Exponents and Roots Quiz

Recall the rule \(\sqrt{\frac{A}{B}} = \frac{\sqrt{A}}{\sqrt{B}}\).

So, we get, \(n = \sqrt{\frac{81}{16}} = \frac{\sqrt{81}}{\sqrt{16}} = \frac{9}{4}\).

And don't forget that you need to take the square root of \(n\): \(\frac{\sqrt{9}}{\sqrt{4}}=\frac{3}{2}\).

The correct answer is B.

You can take a brute force approach: \(\sqrt{36+108} = \sqrt{144} = 12.\)

Or you can get fancy:\(\sqrt{9(4+12)} = \sqrt{(9)(16)} = \sqrt{9}\sqrt{16} = (3)(4) = 12\).

The correct answer is E.

I hope this was a pretty straightforward computation:

\(\sqrt{30 + 42} = \sqrt{72} = \sqrt{(36)(2)} = \sqrt{36}\times{\sqrt{2}} = 6\sqrt{2}\). The correct answer is D.

When we divide exponential expressions with the same base we take the difference of the exponents and keep the base:

\(\frac{A^x}{A^y} = A^{x-y}\).

If we apply this rule to the problem, we get \((0.2)^{5-2} = (0.2)^3 = (0.2)(0.2)(0.2) = 0.008\).

The correct answer is C.

GOOD ENOUGH, simplify the roots: \(\sqrt{63}+\sqrt{28} = \sqrt{(9)(7)}+\sqrt{(4)(7)} = \sqrt{9}\sqrt{7}+\sqrt{4}\sqrt{7} = 3\sqrt{7}+2\sqrt{7} = 5\sqrt{7}\)

BETTER, use estimation: \(63\approx{64}\) and \(28\approx{25}\), so \(\sqrt{63}+\sqrt{28} \approx\sqrt{64}+\sqrt{25} = 8+5 = 13\). Estimation is often the most efficient way to solve a problem because numerical answers to GMAT questions are always in order from least to greatest or greatest to least. Here it's clear the answer choices B, C, D, and E are all larger than 13, so the correct answer is A

Recall the rule \(A\sqrt{x}+B\sqrt{x}=(A+B)\sqrt{x}\).

So, \(\sqrt{5}+3\sqrt(5)=4\sqrt{5}\).

Therefore, \({(4\sqrt{5})}^2=16(5)=80\).

The correct answer is D.

Statements (1) and (2) alone allow for an infinite number of possible values for \(n\).

In particular \(n = 1\) and \(n = -1\) work for both statements. If \(n = 1\) the answer to the question is NO. If \(n = -1\) the answer to the question is YES.

The correct answer is E.

First, convert all values to scientific notation: \(40,000 = {4.0}\times{{10}^4}\) and \(3,000,000 = {3.0}\times{{10}^6}\). If \(x = 4\), then \(A\) is less than \(40,000\), and if \(x = 6\), then \(A\) is greater than \(3,000,000\).

So, we know \(x = 5\). Thus, statement (1) is sufficient. Statement (2) may look a bit intimidating, however, this is a data SUFFICIENCY question, and given some time and/or a calculator, we COULD find \(A\). So, statement (2) is also sufficient.

The correct answer is D.

Statement (1) is insufficient. Consider the following examples \(x = \sqrt[3]{2}\), and \(x = 2\). In the former \(x\) is not an integer, but \({x}^3\) is an integer. In the latter example \(2\) is an integer and \({2}^3\) is also an integer.

A little algebra makes it clear that statement (2) is sufficient :

\(\sqrt[3]{x} = integer\)

\({\sqrt[3]{x}}^3 = {integer}^3\)

\(x = {integer}^3\).

So, \(x\) is the cube of an integer and is clearly an integer.

The correct answer is B.

I tell all of my students that it pays to know powers of two. When bacteria grow, or grasshoppers multiply, or radio isotopes decay on the GMAT, they almost always double or get cut in half.

I recommend knowing powers of two up to \({2}^{10}\). Here it pays of in a different context.

\(0.001 = \frac{1}{1000}\) and \({2}^{10} = 1024\). And \(\frac{1}{1024} < \frac{1}{1000}[/latex].

The correct answer is E.

The fact that the sum of \(a\) and \(b\) is positive doesn't tell us anything about the relative values of \(a\) and \(b\).

So statement (1) is insufficient. If an exponential expression is negative we know that the base must be negative and the exponent odd. However, we still don't know anything about the relative values of \(a\) and \(b\). For example, both \({-2}^{-3}\) and \({-2}^3\) are negative: \(-2\) and \(-8\), respectively. In the former case -3 < -2, so a < b. In the latter case 3 > -2, so a > b.

Combining the two statements, we know that \(b\) must be negative, and because the sum of \(a\) and \(b\) must t be positive, \(a\) is positive. Therefore, \(a>b\).

The correct answer is C.

Statement (1) is insufficient.

Clearly, if \({2}^x = {2}^y\) ,then \(x = y\), but there are several scenarios where \({A}^x = {A}^y\) does not mean that \(x = y\). For instance if \(A = 1\), then \(x\) and \(y\) can take any value because 1 to any power is always 1.

Statement (2) doesn't eliminate the two possibilities mentioned above as 1 and 2 are both positive integers. Further, statement (2) doesn't give us any information that we can combine with statement (1), so statements (1) and (2) together are insufficient.

The correct answer is E.

Statement (1) is sufficient.

Given some time and/or a calculator we COULD find [latex]{3.7}^7\).

Statement (2) is not sufficient. It would be sufficient if we knew that \(x\) is an integer, but no such luck.

The correct answer is A.

If 0 < A < 1, roots and powers of A are counter intuitive. For numbers in this range roots are larger and exponents are smaller. Consequently, a < c, and b < a. Therefore, b < a < c.

The correct answer is B.

A GMAT Classic! The units digit of any power of \(55\) (or \(555\) or \(1,000,000,005\) or any integer ending in \(5\)) has a units digit of \(5\). The units digit sequence for powers of any integer ending in \(7\) is \(7, 9, 3, 1, 7, 9, 3, 1\), and so on - repeating the same pattern until the heat death of the universe. When we divide \(37\) by \(4\) we get a quotient of \(9\) and a remainder of \(1\): i.e. \(37 = (9)(4) + 1\). So, the patter repeats nine times and travels one more step landing on seven. The sum of an integer ending in \(5\) and an integer ending in \(7\) has a units digit of \(2\). The correct answer is B.