Inequalities and Absolute Values

For questions in the Quantitative Comparison format (“Quantity A” and “Quantity B” given), the answer choices are always as follows:

(A) Quantity A is greater.

(B) Quantity B is greater.

(C) The two quantities are equal.

(D) The relationship cannot be determined from the information given.

For questions followed by a numeric entry box , you are to enter your own answer in the box. For questions followed by a fraction-style numeric entry box , you are to enter your answer in the form of a fraction. You are not required to reduce fractions. For example, if the answer is , you may enter or any equivalent fraction.

All numbers used are real numbers. All figures are assumed to lie in a plane unless otherwise indicated. Geometric figures are not necessarily drawn to scale. You should assume, however, that lines that appear to be straight are actually straight, points on a line are in the order shown, and all geometric objects are in the relative positions shown. Coordinate systems, such as xy-planes and number lines, as well as graphical data presentations, such as bar charts, circle graphs, and line graphs, are drawn to scale. A symbol that appears more than once in a question has the same meaning throughout the question.

7y – 3 ≤ 4y + 9

1.

Quantity A y

Quantity B 4

3|x – 4| = 16

2.

Quantity A x

Quantity B

3.If b ≠ 0 and > 0, which of the following inequalities must be true?

Indicate all such inequalities.

• a > b
• b > 0
• ab > 0

4.If 6 < 2x – 4 < 12, which of the following could be the value of x?

(A)4

(B)5

(C)7

(D)8

(E)9

5.If y < 0 and 4x > y, which of the following could be equal to ?

(A)0

(B)

(C)

(D)1

(E)4

3(x – 7) ≥ 9
0.25y – 3 ≤ 1

6.

Quantity A x

Quantity B y

7.If |1 – x| = 6 and |2y – 6| = 10, which of the following could be the value of xy?

Indicate all such values.

• –40
• –14
• –10
• 56

8.If 2(x – 1)³ + 3 ≤ 19, which of the following must be true?

(A)x ≥ 3

(B)x ≤ 3

(C)x ≥ –3

(D)x ≤ –3

(E)x < –3 or x > 3

9.If 3p < 51 and 5p > 75, what is the value of the integer p?

(A)15

(B)16

(C)24

(D)25

(E)26

10. A bicycle wheel has spokes that go from a center point in the hub to equally spaced points on the rim of the wheel. If there are fewer than six spokes, what is the smallest possible angle between any two spokes?

(A)18°

(B)30°

(C)40°

(D)60°

(E)72°

|–x| ≥ 6
xy² < 0 and y is an integer.

11.

Quantity A x

Quantity B –4

12. If > 5 and x < 0, which of the following could be the value of x?

Indicate all such values.

• –6
• –14
• –18

|x³| < 64

13.

Quantity A –x

Quantity B –|x|

14. If |3x + 7| ≥ 2x + 12, then which of the following is true?

(A)x ≤

(B)x ≥

(C)x ≥ 5

(D)x ≤ or x ≥ 5

(E) ≤ x ≤ 5

|3 + 3x| < –2x

15.

Quantity A |x|

Quantity B 4

16. If |y| ≤ –4x and |3x – 4| = 2x + 6, what is the value of x?

(A)–3

(B)

(C)

(D)

(E)10

x is an integer such that –x|x| ≥ 4.

17.

Quantity A x

Quantity B 2

|x| < 1 and y > 0

18.

Quantity A |x| + y

Quantity B xy

|x| > |y| and x + y > 0

19.

Quantity A y

Quantity B x

x and y are integers such that |x|(y) + 9 < 0 and |y| ≤ 1.

20.

Quantity A x

Quantity B –9

p + |k| > |p| + k

21.

Quantity A p

Quantity B k

|x| + |y| > |x + z|

22.

Quantity A y

Quantity B z

b ≠ 0
> 1
a + b < 0

23.

Quantity A a

Quantity B 0

24. If f²g < 0, which of the following must be true?

(A)f < 0

(B)g < 0

(C)fg < 0

(D)fg > 0

(E)f² < 0

25. and . If x is an integer, which of the following is the value of x?

(A)2

(B)3

(C)4

(D)5

(E)6

|x|y > x|y|

26.

Quantity A (x + y)2

Quantity B (x – y)2

27. Which of the following could be the graph of all values of x that satisfy the inequality 4 – 11x ≥ ?

(A)

(B)

(C)

(D)

(E)

–1 < a < 0 < |a| < b < 1

28.

Quantity A

Quantity B

x > |y| > z

29.

Quantity A x + y

Quantity B |y| + z

30. The integers k, l, and m are consecutive even integers between 23 and 33. Which of the following could be the average (arithmetic mean) of k, l, and m?

(A)24

(B)25

(C)25.5

(D)28

(E)32

31.

The number line above represents which of the following inequalities?

(A)x < 1

(B)–6 < 2x < 2

(C)–9 < 3x < 6

(D)1 < 2x < 3

(E)x > –3

32. For a jambalaya cook-off, there will be x judges sitting in a single row of x chairs. If x is greater than 3 but no more than 6, which of the following could be the number of possible seating arrangements for the judges?

Indicate two such numbers.

• 6
• 25
• 120
• 500
• 720

33. Which of the following inequalities is equivalent to for all non-zero values of a, b, and c?

Indicate all such inequalities.

• 
• 
• a > –3bc

|x + y| = 10
x ≥ 0
z < y – x

34.

Quantity A z

Quantity B 10

0 < a < < 9

35.

Quantity A 9 – a

Quantity B – a

For all values of the integer p such that 1.9 < |p| < 5.3,
the function f(p) = p².

36.

Quantity A f(p) for the greatest value of p

Quantity B f(p) for the least value of p

37. If and are reciprocals and < 0, which of the following must be true?

(A)ab < 0

(B) < –1

(C) < 1

(D)

(E)

38. If mn < 0 and , which of the following must be true?

(A)km + ln < (mn)²

(B)kn + lm < 1

(C)kn + lm > (mn)²

(D)k + l > mn

(E)km > –ln

39. If the reciprocal of the negative integer x is greater than the sum of y and z, then which of the following must be true?

(A)x > y + z

(B)y and z are positive.

(C)1 > x(y + z)

(D)1 < xy + xz

(E) > z – y

40. If u and –3v are greater than 0, and , which of the following cannot be true?

(A) < –v

(B) > –3

(C)

(D)u + 3v > 0

(E)u < –3v

41. In the figure above, an equilateral triangle is inscribed in a circle. If the arc bounded by adjacent corners of the triangle is between 4π and 6π long, which of the following could be the diameter of the circle?

(A)6.5

(B)9

(C)11.9

(D)15

(E)23.5

 

Inequalities and Absolute Values Answers

1. (D). Solve the inequality algebraically:

7y – 3	≤	4y + 9
3y – 3	≤	9
3y	≤	12
y	≤	4

Because y could be less than or equal to 4, the relationship cannot be determined from the information given.

2. (D). Solve the inequality by first dividing both sides by 3 to isolate the absolute value. Then solve for the positive and negative possibilities of (x – 4), using the identity that |a| = a when a is positive or zero and |a| = –a when a is negative:

Thus, x could be or , making the two quantities equal or Quantity B greater, respectively. The relationship cannot be determined from the information given.

3. ab > 0 only. If > 0, then both a and b must have the same sign. That is, a and b are either both positive or both negative. The 1st inequality could be true, but is not necessarily true. The relative values of a and b are not indicated by the inequality in the question stem. The 2nd inequality could be true, but is not necessarily true. If a were negative, b could be negative. The 3rd inequality must be true, as it indicates that a and b have the same sign.

4. (C). When manipulating a “three-sided” inequality, perform the same operations on all “sides.” Therefore, the first step to simplify this inequality would be to add 4 to all three sides to get: 10 < 2x < 16. Next, divide all three sides by 2. The result is 5 < x < 8. The only answer choice that fits within the parameters of this inequality is 7.

5. (A). If y is negative, then dividing both sides of the second inequality by y yields < 1. Remember to switch the direction of the inequality sign when multiplying or dividing by a negative (whether that negative is in number or variable form). Next, dividing both sides by 4 changes the inequality to < . The only answer choice less than is 0.

6. (D). Solve each inequality algebraically:

3(x – 7)	≥	9
x – 7	≥	3
x	≥	10

0.25y – 3	≤	1
0.25y	≤	4
y	≤	16

Since the ranges for x and y overlap, either quantity could be greater. For instance, x could be 11 and y could be 15 (y is greater), or x could be 1,000 and y could be –5 (x is greater). The relationship cannot be determined from the information given.

7. –40, –14, and 56 only. Solve each absolute value, using the identity that |a| = a when a is positive or zero and |a| = –a when a is negative:

|1 – x| = 6
+ (1 – x) = 6	or	– (1 – x) = 6
(1 – x) = 6		(1 – x) = –6
–x = 5		–x = –7
x = –5		x = 7
x = –5 or 7
|2y – 6| = 10
+ (2y – 6) = 10	or	– (2y – 6) = 10
(2y – 6) = 10		(2y – 6) = –10
2y = 16		2y = –4
y = 8		y = –2
y = 8 or –2

Since x = –5 or 7 and y = 8 or –2, calculate all four possible combinations for xy:

(–5)(8)	=	–40
(–5)(–2)	=	10
(7)(8)	=	56
(7)(–2)	=	–14

Select –40, –14, and 56. (Do not pick –10, as xy could be 10, but not –10.)

8. (B).     2(x – 1)3 + 3	≤	19
2(x – 1)3	≤	16
(x – 1)3	≤	8

Taking the cube root of an inequality is permissible here, because cubing a number, unlike squaring it, does not change its sign.

x – 1	≤	2
x	≤	3

9. (B). Dividing the first inequality by 3 results in p < 17. Dividing the second inequality by 5 results in p > 15. Therefore, 15 < p < 17. Because p is an integer, it must be 16.

10. (E). In this scenario, if there are n spokes, there are n angles between them. Thus, the measure of the angle between spokes is . Since n < 6, rewrite this expression as . Dividing by a “less than” produces a “greater than” result. Therefore, = greater than 60°. The only answer that is greater than 60° is (E). To verify, note that n can be at most 5, because n must be an integer. Because there are 360° in a circle, a wheel with 5 spokes would have = 72° between adjacent spokes.

11. (B). First, solve the inequality for x, remembering the two cases that must be considered when dealing with absolute value: |a| = a when a is positive or zero and |a| = –a when a is negative:

|–x| ≥ 6
+(–x) ≥ 6	or	–(–x) ≥ 6
–x ≥ 6		x ≥ 6
x ≤ –6
x ≤ –6 or x ≥ 6

Because xy² < 0, neither x nor y equals zero. A squared term cannot be negative, so y² must be positive. For xy² to be negative, x must be negative. This rules out the x ≥ 6 range of solutions for x. Thus, x ≤ –6 is the only range of valid solutions. Since all values less than or equal to –6 are less than –4, Quantity B is greater.

12. –18 only. Solve the absolute value inequality by first isolating the absolute value:

	>	5
|x + 4|	>	10

To solve the absolute value, use the identity that |a| = a when a is positive or zero and |a| = –a when a is negative. Here if (x+4) is positive or zero that leaves:

x + 4	>	10
x	>	6

This, however, is not a valid solution range, as the other inequality indicates that x is negative.

Solve for the negative case, that is, assuming that (x + 4) is negative:

–(x + 4)	>	10
(x + 4)	<	–10
x	<	–14

Note that this fits the other inequality, which states that x < 0.

If x < –14, only –18 is a valid answer.

13. (D). First, solve the absolute value inequality, using the identity that |a| = a when a is positive or zero and |a| = –a when a is negative:

|x3| < 64
+(x3) < 64	or	–(x3) < 64
x < 4		x3 > –64 (Flip the inequality sign when multiplying by –1.)
		x > –4
–4 < x < 4

x could be positive, negative, or zero. If x is positive or zero, the two quantities are equal. If x is negative, Quantity A is greater. The relationship cannot be determined from the information given.

14. (D). Solve |3x + 7| ≥ 2x + 12, using the identity that |a| = a when a is positive or zero and |a| = –a when a is negative:

+(3x + 7) ≥ 2x + 12	or	–(3x + 7) ≥ 2x + 12
x + 7 ≥ 12		–3x – 7 ≥ 2x + 12
x ≥ 5		–7 ≥ 5x + 12
		–19 ≥ 5x
		≥ x

x ≤ or x ≥ 5

15. (B). Solve the absolute value inequality, using the identity that |a| = a when a is positive or zero and |a| = –a when a is negative:

|3 + 3x| < –2x
+(3 + 3x) < –2x	or	–(3 + 3x) < –2x
3 + 5x < 0		–3 – 3x < –2x
5x < –3		–3 < x
x < –
–3 < x <

Since x is between –3 and –, its absolute value is between and 3. Quantity B is greater.

16. (C). The inequality is not strictly solvable, as it has two unknowns. However, any absolute value cannot be negative. Putting 0 ≤ |y| and |y| ≤ –4x together, 0 ≤ –4x. Dividing both sides by –4 and flipping the inequality sign, this implies that 0 ≥ x.

Now solve the absolute value equation, using the identity that |a| = a when a is positive or zero and |a| = –a when a is negative:

|3x – 4| = 2x + 6
+(3x – 4) = 2x + 6	or	–(3x – 4) = 2x + 6
3x – 4 = 2x + 6		–3x + 4 = 2x + 6
x – 4 = 6		4 = 5x + 6
x = 10		–2 = 5x
x = 10 or –

If x = 10 or –, but 0 ≥ x, then x can only be –.

17. (B). If –x|x| ≥ 4, –x|x| is positive. Because |x| is positive by definition, –x|x| is positive only when –x is also positive. This occurs when x is negative. For example, x = –2 is one solution allowed by the inequality: –x|x| = –(–2) × |–2| = 2 × 2 = 4.

So, Quantity A can be any integer less than or equal to –2, all of which are less than 2. Quantity B is greater.

18. (A). The inequality |x| < 1 allows x to be either a positive or negative fraction (or zero). Interpreting the absolute value sign, it is equivalent to –1 < x < 1. As indicated, y is positive.

When x is a negative fraction:

     Quantity A: |x| + y = positive fraction + positive = positive
     Quantity B: xy = negative fraction × positive = negative
     Quantity A is greater in these cases.

When x is zero:

     Quantity A: |x| + y = 0 + positive = positive
     Quantity B: xy = 0 × positive = 0
     Quantity A is greater in this case.

When x is a positive fraction:

     Quantity A: |x| + y = positive fraction + y = greater than y
     Quantity B: xy = positive fraction × y = less than y
     Quantity A is greater in these cases.

In all cases, Quantity A is greater.

19. (B). In general, there are four cases for the signs of x and y, some of which can be ruled out by the constraints of this question:

x	y	x + y > 0
pos	pos	True
pos	neg	True when |x| > |y|
neg	pos	False when |x| > |y|
neg	neg	False

Only the first two cases need to be considered for this question, since x + y is not greater than zero for the third and fourth cases.

If x and y are both positive, |x| > |y| just means that x > y.

If x is positive and y is negative, x > y simply because positive > negative.

In both cases, x > y. Quantity B is greater.

20. (D). If y is an integer and |y| ≤ 1, then y = –1, 0, or 1. The other inequality can be simplified from |x|(y) + 9 < 0 to |x|(y) < –9. In other words, |x|(y) is negative. Because |x| cannot be negative by definition, y must be negative, so only y = –1 is possible.

If y = –1, then |x|(y) = |x|(–1) = – |x| < –9. So, –|x| = –10, –11, –12, –13, etc.

Thus, x = ±10, ±11, ±12, ±13, etc. Some of these x values are greater than –9 and some are less than –9. Therefore, the relationship cannot be determined.

21. (A). In general, there are four cases for the signs of p and k, some of which can be ruled out by the constraints of this question:

p	k	p + |k| > |p| + k
pos	pos	Not true in this case: For positive numbers, absolute value “does nothing,” so both sides are equal to p + k.
pos	neg	True for this case: p + (a positive absolute value) is greater than p + (a negative value).
neg	pos	Not true in this case: k + (a negative value) is less than k + (a positive absolute value).
neg	neg	Possible in this case: It depends on relative values. Both sides are a positive plus a negative.

Additionally, check whether p or k could be zero.

If p = 0, p + |k| > |p| + k is equivalent to |k| > k. This is true when k is negative.

If k = 0, p + |k| > |p| + k is equivalent to p > |p|. This is not true for any p value.

So, there are three possible cases for p and k values. For the second one, use the identity that |a| = –a when a is negative:

p	k	Interpret:
pos	neg	p = pos > neg = k p > k
neg	neg	p + |k| > |p| + k p + –(k) > –(p) + k p – k > –p + k 2p – k > k 2p > 2k p > k
0	neg	p = 0 > neg = k p > k

In all the cases that are valid according to the constraint inequality, p is greater than k. Quantity A is greater.

22. (D). Given only one inequality with three unknowns, solving will not be possible. Instead, test numbers with the goal of proving (D).

For example, x = 2, y = 5, and z = 3.
Check that |x| + |y| > |x + z|: |2| + | 5| > |2 + 3| is 7 > 5, which is true.
In this case, y > z and Quantity A is greater.

Try to find another example such that y < z. Always consider negatives in inequalities and absolute value questions. Consider another example: x = 2, y = –5 and z = 3.
Check that |x| + |y| > |x + z|: |2| + |–5| > |2 + 3| is 7 > 5, which is true.
In this case, z > y and Quantity B is greater.

Either statement could be greater. The relationship cannot be determined from the information given.

23. (B). If is greater than 1, then it is positive. Because |a| is non-negative by definition, b would have to be positive. Thus, when multiplying both sides of the inequality by b, you do not have to flip the sign of the inequality:

      > 1
       |a| > b

To summarize, b > 0 and |a| > b. Putting this together, |a| > b > 0.

In order for a + b to be negative, a must be more negative than b is positive. For example, a = –4 and b = 2 agree with all the constraints so far. Note that a cannot be zero (because = 0 in this case, not > 1) and a cannot be positive (because a + b > 0 in this case, not < 0).

Therefore, a < 0. Quantity B is greater.

24. (B). Neither f nor g can be zero, or f ²g would be zero. The square of either a positive or negative base is always positive, so f ² is positive. In order for f ²g < 0 to be true, g must be negative. Therefore, the correct answer is (B). Answer choices (A), (C), and (D) are not correct because f could be either positive or negative. Answer choice (E) directly contradicts the truth that f ² is positive.

25. (D). Solve the first inequality:

	<	x
	<	x
	<	x
4	<	x

Solve the second inequality:

	<
x	<
x	<
x	<	6

Combining the inequalities gives 4 < x < 6, and since x is an integer, x must be 5.

26. (B). In general, there are four cases for the signs of x and y, some of which can be ruled out by the constraint in the question stem. Use the identity that |a| = a when a is positive or zero and |a| = –a when a is negative:

Note that if either x or y equals 0, that case would also fail the constraint.

The only valid case is when x is negative and y is positive:

     Quantity A: (x + y)² = x² + 2xy + y²
     Quantity B: (x – y)² = x² – 2xy + y²

Ignore (or subtract) x² + y² as it is common to both quantities. Thus:

     Quantity A: 2xy = 2(negative)(positive) = negative
     Quantity B: –2xy = –2(negative)(positive) = positive

Quantity B is greater.

27. (A). First, solve 4 – 11x ≥ for x:

4 – 11x	≥
8 – 22x	≥	–2x + 3
5 – 22	≥	–2x
5	≥	20x
	≥	x
	≥	x

Thus, the correct choice should show the gray line beginning to the right of zero (in the positive zone), and continuing indefinitely into the negative zone. Even without actual values (other than zero) marked on the graphs, only (A) meets these criteria.

28. (A). From –1 < a < 0 < |a| < b < 1, the following can be determined:

     a is a negative fraction,
     b is a positive fraction, and
     b is more positive than a is negative (i.e., |b| > |a|, or b is farther from 0 on the number line than a is).

Using exponent rules, simplify the quantities:

     Quantity A:
     Quantity B:

Dividing both quantities by b would be acceptable, as b is positive and doing so won’t flip the relative sizes of the quantities. It would be nice to cancel a’s, too, but it is problematic that a is negative. Dividing both quantities by a² would be okay, though, as a² is positive.

Divide both quantities by a²b:

     Quantity A:
     Quantity B:

Just to make the quantities more similar in form, divide again by b, which is positive:

     Quantity A:

     Quantity B:

Both quantities are negative, as a and b have opposite signs. Remember that b is more positive than a is negative. (i.e., |b| > |a|, or b is farther from 0 on the number line than a is.) Thus, each fraction can be compared to –1:

     Quantity A: is less negative than –1. That is, –1 < .

     Quantity B: is more negative than –1. That is, < –1.

Therefore, Quantity A is greater.

29. (D). Given only a compound inequality with three unknowns, solving will not be possible. Instead, test numbers with the goal of proving (D). Always consider negatives in inequalities and absolute value questions.

For example, x = 10, y = –9, and z = 8.
Check that x > |y| > z: 10 > |–9| > 8, which is true.
In this case, x + y = 10 + (–9) = 1 and |y| + z = 9 + 8 = 17. Quantity B is greater.

Try to find another example such that Quantity A is greater.
For example, x = 2, y = 1, and z = –3.
Check that x > |y| > z: 2 > |1| > –3, which is true.
In this case, x + y = 2 + 1 = 3 and |y| + z = 1 + (–3) = –2. Quantity A is greater.

The relationship cannot be determined from the information given.

30. (D). The values for k, l, and m, respectively, could be any of the following three sets:

     Set 1: 24, 26, and 28
     Set 2: 26, 28, and 30
     Set 3: 28, 30, and 32

For evenly spaced sets with an odd number of terms, the average is the middle value. Therefore, the average of k, l, and m could be 26, 28, or 30. Only answer choice (D) matches one of these possibilities.

31. (B). The number line indicates a range between, but not including, –3 and 1. However, –3 < x < 1 is not a given option. However, answer choice (B) gives the inequality –6 < 2x < 2. Dividing all three sides of this inequality by 2 yields –3 < x < 1.

32. 120 and 720 only. If x is “greater than 3 but no more than 6,” then x is 4, 5, or 6. If there are 4 judges sitting in 4 seats, they can be arranged 4! = 4 × 3 × 2 × 1 = 24 ways. If there are 5 judges sitting in 5 seats, they can be arranged 5! = 5 × 4 × 3 × 2 × 1 = 120 ways. If there are 6 judges sitting in 6 seats, they can be arranged 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720 ways. Thus, 24, 120, and 720 are all possible answers. Only 120 and 720 appear in the choices.

33. > –3c only. For this problem, use the rule that multiplying or dividing an inequality by a negative flips the inequality sign. Thus, multiplying or dividing an inequality by a variable should not be done unless you know whether to flip the inequality sign (i.e., whether the variable represents a positive or a negative number).

1st inequality: TRUE. Multiply both sides of the original inequality by –3 and flip the inequality sign.

2nd inequality: Maybe. Multiply both sides of the original inequality by b to get the 2nd inequality, but only if b is positive. If b is negative, the direction of the inequality sign would have to be changed.

3rd inequality: Maybe. Multiplying both sides of the original inequality by –3b could lead to the 3rd inequality, but because the inequality sign flipped, this is only true if –3b is negative (i.e., if b is positive).

34. (B). From z < y – x, the value of z depends on x and y. So, solve for x and y as much as possible. There are two cases for the absolute value equation: |x + y| = 10 means that ±(x + y) = 10 or that (x + y) = ±10. Consider these two cases separately.

The positive case:

     x + y = 10, so y = 10 – x.
     Substitute into z < y – x, getting z < (10 – x) – x, or z < 10 – 2x.
     Because x is at least zero, 10 – 2x ≤ 10.
     Putting the inequalities together, z < 10 – 2x ≤ 10.
     Thus, z < 10.

The negative case:

     x + y = –10, so y = –10 – x.
     Substitute into z < y – x, getting z < (–10 – x) – x, or z < –10 – 2x.
     Because x is at least zero, –10 – 2x ≤ –10.
     Putting the inequalities together, z < –10 – 2x ≤ –10.
     Thus, z < –10.

In both cases, 10 is greater than z. Quantity B is greater.

35. (A). The variable a is common to both quantities, and adding it to both quantities to cancel will not change the relative values of the quantities:

     Quantity A: (9 – a) + a = 9
     Quantity B: + a =

According to the given constraint, < 9, so Quantity A is greater.

36. (C). If p is an integer such that 1.9 < |p| < 5.3, p could be 2, 3, 4, or 5, as well as –2, –3, –4, or –5. The greatest value of p is 5, for which the value of f(p) is equal to 5² = 25. The least value of p is –5, for which the value of f(p) is equal to (–5)² = 25. Therefore, the two quantities are equal.

37. (D). If < 0, then the two fractions have opposite signs. Therefore, by the definition of reciprocals, must be the negative inverse of , no matter which one of the fractions is positive. In equation form, this means = –, which is choice (D). The other choices are possible but not necessarily true.

38. (C). In order to get m and n out of the denominators of the fractions on the left side of the inequality, multiply both sides of the inequality by mn. The result is kn + lm > (mn)². The direction of the inequality sign changes because mn is negative. This is an exact match with (C), which must be the correct answer.

39. (D). The inequality described in the question is 0 > > y + z. Multiplying both sides of this inequality by x, the result is 0 < 1 < xy + xz. Notice that the direction of the inequality sign must change because x is negative. Therefore:

(A) Maybe true: true only if x equals –1.

(B) Maybe true: either y or z or both can be negative.

(C) False: the direction of the inequality sign is opposite the correct direction determined above.

(D) TRUE: it is a proper rephrasing of the original inequality.

(E) Maybe true: it is not a correct rephrasing of the original inequality.

40. (D). When the GRE writes a root sign, the question writers are indicating a non-negative root only. Therefore, both sides of this inequality are positive. Thus, you can square both sides without changing the direction of the inequality sign. So u < –3v. Now evaluate each answer choice:

(A) Must be true: divide both sides of u < –3v by 3.

(B) Must be true: it is given that –3v > 0 and therefore, v < 0. Then, when dividing both sides of u < –3v by v, you must flip the inequality sign and get > –3.

(C) Must be true: this is the result after dividing both sides of the original inequality by .

(D) CANNOT be true: adding 3v to both sides of u < –3v results in u + 3v < 0, not u + 3v > 0.

(E) Must be true: this is the result of squaring both sides of the original inequality.

41. (D). Since each of the three arcs corresponds to one of the 60° angles of the equilateral triangle, each arc represents of the circumference of the circle. The diagram below illustrates this for just one of the three angles in the triangle:

The same is true for each of the three angles:

Since each of the three arcs is between 4π and 6π, triple these values to determine that the circumference of the circle is between 12π and 18π. Because circumference equals π times the diameter, the diameter of this circle must be between 12 and 18. Only choice (D) is in this range.