Bunuel wrote:

If \((\frac{5}{4})^{-n} < 16^{-1}\). What is the least integer value of n?

A. 12

B. 13

C. 14

D. 15

E. 16

\((\frac{5}{4})^{-n} < 16^{-1}\)

=> A = \((\frac{4}{5})^n*2^4 < 1\)

For n = 0, A =16; n = 1, A = 16*0.8.

So, as the value of n increases, A decreases and as n is increased by 1, A decreases by 80 %. We need to look for the value of n for which A >1 and A*80% less than 1.

As the power of 2 is 4, I was looking to test multiple of 4 for n while solving. So, I decided to first test with n = 12; from the list of option. As it will bit simplify A.

For n = 12, A = \((\frac{4}{5})^{12}*2^4\) = \((\frac{4^3}{5^3})^4*2^4\) = \((\frac{64}{125})^4*2^4\)

= \((\frac{128}{125})^4\)

A will be slightly greater than 1. But, definitely 80% of A will be just greater than 0.8. So, if n=13, value of A will be around 0.8 and it is the least integer value for which the inequality satisfies.

Hence, B. I might be a bit lucky.

as the question was definitely not an easy one to crack.

_________________

Give a Kudo, If I deserve it.

Target 750+- 17th June 2018 - Diagnostic Test - Kaplan Premier 2015 - 570
- 10th Aug 2018 - Veritas 1 - 650 (Q51,V29)
- 17th Aug 2018 - Veritas 2 - 670 (Q49,V33)
- 24th Aug 2018 - Veritas 3 - 650 (Q49,V30)
- 1st Sep 2018 - Veritas 4 - 680 (Q51,V33)
- 12th Aug 2018 - Veritas 5 - 690 (Q51,V34)

What do you think? Will I achieve my target score?