Exponents - GMAT Quantitative

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Question

is an integer. Is there a real number such that $x^{A}=B$ ?

Statement 1: is negative

Statement 2: is even

Answer

The equivalent question is "does have a real $A \textrm{th }$ root?"

If you know only that is negative, you need to know whether is even or odd; negative numbers have real odd-numbered roots, but not real even-numbered roots.

If you know only that is even, you need to know whether is negative or nonnegative; negative numbers do not have real even-numbered roots, but nonnegative numbers do.

If you know both, however, then you know that the answer is no, since as stated before, negative numbers do not have real even-numbered roots.

Therefore, the answer is that both statements together are sufficient to answer the question, but neither statement alone is sufficient to answer the question.

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Question

Which is the greater quantity, $3^{x}$ or $9^{y}$ - or are they equal?

Statement 1:

Statement 2:

Answer

From Statement 1 alone,

$3^{x} = 3^{2y+2} = 3^{2y} \cdot 3^{2} =\left ( 3^{2} \right ) ^{y}\cdot 9 = 9^{y} \cdot 9>9^{y}$

Now assume Statement 2 alone. We show that this is insufficient with two cases:

Case 1:

$3^{3} = 27$ ; $9^{1} = 9$ ; therefore, $3^{x} >9^{y}$

Case 1:

$3^{-3} = \frac{1}{27}$ ; $9^{-1} = \frac{1}{9}$ ; therefore, $3^{x}<9^{y}$

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Question

Does $\log_{2} \left ( MN \right )$ exist?

Statement 1: and are both negative.

Statement 2: divided by 2 yields an integer.

Answer

A logarithm can be taken of a number if and only if the number is positive. If Statement 1 alone is true, then , being the product of two negative numbers, must be positive, and $\log_{2} \left ( MN \right )$ exists.

Statement 2 is irrelevant; 4 and both yield integers when divided by 2, but $\log_{2} 4 = 2$ and $\log_{2}\left ( -4 \right )$ does not exist.

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Question

Johnny was assigned to write a number in scientific notation by filling the circle and the square in the pattern below with two numbers.

$\bigcirc \times 10 ^{\square }$

Johnny filled in both shapes with numbers. Did he succeed?

Statement 1: He filled in the circle with the number "10".

Statement 2: He filled in the square with a negative integer.

Answer

The number $a \times 10 ^{n}$ is a number written in scientific notation if and only of two conditions are true:

$1 \leq \left | a \right | < 10$

is an integer

By Statement 1 Johnny filled in the circle incorrectly, since it makes .

By Statement 2, Johnny filled in the square correctly, but the statement says nothing about how he filled in the circle; Statement 2 leaves the question open.

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Question

Solve the following rational expression:

$\frac{x^{m-n}10^{n-1}}{10^{(\frac{m}{2}-2)}x^{n}}$

(1)

(2)

Answer

When replacing m=5 in the expression we get:

$\frac{x^{5-n}10^{n-1}}{10^{\frac{1}{2}}x^{n}}$

Therefore statement (1) ALONE is not sufficient.

When replacing m=2n in the expression we get:

$\frac{x^{2n-n}10^{n-1}}{10^{(\frac{2n}{2}-2)}x^{n}}=\frac{x^{n}10^{n-1}}{10^{n-2}x^{n}}=\frac{10^{n-1}}{10^{n-2}}=10$

Therefore statement (2) ALONE is sufficient.

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Question

Is $x^{2} < x$ ?

(1)

(2)

Answer

$x^{2} < x \Leftrightarrow x^{2}-x<0 \Leftrightarrow x(x-1)<0$

From statement 1 we get that and .

So the first term is positive and the second term is negative, which means that is negative; therefore the statement 1 alone allows us to answer the question.

Statement 2 tells us that . If , we have $x^{2}=0.25$ which is less than . Therefore in this case $x^{2}<x$ .

For , we have $x^{2} = 4$ which is greater than . So in this case $x^{2} > x$ .

So statement 2 is insufficient.

Therefore the correct answer is A.

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Question

is a number not in the set $\left{-1, 0, 1 \right}$ .

Of the elements $\left { x^{4}, x^{5}, x^{6}\right }$ , which is the greatest?

Statement 1: is a negative number.

Statement 2: $\left | x\right | > 1$

Answer

Statement 1 alone is inconclusive, as can be demonstrated by examining two negative values of other than .

Case 1: .

Then

$x^{4} = (-2)^{4} = 16$

$x^{5} = (-2)^{5} = -32$

$x^{6} = (-2)^{6} = 64$

$x^{6}$ is the greatest of these values.

Case 2: $x = -\frac{1}{2}$

Then

$x ^{4}=\left ( -\frac{1}{2} \right )^{4} = \frac{1}{16}$

$x ^{5}=\left ( -\frac{1}{2} \right )^{5} = -\frac{1}{32}$

$x ^{6}=\left ( -\frac{1}{2} \right )^{6} = \frac{1}{64}$

$x^{4}$ is the greatest of these values.

Now assume Statement 2 alone. Either or .

Case 1: .

Then $x \cdot x^{4} > 1\cdot x^{4}$ , so $x^{5} > x^{4}$ ; similarly, $x^{6} > x^{5}$ .

$x^{6}$ is the greatest of the three.

Case 2: .

Odd power $x^{5}$ is negative, and even powers $x^{4}$ and $x^{6}$ are positive, so one of the latter two is the greatest. Since , it follows that $x^{2} > 1$ . It then follows that $x^{2} \cdot x^{4} > 1 \cdot x^{4}$ , or $x^{6} > x^{4}$ .

Again, $x^{6}$ is the greatest of the three.

Statement 2 alone is sufficient, but not Statement 1.

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Question

Note: Figure NOT drawn to scale.

Examine the above diagram. True or false: $\overline{AB} \cong \overline{YZ}$ .

Statement 1: $\Delta ACB \sim \Delta ZXY$

Statement 2: $\Delta ACB$ and $\Delta ZXY$ have the same perimeter.

Answer

From Statement 1 alone, it follows by the similarity of the triangles that $\angle C \cong \angle X$ . These are congruent inscribed angles of a circle, which intercept congruent arcs, so $\widehat{AXB} \cong \widehat{YCZ}$ . Since congruent arcs have congruent chords, $\overline{AB} \cong \overline{YZ}$ .

Statement 2 alone only tells us the relative perimeters of the triangles. We have no way of determining the individual sidelengths or angle measures relative to each other, so Statement 2 alone is inconclusive.

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Question

is a nonzero number. Is it negative or positive?

Statement 1: $N < N^{2}$

Statement 2: $N < N^{3}$

Answer

Both statements together are inufficient to produce an answer. For example,

If , then $N^{2} = 4$ and $N^{3} = 8$ .

If $N = -\frac{1}{2}$ , then $N^{2} = \frac{1}{4}$ and $N^{3} = -\frac{1}{8}$ .

In both cases, $N < N^{2}$ and $N < N^{3}$ , but the signs of differ between cases.

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Question

is a nonzero number. Is it negative or positive?

Statement 1: $N \ge N^{2}$

Statement 2: $N >N^{3}$

Answer

All negative numbers are less than their (positive) squares, as are all positive numbers greater than 1. Therefore, if Statement 1 is assumed, $N \in (0,1]$ .

can be determined to be positive.

Statement 2 alone is inconclusive. For example, if $N = \frac{1}{2}$ , then $N^{3} = \frac{1}{8}$ , and if , $N^{3} = -8$ . In both cases, $N >N^{3}$ , but has different signs in the two cases.

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Question

Philip has been assigned to write one number in the circle and one number in the square in the diagram below in order to produce a number in scientifc notation.

$\bigcirc \times 10^{\square }$ .

Did Philip succeed?

Statement 1: Philip wrote in the circle.

Statement 2: Philip wrote in the square.

Answer

A number in scientific notation takes the form

$a \times 10 ^{n}$

where $1 \le |a| < 10$ and is an integer of any sign.

Statement 1 alone proves that Philip entered a correct number into the circle, since $1 \le |-7.5| = 7.5 < 10$ . Statement 2 alone proves that he entered a correct number into the square, since is an integer. But each statement alone is insufficient, since each leaves uinclear whether the other number is valid. The two statements together, however, prove that Philip put correct numbers in both places, thereby writing a number in scientific notation.

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Question

Myoshi has been assigned to write one number in the circle and one number in the square in the diagram below in order to produce a number in scientifc notation.

$\bigcirc \times 10^{\square }$ .

Did Myoshi succeed?

Statement 1: Myoshi wrote in the circle.

Statement 2: Myoshi wrote in the square.

Answer

A number in scientific notation takes the form

$a \times 10 ^{n}$

where $1 \le |a| < 10$ and is an integer of any sign.

Assuming Statement 1 alone, Myoshi did not succeed, since she entered an incorrect number into the circle - $|-24| = 24 \ge 10$ .

Statement 2 alone is inconclusive. Myoshi entered a correct number into the square, since is an integer. But the question is open, since it is not known whether she entered a correct number into the circle or not.

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Question

is a number not in the set $\left{-1, 0, 1 \right}$ .

Of the elements $\left { x^{2}, x^{3}, x^{4}\right }$ , which is the greatest?

Statement 1: is a negative number.

Statement 2: $\left | x\right |>\frac{1}{2}$

Answer

Assume both statements are known. The greatest of the three numbers must be $x^{2}$ or $x^{4}$ , since even powers of negative numbers are positive and odd powers of negative numbers are negative.

Case 1: $x = -\frac{2}{3}$

$x^{2} = \left ( -\frac{2}{3} \right )^{2} = \frac{4}{9} = \frac{36}{81}$

$x^{4} = \left ( -\frac{2}{3} \right )^{4} = \frac{16}{81}$

$x^{2} > x^{4}> x^{3}$

Case 2: ,

then

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

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

$x^{4} > x^{2} > x^{3}$

In both cases, is negative and $\left | x\right |>\frac{1}{2}$ , but in one case, $x^{2}$ is the greatest number, and in the other, $x^{4}$ is. The two statements together are inconclusive.

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Question

is an integer. Is there a real number such that $x^{A}=B$ ?

Statement 1: is negative

Statement 2: is even

Answer

The equivalent question is "does have a real $A \textrm{th }$ root?"

If you know only that is negative, you need to know whether is even or odd; negative numbers have real odd-numbered roots, but not real even-numbered roots.

If you know only that is even, you need to know whether is negative or nonnegative; negative numbers do not have real even-numbered roots, but nonnegative numbers do.

If you know both, however, then you know that the answer is no, since as stated before, negative numbers do not have real even-numbered roots.

Therefore, the answer is that both statements together are sufficient to answer the question, but neither statement alone is sufficient to answer the question.

Compare your answer with the correct one above

Question

Which is the greater quantity, $3^{x}$ or $9^{y}$ - or are they equal?

Statement 1:

Statement 2:

Answer

From Statement 1 alone,

$3^{x} = 3^{2y+2} = 3^{2y} \cdot 3^{2} =\left ( 3^{2} \right ) ^{y}\cdot 9 = 9^{y} \cdot 9>9^{y}$

Now assume Statement 2 alone. We show that this is insufficient with two cases:

Case 1:

$3^{3} = 27$ ; $9^{1} = 9$ ; therefore, $3^{x} >9^{y}$

Case 1:

$3^{-3} = \frac{1}{27}$ ; $9^{-1} = \frac{1}{9}$ ; therefore, $3^{x}<9^{y}$

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Question

Does $\log_{2} \left ( MN \right )$ exist?

Statement 1: and are both negative.

Statement 2: divided by 2 yields an integer.

Answer

A logarithm can be taken of a number if and only if the number is positive. If Statement 1 alone is true, then , being the product of two negative numbers, must be positive, and $\log_{2} \left ( MN \right )$ exists.

Statement 2 is irrelevant; 4 and both yield integers when divided by 2, but $\log_{2} 4 = 2$ and $\log_{2}\left ( -4 \right )$ does not exist.

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Question

Johnny was assigned to write a number in scientific notation by filling the circle and the square in the pattern below with two numbers.

$\bigcirc \times 10 ^{\square }$

Johnny filled in both shapes with numbers. Did he succeed?

Statement 1: He filled in the circle with the number "10".

Statement 2: He filled in the square with a negative integer.

Answer

The number $a \times 10 ^{n}$ is a number written in scientific notation if and only of two conditions are true:

$1 \leq \left | a \right | < 10$

is an integer

By Statement 1 Johnny filled in the circle incorrectly, since it makes .

By Statement 2, Johnny filled in the square correctly, but the statement says nothing about how he filled in the circle; Statement 2 leaves the question open.

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Question

Solve the following rational expression:

$\frac{x^{m-n}10^{n-1}}{10^{(\frac{m}{2}-2)}x^{n}}$

(1)

(2)

Answer

When replacing m=5 in the expression we get:

$\frac{x^{5-n}10^{n-1}}{10^{\frac{1}{2}}x^{n}}$

Therefore statement (1) ALONE is not sufficient.

When replacing m=2n in the expression we get:

$\frac{x^{2n-n}10^{n-1}}{10^{(\frac{2n}{2}-2)}x^{n}}=\frac{x^{n}10^{n-1}}{10^{n-2}x^{n}}=\frac{10^{n-1}}{10^{n-2}}=10$

Therefore statement (2) ALONE is sufficient.

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Question

Is $x^{2} < x$ ?

(1)

(2)

Answer

$x^{2} < x \Leftrightarrow x^{2}-x<0 \Leftrightarrow x(x-1)<0$

From statement 1 we get that and .

So the first term is positive and the second term is negative, which means that is negative; therefore the statement 1 alone allows us to answer the question.

Statement 2 tells us that . If , we have $x^{2}=0.25$ which is less than . Therefore in this case $x^{2}<x$ .

For , we have $x^{2} = 4$ which is greater than . So in this case $x^{2} > x$ .

So statement 2 is insufficient.

Therefore the correct answer is A.

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Question

is a number not in the set $\left{-1, 0, 1 \right}$ .

Of the elements $\left { x^{4}, x^{5}, x^{6}\right }$ , which is the greatest?

Statement 1: is a negative number.

Statement 2: $\left | x\right | > 1$

Answer

Statement 1 alone is inconclusive, as can be demonstrated by examining two negative values of other than .

Case 1: .

Then

$x^{4} = (-2)^{4} = 16$

$x^{5} = (-2)^{5} = -32$

$x^{6} = (-2)^{6} = 64$

$x^{6}$ is the greatest of these values.

Case 2: $x = -\frac{1}{2}$

Then

$x ^{4}=\left ( -\frac{1}{2} \right )^{4} = \frac{1}{16}$

$x ^{5}=\left ( -\frac{1}{2} \right )^{5} = -\frac{1}{32}$

$x ^{6}=\left ( -\frac{1}{2} \right )^{6} = \frac{1}{64}$

$x^{4}$ is the greatest of these values.

Now assume Statement 2 alone. Either or .

Case 1: .

Then $x \cdot x^{4} > 1\cdot x^{4}$ , so $x^{5} > x^{4}$ ; similarly, $x^{6} > x^{5}$ .

$x^{6}$ is the greatest of the three.

Case 2: .

Odd power $x^{5}$ is negative, and even powers $x^{4}$ and $x^{6}$ are positive, so one of the latter two is the greatest. Since , it follows that $x^{2} > 1$ . It then follows that $x^{2} \cdot x^{4} > 1 \cdot x^{4}$ , or $x^{6} > x^{4}$ .

Again, $x^{6}$ is the greatest of the three.

Statement 2 alone is sufficient, but not Statement 1.

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