = 15:12:8

Of course, these are all fractions, so we can’t simplify: there is no way to simplify.

We would like to find the total number of baseball players, a + d + e.

Statement #1 tells us that d = 40 and b = 75.  Thus

total number of baseball players = a + 40 + e

We don’t have enough information to calculate this, and we don’t have enough ratio information to solve.   This statement, alone and by itself, is insufficient.

Statement #2 tells us a = 60.  Thus

total number of baseball players = 60 + d + e

We don’t have enough information to calculate this.  This statement, alone and by itself, is insufficient.

Combined statements: Now we know a = 60, b = 75, and d = 40.

total number of baseball players = 60 + 40 + e= 100 + e

total number of football players = 75 + 40 + f = 115 + f

We know the ratio of these two quantities, baseball to football, is 15:8.  Unfortunately, that would give us only one equation for two unknowns, e & f.  If we can’t solve for these, then we can solve for the total number of baseball players.   Even with both statements combined, we cannot determine the answer.

Both statements combined are insufficient.   Answer = E

6) Statement (1): If four new boys joined the class, the number of boys would increase by 20%.  This means, those four new boys count as 20% of the original boys, which means 2 new boys would be 10% of the original boys, which means there must have been 20 original boys.   If we know how many boys, we can use the ratio to calculate how many girls.   This statement, alone and by itself, is sufficient.

Statement (2): If the number of girls increases by 50%, then after such an increase, the probability that a randomly chosen student would be a boy would be 8/23.  The prompt gives us “ratio information”, and this statement also gives us more “ratio information”.  We have absolutely no “count information”, so we can’t figure out the count or number of anything.   This statement, alone and by itself, is insufficient.

First statement sufficient, second insufficient.  Answer = A

7) We can solve this question by setting up a proportion:

Divide the left fraction, numerator and denominator, by 7:

Now, cancel the factor of 4:

For this multiplication, use the doubling & halving trick — double 15 is 30, and half of 42 is 21:

x = (15)*(42)  = 30*21 = $630

Answer = C

8) The prompt gives us “ratio information”.

Statement #1 also gives us “ratio information”, so there is no way we can calculate a count, such as total number of seats in the hall.  This statement, alone and by itself, is insufficient.

Statement #2: 20 people would be 4% of the audience.  Divide by two — 10 people would have be 2%; now  multiply by five — 50 people would have been 10% of the audience; now multiply by ten — 500 people would be 100% of the audience on Tuesday night.  Since the concert hall was half full, there must be 1000 seats in total.  This piece of information allows us to solve for the answer to the prompt question.   This statement, alone and by itself, is sufficient.

First statement insufficient, second sufficient.  Answer = B

Special Note:

To find out where ratios and proportions sit in the “big picture” of GMAT Quant, and what other Quant concepts you should study, check out our post entitled:

What Kind of Math is on the GMAT? Breakdown of Quant Concepts by Frequency

The post GMAT Quantitative: Ratio and Proportions appeared first on Magoosh GMAT Blog.

We recently issued a call for you, and other students, to submit questions that we could answer in a video series. There were a lot of great submissions, and the ones that we thought useful for a broad swath of students, we answered in our video series. These videos will be released over the next few weeks.

But some of the questions require a longer explanation. So in this post, I want to answer those other questions that came in.

Question 1

Of the 120 passengers on Flight 750, 50% are female. 10% of the passengers sit in first class, and the rest of the passengers sit in coach class. If 1/3 of the passengers in first class are male, how many females are there in coach class?

44

48

50

52

56

Here we have a math problem that you might see on the GMAT. This question would rank as an easy to medium level question on the test. Let’s look at how to solve this problem.

First thing I always do when I approach a problem is wade through the information and write down the facts. So, here is what I’m told:

• 120 people
• 50% female
• 10% of passengers sit in first class
• ⅓ of passengers in first class are male

Now, I will interpret this information and see what else it tells me:

• 50% female means there are 60 males and 60 females on the plane
• 10% of passengers sit in first class means that 12 people are seated in first class. If 12 people are in first class, there are 108 people in coach.
• ⅓ of passengers in first class are male, which means that 4 males and 8 females are seated in first class

Now our job is to find out how many females are seated in coach. This may seem like a tall order, but if we use a Double Matrix this will be fairly simple. First we need to make a table with what we know:

Alright now we can start filling in our boxes with the other information and we will be able to solve for how many females are sitting in coach.

Now with some simple subtraction (108 – 56 = 52 or 60 – 8 = 52), we have our answer. You could’ve used a Venn diagram to find the solution, but I find the Double Matrix more straightforward. If you don’t, spend some time learning about the Double Matrix because it is an important tool have in your math toolbox.

Question 2

What is a ratio?

This is a question that we have addressed in a few other posts (here, here, and here). And I am not sure if I can provide a better explanation than some of these other resources (Wikipedia or Khan Academy), but I will give it a shot.

Simply stated, a ratio expresses a relationship between two things. These two things are connected for some reason and exert some influence on each other. Whether it is the relationship between miles and time, the relationship between a distance on a map to an actual distance, or the relationship of teachers to students in a school, they all tell us something about the relationship.

For example, when we talk about the speed limit, we are talking about a relationship, one of the fundamental relationships in the universe—time and distance. So speed is really just a ratio that tells us how far we have traveled, or could travel, in a certain amount of time.

A ratio can be expressed two different ways—with a colon (:) or with a fraction bar (/). As we know with speed, we usually say “per” to indicate the fraction bar or the colon. So when your friend turns to you and says, “We’re going a 165 miles per hour,” you could write this as:

165 miles : 1 hour

165 miles / 1 hour

One crucial point to know is that a ratio doesn’t necessarily tell you totals. Let me show you what I mean. If I told you that a school has 1 teacher for every 32 students, I haven’t told you anything that helps you know how many students are at the school and how many teachers are at the school.

But with another piece of information, you’d be able to figure out all this information. So, if I told you that there are 12 teachers at the school, then we can determine how many students are at the school. If 1 teacher has 32 students, then 2 teachers have 64 students, and 3 teachers have 96 students, and…. Hopefully, you can see that we are just counting by 32 until we have counted 12 teachers. That is the same as a simple multiplication formula: 12 * 32 = 384.

So this is not the end all for your understanding of ratios—this is the beginning! Take the time to review some of the other resources to deepen your understanding of ratios.

The post Left Overs Ratios and a Double Matrix from the QA Webinar appeared first on Magoosh GMAT Blog.

First, a few practice problems.  Remember: no calculator!

1) If $5,000,000 is the initial amount placed in an account that collects 7% annual interest, which of the following compounding rates would produce the largest total amount after two years?

(A) compounding annually

(B) compounding quarterly

(C) compounding monthly

(D) compounding daily

(E) All four of these would produce the same total

2) If A is the initial amount put into an account, R is the annual percentage of interest written as a decimal, and the interest compounds annually, then which of the following would be an expression, in terms of A and R, for the interest accrued in three years?

3) At the beginning of January 2003, Elizabeth invested money in an account that collected interest, compounding more frequently than a year. Assume the annual percentage rate of interest remained constant.  What is the total amount she has invested after seven years?

Statement #1: her initial investment was $20,000

Statement #2: the account accrued 7% annual interest

4) Sarah invested $38,700 in an account that paid 6.2% annual interest, compounding monthly.  She left the money in this account,  collecting interest for a full three-year period.  Approximately how much interest did she earn in the last month of this period?

(A) $239.47

(B) $714.73

(C) $2793.80

(D) $7,888.83

(E) $15,529.61

Solutions to these will be given at the end of the article.

Simple Interest

In grade school, you learn about simple interest, largely because we want to teach little kids something about the idea of interest, and that’s the only kind of interest that children can understand.  No one anywhere in the real world actually uses simple interest: it’s a pure mathematical fiction.

Here’s how it works.  There’s an initial amount A, and an annual percentage P.  At the end of each year, you get interest in the amount of P percent of A — the same amount every year.  Suppose the initial amount is $1000, and the annual percentage is 5%.  Well, 5% of $1000 is $50, so each year, you would get the fixed value of $50 in interest.  If you plotted the value of the account (principle + interest) vs. time, you would get a straight line.

Again, this is a fiction we teach children, the mathematical equivalent of Santa Claus.  This never takes place in the real world.

Compound Interest, compounding annually

With compound interest, in each successive year or period, you collect more interest not merely on the principle but on all the interest you have accrued up to that point in time.   Interest on interest: that’s the big idea of compound interest.

Here’s how it plays out.  Again, there’s an initial amount A, and an annual percentage P, and we also have to know how frequently we are compounding.  For starters, let’s just say that we are compounding annually, once a year at the end of the year.   In fact, let’s say we have $1000, and the annual interest rate is 5%.  Well, in the first year, we would earn five percent on $1000, and gain $50 in interest.   The first year is exactly the same as the simple interest scenario.  After that first year, we now have $1050 in the account, so at the end of the second year, we gain 5% of $1050, or $52.50, for a new total of $1102.50.  Now, that’s our new total, so at the end of the third year, we gain 5% of $1102.50, or $55.12 (rounded down to the nearest penny), for a new total of $1157.62.  At the end of three years, the simple interest scenario would give us $1150, so the compound interest gains us an extra $7.62 —- not much, but then again, $1000 is not a lot to have invested.  You can see that, with millions or billions of dollars, this would be a significant difference.

Here, I was demonstrating everything step-by-step for clarity, but if we wanted to calculate the total amount after a large number of years, we would just use a formula.  We know that each year, the amount increased by 5%, and we know that 1.05 is the multiplier for a 5% increase.  After 20 years, the amount in the account would have experienced twenty 5% increases, so the total amount would be

We don’t have to do it step-by-step: we can just jump to the answer we need, using multipliers.  Of course, for this exact value, we would need a calculator, and you don’t get a calculator on the GMAT Quantitative section.  Sometimes, though, the GMAT lists some answers in “formula form”, and you would just have to recognize this particular expression, , as the right formula for this amount.

If we graph compound interest against time, we get an upward curving graph (purple), which curves away from the simple interest straight line (green):

The curve of the graph, that is to say, the multiplying effect of the interest, gets more pronounces as time goes on.

BIG IDEA #1: as long as there is more than one compounding period, then compound interest always earns more than simple interest.  

Other compounding periods

A year is a long time to wait to get any interest.   Historically, some banks have compounded over shortened compounding period.  Here is a table of common compounding periods:

Technically, the fraction for “compounding daily” would be 1/365 in a non-leap year and 1/366 in a leap year; alternatively, one could use 1/365.25 for every year.

Now, how does this work?  Let’s say the bank gives 5% annual, compounding quarterly.  It would be splendid if the bank wanted to give you another 5% each and every quarter, but that’s not how it works.  The bank takes the percentage rate of interest and multiplies it by the corresponding fraction.  For 5% annual, compounding quarterly, we would multiply (5%)*(1/4) = 1.25%.  That’s the percentage increase we get each quarter.   The multiplier for a 1.25% increase is 1.0125.  Suppose we invest $1000 initially and keep the money in this account for seven years: that would be 7*4 = 28 compounding periods, so there are twenty-eight times in that period in which the account experiences a 1.25% increase.  Thus, the formula would be

For compounding quarterly, we divide the annual rate by four and compound four times each year.  For compounding monthly, we divide the annual rate by twelve and compound twelve times a year.   Similarly, for daily or any other conceivable compounding period.

How do the amounts of interest accrued compare for different compounding periods?  To compare this, let’s pick a larger initial value, $1,000,000, and collect over a longer period, 20 years.  Below are the total amounts, after twenty years, on an initial deposit of one million dollars compound at 5% annual:

• simple interest = $2,000,000
• compounding annually = $2,653,297.71
• compounding quarterly = $2,701484.94
• compounding monthly = $2,712,640.29
• compounding daily = $2,718,095.80
• compounding hourly = $2,718,274.07

As we go down that list, notice the values keep increasing as we decrease the size of the compounding period (and, hence, increase the total number of compound periods).  This leads to:

BIG IDEA #2: We always get more interest, and larger account value overall, when the compounding period decreases; the more compounding periods we have, the more interest we earn. 

Admittedly, the difference between “compounding daily” and “compounding hourly” only turn out to be a measly $178 on a million dollar investment over a twenty year period. A infinitesimally small difference, but technically, it is still an increase to move from “compounding daily” to “compounding hourly.”

Both ways

Notice that so far, we have been talking about investments and interest that you earn.   All of this, everything in this article, works equally well for debt and interest that you have to pay.  Just as the compounding effect, over time, magically multiplies an investment, so the same compounding effect will sink you deeper and deeper into debt.  This, in a nutshell, is the lose-lose proposition of credit card debt.

A little more than you need to know …

You may have noticed that, as the compounding periods get smaller and smaller, the amount increase in every diminishing steps.  In fact, as you may suspect, the total amount you possibly could earn from decreasing the compound period reaches a ceiling, a limit.  This limit is called “compounding continuously.”  The mathematics of this involves the special irrational number e, named for Leonard Euler (1707 – 1783), who is often considered the single greatest mathematician of all times.

e = 2.71828182845904523536028747135266249775724709369995 …

The formula for calculating continuously compound interest involves e, and is more complicated than anything you need to understand for the GMAT.   I will simply point out, in the example above, with one million dollars invested at 5% annual for 20 years, the limit of continuously compounding would be 1 million times e, which is $2,718,281.83.  You do not need to understand why that is or how this was calculated.

Most banks use monthly compounding interest, for accounts and for mortgages: this make sense for accounts with monthly statement or payments.  Credit cards tend to use continually compounding interest, because charges or payments could occur at any point, at any time of any day of the month.    In addition to understanding the mathematics of compound interest, it’s good to have a general idea of how it works in the real world: after all, the history or logic of compound interest would be a very apt topic for a Reading Comprehension passage or a Critical Reasoning prompt on the GMAT!

Summary

If you had any “aha’s” while reading this article, you may want to go back a take another look at the four practice problems above.  If you would like to express anything on these themes, or if you have a question about anything I said in this article, please let us know in the comments section.

Practice problem explanations

1) The smaller the compounding period is, the greater the number of times the interest will be compounded.  Of course, if we compound monthly instead of quarterly, then we are compounding by 1/12 of the annual rate each time, instead of 1/4.  The number of times we compound goes up, but the percentage by which we compound each time goes down.  Naively, you may think that those two would cancel out, but they don’t.  As discussed above, as the compounding period gets smaller, the total amount of interest earned goes up.  Therefore, we will get the most with the smallest compound period, daily.   Answer = (D)

2) Notice that, since R is the annual percent as a decimal, we can form a multiplier simply by adding one: (1 + R).  That’s very handy!  We will explore two different methods to get the answer.

Method One: Step-by-step

Starting amount = A

After one year, we multiply by the multiplier once

That’s the total amount at the end of the first year. The amount A is the original principle, and AR is the interest earned.

At the end of the second year, that entire amount is multiplied by the multiplier.  We need to FOIL.

That’s the total amount at the end of the second year. The amount A is the original principle, and the rest is the interest earned.

At the end of the third year, this entire amount is again multiplied by the multiplier.

That’s the total amount at the end of the third year. The amount A is the original principle, and the rest is the interest earned.

Answer = (C)

Method Two: some fancy algebra

Over the course of three years, the initial amount A is multiplied by the multiplier (1 + R) three times.  Thus, after three years,

Now, if you happen to know it offhand, we can use the cube of a sum formula:

Thus,

and

Answer = (C)

3) In order to determine the total amount at the end of an investment, we would need to know three things: (a) the initial deposit; (b) the annual percentage rate; and (c) the compounding period.

Statement #1 tells us the initial deposit but not the annual percentage rate. Insufficient.

Statement #2 tells us the annual percentage rate but not the initial deposit. Insufficient.

Together, we know both the initial deposit and the annual percentage rate, but we still don’t know the compounding period.  All we know is that it’s less than a year, but quarterly compounding vs. monthly compounding vs. daily compounding would produce different total amounts at the end.  Without knowing the exact compounding period, we cannot calculate a precise answer.  Even together, the statements are insufficient.

Answer = (E)

4) Without a calculator available, this is a problem screaming for estimation.  The problem even uses the magic word “approximately” to indicate that estimation is a good idea, and the answer choices are spread far apart, making it easier to estimate an individual answer.

Let’s round the deposit up to $40,000, and the percentage down to 6% annual.  Compounding monthly means each month, Sarah will accrue 6/12 = 0.5% in interest.  Well, 1% of $40,000  is $400.  Divide by 2: then 0.5% of $40,000 would be $200.  That would be the simple interest amount, as well as the interest in the first month.  We expect the amount in the last month to be a little more than this, but certain not even as large as double this amount.  The only possible answer is (A).

The post Compound Interest on the GMAT appeared first on Magoosh GMAT Blog.

Success on the GMAT depends on many skills, and one skill that can help a student make enormous progress is the skill of asking excellent questions.  In order to discuss this skill, first let’s look at a relatively straightforward practice problem.

(A) 14

(B) 26

(C) 28

(D) 32

(E) 38

I will show a full solution at the bottom of this article.  Right now, let’s just suppose that you got this question wrong and wanted to ask about it.  The correct answer is the lowest of the five answer choices, so for example, if you went for the trap answer of 32, you fell right into one of the mistakes that the question was specifically designed to elicit.  Questions writers regular incorporate trap answers like big butterfly nets, and simply let hordes of test-takers run into them.  The more you, as a student, can recognize these traps, the more you can set yourself apart from the  mass of ordinary test takers.  I will discuss this more in the solutions below.

Asking poor questions

Suppose you did this questions, got it wrong, and didn’t have access to an immediate explanation for the question.  Suppose you wanted to get an explanation from an expert, either on a blog or on a forum such as GMAT Club.  How would you ask your question?

Here’s what many people might say:

1) I don’t understand.  Please help.

2) I don’t get this.

3) Please explain.

I see questions of this ilk often.  These are abysmally poor questions.  If I were to assign a grade to such a question, I would give an F.  Asking a questions of this sort does not indicate any willingness, on the part of the student, to engage the material and put significant energy into the process of coming to a deeper understanding.

Notice, that questions like this require almost no effort to write.  One could write one of these questions and post it in under 15 seconds.  Right there, that’s an indication that it’s not necessarily going to help you.  You are not going learn the most and make the most progress unless you put a significant amount of effort into your own cause.

Another problem with these question is: think about it from the perspective of the expert who gets such a question.  I certainly understand how to solve this problem myself, but if  student tells me simply “I don’t understand. Please help,” then I don’t know where that student is stuck.  Does that student not know how to factor quadratics at all?  Did the student fall for the trap answer of (D)?  Did the student make a simple arithmetic mistake?  Does the student not understand algebra in the least?  Given that meager question, I have absolutely no idea what would best help the student.

The presumption of such a question is that the teacher, not the student is responsible for the process of education.  This is incorrect.  Education is not something a teacher or test prep company does to you.  Education is not a spectator sport.  Education is primarily something you do to yourself, by yourself, for yourself, and teachers or test prep companies simply provide the support and resources to enable you to educate yourself.

The art of asking excellent questions

The habits of excellence can set a student apart from others.  Excellence always takes more work, and many students avoid it for that very reason.  Many people don’t realize that, in choosing to exert a minimum of effort in many things related to their studies, they are implicitly choosing mediocrity.  The great Law of Mediocrity is: if you put in about the same effort as most people, then you probably will get about the same results as most people.  The corollary is: if you want results that stand out, you need to put in outstanding effort.  The habits of excellence require outstanding effort, and one of these habits is asking excellent questions.

Here are some examples of excellent questions:

4) When I looked at the problem, I saw right away that we could solve the first equation by subtracting 8 from both sides and solving x² = 9 to get x = 3.  I didn’t know how to solve that other equation, so I am glad I didn’t have to deal with it.  From there, I simply plugged 3 into 3x + 23 to get 9 + 23 = 32.  I chose (D), but that does not agree with the OA.  What did I do wrong?

5) I began by moving the 8 to the other side, and then I solved x² = 25 to get x = ±5.  I then factored the quadratic: x² – 2x – 15 = (x + 3)(x – 5) = 0, x = –3 or x = +5, so because of the combined constraints of both equations, we see x must equal +5, and therefore 3x + 23 = 38, which would be (E), but that’s not the OA given. 

Both of these are much much better questions.  Notice, these two students made vastly different mistakes.  The first needs to learn some basic algebra facts, while the second appears to have made a very superficial arithmetic error but really knows her algebra well.  Notice that an excellent question tells a mini-story of your entire experience of the problem.  “When I first saw the problem, I thought this or tried this. Then when I did this, I thought this ...”  An excellent question makes clear all the steps, the entire thought process.  First of all, this is tremendously helpful to the expert answering the question: when we read the question, we can fine-tune our answer to exactly what the student needs, and thus provide an answer that gives the student the most help.  More importantly, though, formulating this questions involves self-reflection on the part of the student.  Self-reflection is the very substance of education.  A student who is not actively reflecting on his process and his progress will not be learning nearly as fast as he could.  When a student puts in the effort to formulate an excellent question, this very process forces her to consider all the nuances of the question; she may thereby answer her own question, which is ideal, but if she doesn’t and asks her question, her mind will be primed to receive the focused response that an expert will give her.  Learning is not simply about getting all the information, because many students get all the information in front of them and simply can’t remember it all or apply it all.  Learning also has to involve preparing one’s mind to absorb the information one will receiving, and asking an excellent question, detailed and well thought-out, is one of the best ways to guarantee that you will retain and assimilate the answer you receive.

Summary

One of the habits of excellence is asking excellent questions.  Many students don’t do this, and are mystified about why they continue to struggle and cannot retain or apply the valuable information they have learned.  I urge you to embrace the habit of asking excellent questions and all the habits of excellence.  If you have experiences with your own practice of asking excellent questions, we would love to hear from you in the comments section.

Practice problem explanation

1) In the first equation, we need to subtract 8 from both sides, which gives us:

When we ourselves take a square root, we need to remember to include both the positive and the negative root.  See this GRE post for details.  We get two roots from this:

From this equation alone, we cannot determine unique value of x.  We have to factor the second equation.

The only root that is common to both equations is x = –3, so this must be the value of x.  This means

3x + 23 = 3(–3) + 23 = –9 + 23 = 23 – 9 = 14

Answer = (A) 

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