GRE_Math_Bible_eBook

Inequalities 231

Miscellaneous Properties of Positive and Negative Numbers

1.The product (quotient) of positive numbers is positive.

2.The product (quotient) of a positive number and a negative number is negative.

3.The product (quotient) of an even number of negative numbers is positive.

4.The product (quotient) of an odd number of negative numbers is negative.

5.The sum of negative numbers is negative.

6.A number raised to an even exponent is greater than or equal to zero.

Example: If xy2z < 0 , then which one of the following statements must also be true?

I. xz < 0

II. z < 0

III. xyz < 0

Since a number raised to an even exponent is greater than or equal to zero, we know that y2 is positive (it cannot be zero because the product xy2z would then be zero). Hence, we can divide both sides of the

Therefore, I is true, which eliminates (A), (C), and (E). Now, the following illustrates that z< 0 is not necessarily true:

−1 22 3 = −12 < 0

This eliminates (D). Hence, the answer is (B).

Absolute Value

The absolute value of a number is its distance on the number line from 0. Since distance is a positive number, absolute value of a number is positive. Two vertical bars denote the absolute value of a number:

–5 –4 –3 –2 –1 0 1 2 3 4 5

Students rarely struggle with the absolute value of numbers: if the number is negative, simply make it positive; and if it is already positive, leave it as is. For example, since –2.4 is negative, − 24 = 2.4 and since 5.01 is positive 5.01 = 5.01.

Further, students rarely struggle with the absolute value of positive variables: if the variable is positive, simply drop the absolute value symbol. For example, if x > 0, then x = x .

However, negative variables can cause students much consternation. If x is negative, then x = −x .

This often confuses students because the absolute value is positive but the –x appears to be negative. It is actually positive—it is the negative of a negative number, which is positive. To see this more clearly let x = –k, where k is a positive number. Then x is a negative number. So x = −x = −(−k) = k . Since k is

positive so is –x. Another way to view this is x = −x = (−1) x = (–1)(a negative number) = a positive number.

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Example: If x = – x , then which one of the following statements could be true?

Statement I could be true because – 0 = −(+0) = −(0) = 0. Statement II could be true because the right side of the equation is always negative [ – x = –(a positive number) = a negative number]. Now, if one side of an equation is always negative, then the other side must always be negative, otherwise the opposite sides of the equation would not be equal. Since Statement III is the opposite of Statement II, it must be false. But let’s show this explicitly: Suppose x were positive. Then x = x , and the equation x = – x becomes x =

–x. Dividing both sides of this equation by x yields 1 = –1. This is contradiction. Hence, x cannot be positive. The answer is (D).

Now, the only numbers at which the expression can change sign are –5 and –1. So –5 and –1 divide the number line into three intervals. Let’s set up a number line and choose test points in each interval:

inequality. That is, x < –5. When x = –3, x 2 > −6x − 5 becomes 9 > 13. This is false. Hence, no numbers in Interval II satisfy the inequality. When x = 0, x 2 > −6x − 5 becomes 0 > –5. This is true. Hence, all numbers in Interval III satisfy the inequality. That is, x > –1. The graph of the solution follows:

–5 –1

Note, if the original inequality had included the greater-than-or-equal symbol, ≥ , the solution set would have included both –5 and –1. On the graph, this would have been indicated by filling in the circles above –5 and –1. The open circles indicate that –5 and –1 are not part of the solution.

Inequalities 233

Summary of steps for solving higher order inequalities:

1.Replace the inequality symbol with an equal symbol.

2.Move all terms to one side of the equation (usually the left side).

3.Factor the equation.

4.Set the factors equal to 0 to find zeros.

5.Choose test points on either side of the zeros.

6.If a test point satisfies the original inequality, then all numbers in that interval satisfy the inequality. Similarly, if a test point does not satisfy the inequality, then no numbers in that interval satisfy the inequality.

The answer is (C).

Like Inequalities Can Be Added

If x < y and w < z, then x + w < y + z

Example: If 2 < x < 5 and 3 < y < 5, which of the following best describes x – y ?

(A)–3 < x – y < 2

(B)–3 < x – y < 5

(C)0 < x – y < 2

(D)3 < x – y < 5

(E)2 < x – y < 5

Multiplying both sides of 3 < y < 5 by –1 yields –3 > –y > –5. Now, we usually write the smaller number on the left side of the inequality. So –3 > –y > –5 becomes –5 < –y < –3. Add this inequality to the like inequality 2 < x < 5:

2 < x < 5

(+) − 5 < −y < −3

− 3 < x − y < 2

The answer is (A).

Inequalities 235

8.If x < y < –1, then which one of the following expressions is positive?

(A)–x2

(B)y

(C)x2y

x2

(D)y 2

(E)y – x2

9.If x + 3 is positive, then which one of the following must be positive?

(A)x – 3

(B)(x – 3)(x – 4)

(C)(x – 3)(x + 3)

(D)(x – 3)(x + 4)

(E)(x + 3)(x + 6)

10.If (x + 1)2 – 2x > 2(x + 1) + 2, then x cannot equal which one of the following?

(A)–5

(B)–3

(C)0

(D)3

(E)5

11.If x5 + x2 < 0, then which one of the following must be true?

(A)x < –1

(B)x < 0

(C)x > 0

(D)x > 1

(E)x4 < x2

12.In ∆PQR, PQ = x, QR = x + 3, and PR = y. If x = y + 3, then which one of the following is true?

(A)P < Q < R

(B)Q < R < P

(C)R < P < Q

(D)P < R < Q

(E)Q < P < R

13.If 5 < x < 10 and y = x + 5, what is the greatest possible integer value of x + y ?

(A)18

(B)20

(C)23

(D)24

(E)25

15.If (x – y)3 > (x – y)2, then which one of the following must be true?

(A)x3 < y2

(B)x5 < y4

(C)x3 > y2

(D)x5 > y4

(E)x3 > y3

Inequalities 237

21.If it is true that 1/55 < x < 1/22 and 1/33 < x < 1/11, then which of the following numbers could x equal?

I.1/54

II.1/23

III.1/12

(A)I only

(B)II only

(C)III only

(D)I and II only

(E)II and III only

238GRE Math Bible

Answers and Solutions to Problem Set M

Easy

1.The product ab is positive when both a and b are positive or when both a and b are negative; in either case, a/b is positive. Hence, choice (A) is always positive, and the answer is (A).

2.Multiplying the given inequality x/15 > y/25 by 75 yields 5x > 3y.

Now, subtracting 3y and 5x from both columns yields

Since we know that 5x > 3y, Column B is greater than Column A and the answer is (B).

Medium

3. We are given the inequality 3x + 7y > 7x + 3y. Subtracting 3x + 3y from both sides of the inequality yields 4y > 4x, and dividing both sides of this inequality by 4 yields y > x.

Since y > x, Column A is greater than Column B and the answer is (A).

4.If p = 2, then both columns equal 4. If p ≠ 2, then the columns are unequal. This is a double case, and therefore the answer is (D).

5.Canceling z from both sides of the inequality x + z > y + z yields x > y. Adding –z to both sides yields x – z > y – z. Hence, I is true.

If z is negative, multiplying the inequality x > y by z would flip the direction of the inequality resulting in the inequality xz < yz. Hence, II may not be true.

If z is negative, dividing the inequality x > y by z would flip the direction of the inequality resulting in the inequality x/z < y/z. Hence, III may not be true.

The answer is (A) since we are asked for statements that MUST be true.

6. We are given that a = 3 + b. This equation indicates that a is 3 units larger than b, so a is greater than b + 2.5. Hence, I is true, and II is false.

Similarly, a is greater than 2 + b. Hence, III is true.

Hence, the answer is (E): I and III are true.

Inequalities 239

7. We are given that x + x = 4. If x is negative or zero, then x equals –x and x + x equals –x + x = 0.

This conflicts with the given equation, so x is not negative nor equal to 0. Hence, x is positive and therefore

xequals x. Putting this in the given equation yields

x+ x = 4

x + x = 4 2x = 4 x = 2

Now, select the answer-choice that results in an odd number when x = 2.

Choice (A): x2 + 3x = 22 + 3(2) = 4 + 6 = 10, an even number. Reject.

Choice (B): x2 + 3x + 2 = 22 + 3(2) + 2 = 4 + 6 + 2 = 12, an even number. Reject.

Choice (C): x2 + 4x = 22 + 4(2) = 4 + 8 = 12, an even number. Reject.

Choice (D): x2 + 4x + 2 = 22 + 4(2) + 2 = 4 + 8 + 2 = 14, an even number. Reject.

Choice (E): x2 + 4x + 3 = 22 + 4(2) + 3 = 4 + 8 + 3 = 15, an odd number. Accept.

The answer is (E).

8. From the inequality x < y < –1, we have that x and y do not equal zero. Since the square of a nonzero number is positive, x2 is positive.

Hence, –x2 must be negative, and therefore choice (A) is negative.

Since we are given that y < –1, y is negative. Hence, choice (B) is negative.

Multiplying the inequality y < –1 by the positive value x2 yields x2y < –x2, a negative value. Hence, choice

(C) is also negative.

and choice (D) is positive.

Since y and –x2 are negative, their sum y + (–x2) = y – x2 must also be negative. Hence, choice (E) is negative.

The only choice that is positive is (D), so the answer is (D).

9. We are given that x + 3 > 0.

Subtracting 3 from both sides yields x > –3.

In case x equals 2, x – 3 is negative. Hence, reject choice (A).

(x – 3)(x – 4) is negative for the values of x between 3 and 4 [For example, when x equals 3.5, the expression is negative]. The known inequality x > –3 allows the values to be in this range. Hence, the expression can be negative. Reject choice (B).

(x – 3)(x + 3) is negative for the values of x between 3 and –3 [For example, when x equals 0, the expression is negative]. The known inequality x > –3 allows the values to be in this range. Hence, the expression can be negative. Reject choice (C).

(x – 3)(x + 4) is negative for the values of x between 3 and –4 [For example, when x equals 0, the expression is negative]. The known inequality x > –3 allows the values to be in this range. Hence, the expression can be negative. Reject choice (D).

(x + 3)(x + 6) is negative only for the values of x between –3 and –6 [For example, when x equals –4, the expression is negative]. But the known inequality x > –3 does not allow the values to be in this range. Hence, the expression cannot be negative. Hence, accept choice (E).

240GRE Math Bible

10. We are given the inequality

Square rooting both sides of the inequality yields two new inequalities: x – 1 > 2 or x – 1 < –2. Adding 1 to both sides of the solutions yields x > 3 and x < –1. Hence, x is either less than –1 or x is greater than 3. In either case, x does not equal 0. Hence, the answer is (C).

11.Subtracting x2 from both sides of the given inequality yields x5 < –x2. Dividing the inequality by the positive value x2 yields x3 < –1. Taking the cube root of both sides yields x < –1. The answer is (A).

12.Subtracting 3 from both sides of the equation x = y + 3 by 3 yields x – 3 = y. Hence, PR = x – 3. Since

x – 3 < x < x + 3, we have the inequality PR < PQ < QR (since PR = y = x – 3, PQ = x, and QR = x + 3, given). Since the angle opposite the greater side is greater, in the triangle Q < R < P. Hence, the answer is (B).

13.Adding x to both sides of the equation y = x + 5 yields x + y = x + (x + 5), or x + y = 2x + 5. Hence, the greatest possible value of x + y is the maximum possible value of 2x + 5. Now, let’s create this expression out of the given inequality 5 < x < 10. Multiplying the inequality by 2 yields 10 < 2x < 20. Adding 5 to each part of the inequality yields 10 + 5 < 2x + 5 < 20 + 5, or 15 < 2x + 5 < 25. So, 2x + 5 is less than 25. The greatest possible integer value of 2x + 5 is 24. Hence, the answer is (D).

14.Factoring Column B yields

Since a > b, a – b > 0. Hence, we can cancel a – b, a positive value, from both columns:

Since a > b > 0, both a and b are positive. Hence, the Least Common Denominator of the fractions, (a + b)(a2 + b2), is positive. So, we can multiply both columns by the LCD to clear the fractions:

Since both a and b are positive, 2ab is positive. Hence, Column B is greater than Column A, and the answer is (B).