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? Once you’ve got these questions down pat, head over to our full GMAT practice test to practice your skills.

As always, remember there’s 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:

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

- use {3, 7}, product = 21, abcd could be 3721 or 7321
- use {3, 9}, product = 27, abcd could be 3927 or 9327
- 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 36 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)**

Hi,

In the 8th solution, while we consider AABC, ABBC, ABCC, why do we not take their permutations into account (AACB, CBBA, BACC) ?

Firas

Hi Firas! Those would be different numbers and are counted by our Fundamental Counting Principle approach. The A’s and B’s just remind us where the duplicate pair is in the number. So, if you had 1123, that’s a good choice. 2331 is another choice. We absolutely count those separately, but they’re different numbers that have the duplicate in a different place. If we were just going to list all of the four digit numbers, it would be 9*9*8*7, but we have more limitations than that. We have to duplicate one of the numbers, and that could go in the first two spots, the middle two, or the last two. The letters just help us differentiate these. We wouldn’t mix them up because as you can see, the permutations refer to different numbers that end up getting counted. We count all the numbers where the duplicates are in the first two spots, all of the numbers where the duplicates are in the middle, and all of the numbers where the duplicates are in the last two spots. I hope that helps clarify!

Hi, I think you missed out some equilateral triangles from the solution to Q6. LDH, KCG and so on would be distinct equilateral triangles too. So i get an answer of 60 instead of 52?

Hi Nick,

I don’t believe you’d count the equilateral triangles twice, since orienting an equilateral triangle to a new vertex doesn’t change the altitude of the triangle. It also doesn’t truly change the orientation of one angle measurement to another, since all angle measurements are the same. That’s why you get only 52– an equilateral triangle looks exactly the same form any orientation.

Hi,

For question 8, wouldn’t “00” be allowed as possibilities for case 2 and 3 for “B?” That would create 10 possible scenarios for case 2 and 3.

Although not listed as an answer choice, I derived a sum of 2088.

Hi Be! The issue is that the leftmost digit has to be taken from the numbers 1 through 9. Then the other digits can all be taken from the numbers 0 through 9. We already selected a number to go in the leftmost digit, so there are 9 options left. Meaning, in Case 2, we have 9 options for A from the 9 numbers from 1 through 9. We have 9 options for B from the 10 numbers from 0 through 9. We already selected a number for A, which is why there are only 9 options for B. We are taking from a pool of numbers that includes 0, but we’re one number down because one got used on A. So yes, BB could be 00 and so could CC, but we don’t have 10 options given that A already took one in Case 2 and both A and B took options in Case 3, leaving 8 options for C, even though it technically could have been one of 10 different numbers. I hope that helps 🙂

Hi,

In the solution for Problem 5) there’s a mistake. You’re totalling 12 and 24 up to 35, which should be 36 instead.

Regards,

Sebastian

I just corrected the table and text. Thanks for bringing that to my attention, Sebastian. 🙂