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\(\frac{\left(3^{-1}+4^{-1}\right)^{-2}}{\frac{4^2}{7^2}}=\)

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by Gmat_mission » Sun May 05, 2019 10:31 am

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Answers

A

B

C

D

E

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Difficulty—

$$\frac{\left(3^{-1}+4^{-1}\right)^{-2}}{\frac{4^2}{7^2}}=$$ A. \(\frac{12}{9}\)

B. \(\frac{49}{144}\)

C. \(\frac{12}{7}\)

D. \(7\)

E. \(9\)

[spoiler]OA=E[/spoiler]

Source: Veritas Prep
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by Scott@TargetTestPrep » Sun May 12, 2019 6:12 pm
Gmat_mission wrote:$$\frac{\left(3^{-1}+4^{-1}\right)^{-2}}{\frac{4^2}{7^2}}=$$ A. \(\frac{12}{9}\)

B. \(\frac{49}{144}\)

C. \(\frac{12}{7}\)

D. \(7\)

E. \(9\)

[spoiler]OA=E[/spoiler]

Source: Veritas Prep
Simplifying the numerator we have:

(1/3 + 1/4)^-2

(4/12 + 3/12)^-2

(7/12)^-2 = (12/7)^2 = 12^2/7^2

Now we have:

(12^2/7^2)/(4^2/7^2)

12^2/4^2 = (12/4)^2 = 9

Answer: E

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