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a^n^2+n-1/^a(n-1)(n+2)=?

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by Max@Math Revolution » Fri Feb 15, 2019 12:02 am

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Answers

A

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C

D

E

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Difficulty—

[GMAT math practice question]

$$\frac{a^{n^2+n-1}}{a^{\left(n-1\right)\left(n+2\right)}}=?$$

1) n=5
2) a=5
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Source: — Data Sufficiency |
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by deloitte247 » Sun Feb 17, 2019 6:00 am
Statement 1
n = 5
$$\frac{a^{n^2+n-1}}{a^{\left(n-1\right)\left(n-2\right)}}=\frac{a^{5^2+5-1}}{a^{\left(5-1\right)\left(5+2\right)}}$$
$$=\frac{a^{25+5-1}}{a^{\left(4\right)\left(7\right)}}$$
$$=\frac{a^{29}}{a^{\left(7\right)}}$$
$$=\frac{a^{29}}{a^{28}}=a^{29-28}=a^1$$
value of a is unknown, hence statement 1 is INSUFFICIENT.

Statement 2
when a = 5
$$\frac{a^{n^2+n-1}}{a^{\left(n-1\right)\left(n+2\right)}}=\frac{5^{n^2+n-1}}{5^{\left(n-1\right)\left(n+2\right)}}$$
The value of n is unknown, hence statement 2 is INSUFFICIENT.

Both statements together
n = 5 $$\frac{a^{n^2+n-1}}{a^{\left(n-1\right)\left(n+2\right)}}=\frac{5^{5^2+5-1}}{5^{\left(5-1\right)\left(5+2\right)}}$$
$$=\frac{5^{25+4}}{5^{\left(4\right)\left(7\right)}}$$
$$=\frac{5^{29}}{5^{28}}=5^{29-28}=5^1=5$$
both statements combined together are SUFFICIENT.

$$answer\ is\ Option\ C$$
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by Max@Math Revolution » Sun Feb 17, 2019 5:14 pm
=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

$$\frac{a^{n^2+n-1}}{a^{\left(n-1\right)\left(n+2\right)}}=\frac{a^{n^2+n-1}}{a^{n^2+n-2}}=a^{n^2+n-1-n^2-n+2}=a^{2-1}=a$$

Asking for the value of $$\frac{a^{n^2+n-1}}{a^{\left(n-1\right)\left(n+2\right)}}$$

is equivalent to asking for the value of a. Thus, condition 2) is sufficient, but condition 1) is not sufficient as it gives us no information about the value of a.

Therefore, the answer is B.
Answer: B
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