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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.

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

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!

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)**

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1) The dark purple region on the number line above is shown in its entirety. This region is delineated by which of the following inequalities?

(A) 10 < |x + 10| < 80

(B) 10 < |x – 100| < 80

(C) |x – 20| < 70

(D) |x – 20| < | x – 90|

(E) |x – 55| < 35

2) If |x| < 20 and |x – 8| > |x + 4|, which of the following expresses the allowable range for x?

(A) –12 < x < 12

(B) –20 < x < 2

(C) –20 < x < –12 and 12 < x < 20

(D) –20 < x < –8 and 4 < x < 20

(E) –20 < x < –4 and 8 < x < 20

3) If |(x – 3)^{2} + 2| < |x – 7| , which of the following expresses the allowable range for x?

(A) 1 < x < 4

(B) 1 < x < 7

(C) – 1 < x < 4 and 7 < x

(D) x < – 1 and 4 < x < 7

(E) – 7 < x < 4 and 7 < x

Explanations to these will follow the discussion.

I want to make clear that this question type is very tricky, something that might pose challenges even to students relatively strong in math. I also want to make clear that we could send 50 people to take 50 separate GMATs, and it may be that not one of them would see such a question on their GMAT. This is a very infrequent question genre, and it would only be asked if you were acing the Quant sections and the CAT were throwing everything including the kitchen sink at you!

If you find what I say in this blog article helpful, that’s great: it may give you some insight into other, easier questions involving absolute values and/or inequalities. If you are still confused, don’t worry: you are rather unlikely to see this question type on test day.

Fundamentally, the **absolute value is about distance, and thinking about it geometrically is often the key to difficult absolute value questions**. The naïve understanding of absolute value is that it “makes things positive.” While this is undeniably true, it is not always the most mathematically productive way to understand the concept. It’s undeniably true that |+5| = +5 and that |-5| = +5. We could say that both are true because the absolute value “makes everything positive,” or we could say that these two numbers, +5 and –5, have a distance of 5 from zero on the number line.

This geometric interpretation about distance becomes considerably more important when we are dealing with algebra. The expression |x| is the distance from x to zero. The expression |x – 6| is the distance from x to +6. The expression |x + 2| is the distance from x to – 2: to understand this, remember that x + 2 = x – (– 2). In general, the expression |x – c| is the distance from any x to the fixed point x = c.

This distance interpretation replaces a whole lot of complicated calculations. For example, if we had to solve the inequality |x – 10| ≥ |x – 20|, that’s just saying that the distance from x to 20 is smaller than the distance from x to 10—in other words, *x is closer to 20 than to 10,* and because of the “or equal” part, it could be an equal distance also. Well, the point x = 15 is the only point on the number line that is equidistant from x = 10 and x = 20. If x has to be either an equal distance from 10 and 20, or closer to 20, then it has to be at the point x = 15 or to the right. Any point between 15 and 20 is closer to 20 than to 10, and any point to the right of 20 has to be closer to 20 than to 10. Thus, the solution is 15 ≤ x.

What region is denoted by the inequality |x – 25| > 10? Well, all we are saying is that we have to be more than 10 units away from the point x = 25. If we go 10 units to the left, we get to x = 15: we can’t be here, at a distance equal to 10, but we could be the left of x = 15, at a distance of more than 10. If we go 10 units to the right, we get to x = 35: we can’t be here, at a distance equal to 10, but we could be the right of x = 35, at a distance of more than 10. Thus, the solution includes the pair of regions x < 15 and 35 < x.

Notice that x = 25 is the point of symmetry of this diagram, since everything is about distance from that point.

The perspective above may give you some insight into this problem. If you had any “aha” moments reading this, then you might want to give the practice problems another try before looking at the solutions below. The solutions may afford you a few more insights into problem solving.

1) Step one: find the midpoint of the region. The midpoint, halfway between and 20 and 90, is 55. In other words, 20 and 90 have the same distance from 55, a distance of 35. These endpoints are not included, but the region includes all the points that have a distance from from x = 55 that is less than 35. Translating that into math, we get the following:

|x – 55| < 35

Answer = **(E)**

2) Some folks might think this involves a sophisticated calculation, but much of this can be done with simple spatial analysis. Look at the second inequality, the more complicated one: |x – 8| > |x + 4|. All this says is that we are looking for points such that the distance to x = 8 is greater than the distance from x = –4; in other words, we want all the points that are closer to x = –4 and farther from x = 8.

The midpoint between x = –4 and x = 8 is the point x = 2. This point is not included because it’s equidistant from both points, but everything to the left of this point on the number line is closer to x = –4 than it is to x = 8. That entire complicated inequality simplifies to x < 2.

Combine that with the first inequality, |x| < 20, which in the negative realm means that x must be greater than –20. Thus, the allowed region is –20 < x < 2.

Answer = **(B)**

3) For this one, we need to be clever in a few ways. First of all, the expression inside the absolute value on the left is of the form [a square] + [a positive number]. Anything squared is either zero or positive, because something squared can never be negative. When we add a positive number, we are guaranteed that it is always positive. Thus, the absolute values around it are entirely superfluous, because it’s always positive anyway. We can remove those absolute value signs on the left with changing the mathematical meaning of the statement one bit.

The other expression, (x – 7), could be positive or negative, depending on the value of x, so it may equal +(x – 7) or it may equal –(x – 7). We have to investigate either case.

Think about the graph of a parabola.

This is an upward opening parabola, one that goes up on both sides. As you may know,

Furthermore, the vertex of the parabola is on this axis of symmetry. For this particular parabola, we get x = –(–7)/(2*1) = 7/2 as the axis of symmetry. Plug this value in to find the height of the vertex.

We don’t have to solve that last expression. The fraction 49/4 is between 12 and 13, so when this is subtracted from 18, we get something between 5 and 6. In other words, the vertex of this parabola starts *above* the x-axis and continues *up*. In other words, it is never negative, never less than zero. This means that the Case I inequality has **no solutions**.

We can solve the equation to find the boundary points. We will have to factor the quadratic.

Think about the graph of this parabola.

This is also an upward opening parabola, also one that goes up on both sides. It equals zero at x = 1 and x = 4, so it must have its vertex between them, and it must be negative between those two points and positive to the left of x = 1 and to the right of x = 4. Thus, the solution of the Case II inequality is **1 < x < 4**.

Since Case I had no solution, the Case II solution is the whole shebang.

Answer = **(A) **

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]]>Full solutions will come at the end of this article.

First of all, let’s review what should be more familiar — the arithmetic of equations. Suppose A = B and P = Q. The soundbyte is: you can combine them in almost any way imaginable to get a new valid equation. You can add them, in either order (A + P = B + Q) or (A + Q = B + P). You can subtract then, either one from the other, in either order (four subtraction equations: e.g. A – P = B – Q). You can multiply them in either order (A* P = B * Q) or (A * Q = B * P). You can divide either one by either other (assuming you are not dividing by zero), in either order (four division equations: e.g. A/P = B/Q). You can raise either one to the power of the other, in either order (four exponentiation equation: e.g. A^P = B^Q).

(Note: for the GMAT, you are responsible for know what happens when you raise a base, say 3, to the power of integers or to the 1/2. Technically, you can raise any positive base to any power — say, 3 to the pi — but that’s more complicated than the GMAT expects you to know. For negative bases, things get dicier. For the GMAT, just worry about negative bases to integer powers, (–2)^3 or (–2)^(–1). Non-integral powers of negatives leads into complex numbers — again, beyond what the GMAT expects you to know).

Finally, the above may seem simple, and if A & B & P & Q are just individual numbers, then those equations are pretty much “duh”-simple. Things get considerably more interesting if some or all of those letters are not individual numbers but algebraic expressions. Even with four different algebra expressions, all this still holds for equations.

Everything gets trickier with inequalities. First of all, an equation, such as A = B, is inherently symmetrical and “two-sided”, but an inequality is more a one-sided, unidirectional thing. With any arithmetic of inequalities, we must consider the direction of the inequality. Furthermore, equations are very intuitive, but some of the arithmetic of inequalities is quite unintuitive.

Adding inequalities is not so bad: you can add inequalities with the same direction of inequality. Thus, if A > B, and P > Q, then it must be true that A + P > B + Q. That always works, and it is in many ways what you’d expect.

This is the one that’s much trickier. If A > B, and P > Q, then naïvely one might expect that (A – P) would be greater than (B – Q), but that’s not necessarily the case. For example, suppose we have 10 > 5 and 2 > 1 — then we could subtract them in the same direction of inequality, and we’d get 8 > 4, which still works. BUT, suppose 10 > 5, and 100 > 1, both true — now, if we subtract in the same direction of inequality, we find that (–90) is *not* greater than 4. If we subtract two inequalities in the same direction of inequality, we may get another true statement, but we are not guaranteed that this will lead to a true statements, so it’s a very unwise move in problem solving.

Here’s the real head-scratcher: we can’t subtract inequalities with the *same* direction of inequality, but we CAN subtract inequalities with the *opposite* direction of inequality — in other words, if A > B, and P > Q, then it must be true that (A – Q) > (B – P). That may be perplexing symbolically, so think about it in real world terms. Suppose Ann has more money than Bob. Suppose I take less from the richer person, Ann, and more from the poorer person, Bob. When I am done, Ann still will have more than Bob — in fact, a greater difference than existed between them before!

Everything gets much hairier with multiplying or dividing inequalities when you consider — one or both sides could be negative. Hmmm. If we multiply or divide both sides of inequality by a single negative number, that’s perfectly legal, but we must remember to reverse the direction of inequality. What happens if we were to multiply or divide inequalities and negatives were involved? For example, if we know that x > –6 and y > –7, then what can we say about the product xy? As it turns out, we could pick an x and a y that would satisfy the original inequalities and make the product xy equal absolutely any number on the number line. This mindboggling bit of math is just to demonstrate why the GMAT is not going to touch multiplying or dividing general inequalities with a ten-foot pole!

Let focus, though, on a special case that could, in very advance problems, appear on the GMAT. Suppose we know that all four numbers or expressions are *positive*: A > B > 0 and P > Q > 0. Then, as with addition, we can multiply inequalities with the same direction: A*P > B*Q must be true. And, as with subtraction, we can divide inequalities with the opposite direction: A/Q > B/P. Again, remember the caveat: everything must be positive for these patterns to work. If anything can be negative, things get much more complicated, so complicated that the GMAT won’t ask about them.

Absolute value inequality is a sizeable topic, with some mind-bending twists and turns, but chances are very good the GMAT is not going to probe this topic all that deeply. In fact, probably most of the absolute value inequalities on the GMAT will be of the form: |x – p|, where p is some given fixed number.

Here, we must remember a few key mathematical facts. First of all, subtraction gives us distance on the number line. Technically, subtraction gives **signed distance**. What do I mean by that? Well, 5 – 2 = +3, indicating that it’s a signed distance of positive 3, i.e. three units to the right, from 2 to 5; by contrast, 2 – 5 = –3 indicating that it’s a signed distance of negative 3, i.e. three units to the left, from 5 to 2.

For ordinary distance, distance in the geometric sense of the word, we don’t care about sign or direction — the distance between two points is just a positive number and is the same, whether from A to B or from B to A. That’s where absolute value comes in. **The expression |p – q| is the distance between numbers p & q on the number line**. That is a HUGE idea.

Thus, |x – 5| is the distance between variable point x and fixed point 5. The expression |x – 5| < 2 indicates the set of all points x that have a distance to the point 5 of less than two. Immediately, just thinking about this logic, and without any further calculations, we can see that |x – 5| < 2 is entirely equivalent to 3 < x < 7. Recognize that this is exactly the kind of math the GMAT adores: with simple logic, you can jump to conclusions without having to do any calculations. The GMAT goes crazy for math of this sort. I can’t emphasize enough how important this particular set of logically interconnected ideas is. If the GMAT asks you anything at all about absolute value inequalities, it is highly likely it would be something of this genre, and unlikely that it would be anything else in this extensive topic.

If you had some insights reading this article, I encourage you to take another look at the practice questions at the top before reading the solutions below. Here’s a similar question from inside Magoosh.

5) http://gmat.magoosh.com/questions/960

If you have anything to add, or any questions, please let us know in the comment section below.

1) In this problem, the tempting incorrect answer would be **(A)**, or somehow would involve statement #1 as sufficient, but it’s not. This gets into an idea discussed in this blog —- we can’t subtract inequalities with the same direction of inequality, but we can subtract inequalities with the opposite direction of inequality.

Statement #1: So the prompt inequality and this statement’s inequality would be true for a + b = 15, c + d = 7, b = 6, d = 2

prompt inequality: 15 > 7 — true

statement #1 inequality: 6 > 2 — true

prompt question: 9 > 5 — a > c, an answer of “yes”

But, both inequalities would still be true for a + b = 20, c + d = 18, b = 15, d = 3

prompt inequality: 20 > 18 — true

statement #1 inequality: 15 > 3 — true

prompt question: 5 < 15 — a < c, an answer of “no”

We can make different choices consistent with all the given statements that would produce either a “yes” or “no” answer to the prompt question. This means that, given this statement, we are unable to determine a unique definitive answer to the prompt question. This statement, alone and by itself, is **insufficient**.

Statement #2: Here, we are allowed legitimately to subtract the inequalities, because the directions of inequality are in opposite order. When we subtract (b < d) from (a + b > c + d), we get a > c, a definitive “yes” answer.

Another way to think about it: (a + b) = LARGER, and (c + d) = SMALLER, so of course, the former is greater than the latter. Now, say that b = tiny, and d = bigger, so of course, b < d. Now, think about the two differences we will compare:

(i) a = (a + b) – b = LARGER – tiny

(ii) c = (c + d) – d = SMALLER – bigger

Obviously, it start with something LARGER, and subtract something tiny, the result will be greater than starting with something SMALLER and subtracting something bigger. Therefore, a > c, a definitive answer to the prompt question.

This statement allows us to give a definitive “yes” answer to the prompt, so this statement, alone and by itself, is **sufficient**.

Answer = **B**

2) Statement #1: if p = 2 and q = 1, then this statement’s equation is true, 4 > 1, and p > q, so the answer to the prompt is “yes”.

But, if p = –2 and q = –1, then it’s still true that the square of p is larger than the square of q, 4 > 1, but now it’s true that p < q, so the answer to the prompt question is “no.”

We can make different choices consistent with all the given statements that would produce either a “yes” or “no” answer to the prompt question. This means that, given this statement, we are unable to determine a unique definitive answer to the prompt question. This statement, alone and by itself, is **insufficient**.

Statement #2: if p = 2 and q = 1, then this statement’s equation is true, 8 > 1, and p > q, so the answer to the prompt is “yes”.

Switching to negatives won’t make a difference, because the cube of a negative is still negative. The values p = –2 and q = –3 satisfy the statement inequality, and it’s still true that p > q

What about fractions? If p = 1/2 and q = 1/3, then p cubed is still larger than q cubed, (1/8) > (1/27), and p is still greater than q. No matter what numbers we pick, the inequality in statement #2 directly implies the prompt inequality. The mathematical way to say this is: cubing, or taking a cube-root, preserves the order of any inequality.

This statement gives us a definitive answer of “yes” to the prompt question, so this statement, alone and by itself, is **sufficient**.

Answer = **B**

3) Think about the **distance interpretation** of absolute value inequalities. We want to know: is x further than two units away from 6 on the number line?

Statement #1: the x-values allowed by this statement are x’s that are more than three units from 4. Here’s a picture of these values, in green:

Notice, the endpoints, 1 & 7, are not allowed, because they are exactly three units from 4, and exactly 3 is not greater than 3. Most of these points are further than two from the point 6, but some, such as the point 7, is closer than two units to six. Thus, given this constraint, we could find many points that produce a “yes” answer to the prompt, but some that produce a “no” answer. Different choices give different answers. This means that, given this statement, we are unable to determine a unique definitive answer to the prompt question. This statement, alone and by itself, is **insufficient**.

Statement #2: the x-values allowed by this statement are x’s that are more than one unit from 8. Here’s a picture of these values, in green:

Again, notice the endpoints are not included. Many of these points are further than two from the point 6, but some, such as the point 7, is closer than two units to six. In fact, the point 6 itself — which is a distance of zero units from 6 — is allowed by this statement! Thus, given this constraint, we could find many points that produce a “yes” answer to the prompt, but some that produce a “no” answer. Different choices give different answers. This means that, given this statement, we are unable to determine a unique definitive answer to the prompt question. This statement, alone and by itself, is **insufficient**.

Combined statements: Now, we combine the constraints of the individual statements. Now, the allowed points must be both more than three units from 4 and more than one unit from 8. Shown in green are the points that simultaneously satisfy both constraints:

Now, all the green points are more than two units away from 6, and it absolutely impossible to pick a value of x that simultaneously satisfies the constraints of both statements and is closer than two units to 6. The combined statements allow for a definitive “yes” answer to the prompt. Together, the statements are **sufficient**.

Answer = **C**

4) Statement #1: this gets at the fundamental meaning of the inequality. If y is one more than x, it must be greater than x. To add one to a number means to move it one unit to the right on the number line, so y must be one unit right of x, which means it is greater than x. This allows us to determine a definite “yes” to the prompt question. This statement, alone and by itself, is **sufficient**.

Statement #2: There are a couple ways to think about this one. One is to treat different categories of numbers.

(i) if x is negative, then by squaring, y will be positive, and y > x.

(ii) if x = 0, y = 1, and y > x

(iii) if x is a fraction between 0 and 1, then its square will also be between 0 and 1, and adding one to this will produce a number greater than 1, between 1 and 2. Therefore, y > x.

(iv) if x = 1, y = 2, and y > x

(v) if x > 1, then squaring x makes it bigger, and adding one makes that even bigger, so y > x

Thus, for all possible values of x, y > x. Thus, the prompt gives a definitive “yes” answer.

A totally different way to think about it: using coordinate geometry. The graph of y = (x^2) is a parabola that passes through the origin. The graph of y = (x^2) + 1 is this same parabola shift up one in the y-direction, passing through (0, 1) instead of the origin. Now, compare this shifted parabola to the line y = x. One of the special properties of the line y = x is that all points above this line, regardless of quadrant, have the property y > x. Think about the graph:

The parabola is always above the line y = x, so every point on the parabola must satisfy y > x.

Either way, this allows us to determine a definite “yes” to the prompt question. This statement, alone and by itself, is **sufficient**.

Both statements are individually sufficient.

Answer = **D**

To find out where inequalities 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

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