Inequalities - GMAT Quantitative Reasoning

Card 0 of 19

Question

True or false:

Statement 1:

Statement 2: $x^{2}< 81$

Answer

Assume Statement 1 alone. can be rewritten as .

Assume Statement 2 alone. It can be rewritten as

$x^{2}< 81$

the solution set of which is

From either statement alone, it follows that .

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Question

The variables stand for integer quantities.

Order from least to greatest:

$A^{P}, B^{Q}, C^{R}, D^{S}$

Statement 1:

Statement 2:

Answer

Statement 1 alone only tells us the common base; without knowing the order of the exponents, we cannot order the powers. Similarly, Statement 2 only gives us the order of the exponents; we cannot order the powers without any information about the bases.

Assume both statements are true. Since , it holds that

$N^{2}= N \cdot N > N$

$N^{3}= N^{2} \cdot N > N^{2}$

et cetera, and

$N^{0} = 1 < N^{1}$

$N^{-1} = N^{0} \div N < N^{1} \div N = N^{0}$

and so forth in both directions. That is,

$...<N^{-2} <N^{-1}< N^{0} <N ^{1}< N^{2} < ...$

Therefore, since , it follows that

$N^{P}< N^ {Q} <N ^{R}< N^{S}$ ,

and the ordering is determined.

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Question

Data sufficiency question- do not actually solve the question

Is $\small xy< 12$ ?

1. $\small x^2=64; 1\leq y\leq 1.5$

2. $\small x+y=6$

Answer

From statement 1, we can conclude that $\small xy\leq 12$ but not $\small xy< 12$ . From the second statement, we can conclude that the greatest product will result from $\small 3+3=6$ or 9, which is less than 12.

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Question

How many solutions does the equation $\left | x - A \right | = B$ have?

Statement 1:

Statement 2:

Answer

If we only know that , then the above statement becomes $\left | x - 8 \right | = B$ , and it can have zero, one, or two solutions depending on the value of . For example:

If , the equation is $\left | x - 8 \right | = -5$ , which has no solution, as an absolute value cannot be negative.

If , the equation is $\left | x - 8 \right | = 0$ , which requires that , or , since only 0 has absolute value 0; this means the equation has one solution.

If we only know that , then the equation becomes $\left | x - A \right | = -8$ , which has no solution regardless of the value of ; this is because, as stated before, an absolute value cannot be negative.

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Question

True or false: is a positive number.

Statement 1: $N = \left | N \right |$

Statement 2: $N > N^{2}$

Answer

If is negative, then and $N ^{2} > 0 > N$ . Therefore, either Statement 1 or Statement 2 alone proves nonnegative. However, if , then $N = \left | N \right |$ , but $N > N^{2}$ is false.

Therefore, Statement 2 proves positive, but Statement 1 only proves nonnegative.

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Question

True or false:

Statement 1: $x^{8} >256$

Statement 2: $x^{10} >1,024$

Answer

Both statements together provide insufficient information. For example,

If , then:

$x^{8}= 3^{8}= 6,561 > 256$

$x^{10}= 3^{10}= 59,049 > 1,024$

If , then

$x^{8}= (-3)^{8}= 6,561 > 256$

$x^{10}= \left (-3 \right )^{10}= 59,049 > 1,024$

Both values fit the conditions of both statements, but only one is greater than . The question is not answered.

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Question

True or false:

Statement 1: $x^{5} >100,000$

Statement 2: $x^{7} >10,000,000$

Answer

Assume Statement 1 alone. Then, since an odd (fifth) root of a number assumes the sign of that number, and an odd root function is an increasing function, we can simply take the fifth root of each side:

$x^{5} >100,000$ ,

$\sqrt[5]{x^{5}} >\sqrt[5]{100,000}$

Assume Statement 2 alone.By a similar argument,

$x^{7} >10,000,000$ ,

$\sqrt[7]{x^{7}} >\sqrt[7]{10,000,000}$

From either statement alone, it follows that is true.

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Question

True or false:

Statement 1: $x^{2}< 144$

Statement 2: $x^{3}< 1,728$

Answer

The absolute value of a number is its unsigned value - that is, if the number is nonnegative, it is the number itself, and if the number is negative, it is the corresponding positive.

From Statement 1 alone, since $x^{2}< 144$ , then it follows that - this is equivalent to saying .

Statement 2 alone provides insufficient information. For example:

If , then $x^{3}= 11^{3} = 1,331 < 1,728$

if , then $x^{3}= (-13)^{3} =- 2,197 <- 1,728$

Both numbers fit the condition of Statement 2, but

and

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Question

True or false:

Statement 1: $x^{2} > 49$

Statement 2: $x^{3}$ is positive.

Answer

Assume Statement 1 alone. From $x^{2} > 49$ , it can be determined that either or , but nothing more; the statement alone provides insufficient information. Statement 2 taken alone provides insufficient information as well, since the positive numbers include numbers both less than and greater than 7.

Assume both statements are true. From Statement 1, either or , but Statement 2 gives us that . Therefore, , and the question is answered.

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Question

True or false:

Statement 1: $x^{2} > 121$

Statement 2: $x^{3}> 1,331$

Answer

Assume Statement 1 only. Both and 12 make the statement true, since $12^{2}= (-12)^{2}= 144 > 121$ . But one is less than 11 and one is not.

Assume Statement 2 only. Then, since an odd (third) root of a number assumes the sign of that number, and an odd root function is an increasing function, we can simply take the cube root of each side:

$x^{3}> 1,331$

$\sqrt[3]{x^{3}}>\sqrt[3]{ 1,331}$

or

.

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Question

True or false:

Statement 1:

Statement 2: $x^{2}> 25$

Answer

makes both statements true, since and $6^{2}= 36 > 25$ .

makes both statements true, since and $(-6)^{2}= 36 > 25$ .

One of the two values is less than 5, and one is greater than 5. The statements together provide insufficient information.

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Question

True or false:

Statement 1: $x^{5}< 0$

Statement 2: $3 ^{x}> 6^{x}$

Answer

Assume Statement 1 alone. Since the fifth (odd) power of a number assumes the same sign as the number itself, and $x^{5}$ have the same sign, and $x^{5}< 0$ implies that .

Assume Statement 2 alone. Since $3^{x}$ and $6^{x}$ are both positive, we can divide both sides by $3^{x}$ to yield the statement

$3^{x}>6^{x}$

$6^{x} < 3^{x}$

$\frac{6^{x}}{3^{x}} <\frac{ 3^{x}}{3^{x}}$

$\left (\frac{6}{3} \right )^{x} <1$

$2^{x}< 1$

Since $f(x) = 2^{x}$ increases as does, and since $2^{0}= 1$ , it follows that .

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Question

True or false:

Statement 1: $2^{x}<4^{x}$

Statement 2: $x^{3}> 0$

Answer

Assume Statement 1 alone. Since $2^{x}$ and $4^{x}$ are both positive, we can divide both sides by $2^{x}$ to yield the statement

$2^{x}<4^{x}$

$4^{x} > 2^{x}$

$\frac{ 4^{x}}{2^{x}} >\frac{ 2^{x}}{2^{x}}$

$\left (\frac{ 4}{2} \right )^{x} >1$

$2^{x}> 1$

Since $f(x) = 2^{x}$ increases as increases, and since $2^{0}= 1$ , it follows that .

Assume Statement 2 alone. Since the cube root of a number assumes the same sign as the number itself, $x^{3}> 0$ implies that .

From either statement alone it follows that .

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Question

is a whole number.

True or false: is odd.

Statement 1: $2^{x} > 0$

Statement 2: $(-2)^{x} > 0$

Answer

Statement 1 alone is a superfluous statement, since a positive number raised to any power must yield a positive result.

Statement 2 alone answers the question, since a negative number raised to a whole number exponent yields a positive result if and only if the exponent is even. Since Statement 2 states that $(-2)^{x}$ is positive, is even, not odd.

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Question

The variables stand for integer quantities.

Order from least to greatest:

$A^{P}, B^{Q}, C^{R}, D^{S}$

Statement 1:

Statement 2:

Answer

We show the two statements provide insufficient information by assuming both statements to be true and showing that the ordering is different depending on the common exponent. In both cases, we let .

If , then the expressions become

$A^{P}= 2^{1}= 2$

$B^{Q}= 2^{2} = 4$

$C^{R}= 2^{3}= 8$

$D^{S}= 2^{4}= 16$

The correct ordering is $A^{P}<B^{Q}<C^{R}< D^{S}$ .

If , then the expressions become

$A^{P}= \left ( -2\right )^{1}= -2$

$B^{Q}= \left ( -2\right )^{2} = 4$

$C^{R}= \left ( -2\right )^{3}=- 8$

$D^{S}= \left ( -2\right )^{4}= 16$

The correct ordering is $C^{R}< A^{P}< B^{Q}<D^{S}$ .

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Question

The variables stand for integer quantities.

Order from least to greatest:

$A^{P}, B^{Q}, C^{R}, D^{S}$

Statement 1:

Statement 2:

Answer

We show the staements together provide insufficient information by examining two situations.

Suppose .

If , the four expressions become:

$A^{P}= 1^{2}= 1$

$B^{Q}= 2^{2} = 4$

$C^{R}=3^{2}= 9$

$D^{S}= 4^{2}= 16$

and $A^{P} < B^{Q}<C^{R}< D^{S}$

If , the four expressions become:

$A^{P}= 1^{-2}= 1$

$B^{Q}= 2^{-2} = \frac{1}{4}$

$C^{R}=3^{-2}= \frac{1}{9}$

$D^{S}= 4^{-2}= \frac{1}{16}$

In ascending order, the expressions are $D^{S}<C^{R}< B^{Q}<A^{P}$ .

Since the orderings are different in the two cases, the two statements together give insufficient information as to their correct ordering.

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Question

The variables stand for integer quantities.

Order from least to greatest:

$A^{P}, B^{Q}, C^{R}, D^{S}$

Statement 1:

Statement 2:

Answer

Statement 1 only gives us the order of the bases; we cannot order the powers without any information about the exponents. Similarly, Statement 2 alone only tells us the common exponent; without knowing the order of the bases, we cannot order the powers.

Assume both statements are true. Let's look at $A^{P}$ and $B^{Q}$ .

Since and are both positive, and , we can apply the multiplication property of equality:

$B\cdot B>A \cdot A$

$B^{2}> A^{2}$

Similarly, $B^{3}> A^{3}, B^{4}> A^{4},$ etc.

So, if and ,

$A^{P}< B^{P}$ .

Since from Statement 1, it follows that

$A^{P}< B^{P}< C^{P}< D^{P}$ ,

which, from Statement 2, can be rewritten as

$A^{P}< B^{Q}< C^{R}< D^{S}$ .

The order has been determined.

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Question

is a real number. True or false: is positive.

Statement 1: $N^{2}+ 10N + 24 < 0$

Statement 2:

Answer

Assume Statement 1 alone. If is positive, then $N^{2} > 0$ and ; since $N^{2}+ 10N + 24$ is the sum of three positive numbers, then $N^{2}+ 10N + 24 > 0$ , and $N^{2}+ 10N + 24 < 0$ is a false statement. Therefore, cannot be positive.

Assume Statement 2. If is positive, then so is , and the inequality can be rewritten as

Consequently,

,

a contradiction since is positive. Therefore, is not positive.

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Question

is a real number. True or false: is positive.

Statement 1: $2^{N} > 8$

Statement 2: $N^{2} > 9$

Answer

Assume Statement 1 alone. $2^{N} > 8$ , or $2^{N} > 2^{3}$ . For any base , if $a^{b}> a^{c}$ , then , so it follows that . is therefore positive.

Assume Statement 2 alone. Both and are solutions:

$4^{2} = 16 > 9$

and

$(-4)^{2} = 16 > 9$

The sign of cannot be determined.

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