Question from OG - 2nd Edition

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Question from OG - 2nd Edition

by shanice » Sat Apr 07, 2012 9:48 am
What is the largest integer n such that 1/2^n > 0.01?

A)5
B)6
C)7
D)10
E)51

Answer is B.

Someone please help with above question. I need detailed explanation.

Thank you in advance.

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by neelgandham » Sat Apr 07, 2012 10:09 am
What is the largest integer n such that 1/2^n > 0.01?
Before answering the question, I would like to rephrase the question.
1/2^n > 0.01
1/2^n > 1/100
Since 2^n is positive, I can multiply both the sides with 100*(2^n) without changing the inequation. The resultant inequation is as below.
2^n < 100. Now, the question can be rephrased to: What is the largest integer n such that 2^n < 100?

A)5 -> 2^n = 2^5 = 32<100, Option A can be the answer
B)6 -> 2^n = 2^6 = 64<100, Option B can be the answer
C)7 -> 2^n = 2^7 = 128>100. Oops, I will stop it here. Since I know that the value of 2^10 and 2^51 is greater than 100. I am, now, sure that n = 6.
D)10
E)51
Anil Gandham
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by Shalabh's Quants » Sun Apr 08, 2012 5:39 am
shanice wrote:What is the largest integer n such that 1/2^n > 0.01?

A)5
B)6
C)7
D)10
E)51

Answer is B.

Someone please help with above question. I need detailed explanation.

Thank you in advance.
Lets make this eqn as 1/2^n>1/100

=>1/2^n>1/10^2

=>2^n<10^2 .... Reciprocating the identity.

=>2^n<(2.5)^2

=>2^n<2^2.5^2

=>2^(n-2)<5^2........ dividing both sides by 2^2.

=>2^(n-2)<25

The closest value but less than 25 for 2^(n-2)is 16 as next value would be 32.

=>2^(n-2)=16=2^4

=> It means n-2=4 => n=6.
Shalabh Jain,
e-GMAT Instructor