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OG Negative Exponents Q

Expert replies
by AbeNeedsAnswers » Wed Jul 19, 2017 10:04 pm

Timer

00:00

Answers

A

B

C

D

E

Stats

Difficulty

The value of [2^(−14) + 2^(−15) + 2^(−16) + 2^(−17)]/ 5 is how many times the value of 2^(−17)?

A. 3/2
B. 5/2
C. 3
D. 4
E. 5

C
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Source: — Problem Solving |

by Jay@ManhattanReview » Thu Jul 20, 2017 12:16 am
AbeNeedsAnswers wrote:The value of [2^(−14) + 2^(−15) + 2^(−16) + 2^(−17)]/ 5 is how many times the value of 2^(−17)?

A. 3/2
B. 5/2
C. 3
D. 4
E. 5

C
Say the value of [2^(−14) + 2^(−15) + 2^(−16) + 2^(−17)]/ 5 is x times the value of 2^(−17).

We then need to get the value of x.

=> [2^(−14) + 2^(−15) + 2^(−16) + 2^(−17)] / 5 = 2^(−17)*x

Taking 2^(−17) common from the RHS. We get,

2^(−17)[[2^(3) + 2^(2) + 2^(1) + 1] / 5] = 2^(−17)*x

=> [2^(3) + 2^(2) + 2^(1) + 1] / 5 = x; 2^(−17) gets cancelled

= [8 + 4 + 2 + 1] /5 = x

= [15] /5 = x

[spoiler]x = 3[/spoiler].

The correct answer: C

Hope this helps!

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by Matt@VeritasPrep » Sun Jul 23, 2017 5:19 pm
Let's rephrase the question first. We're asked to solve for x in the following:

(2�¹� + 2�¹� + 2�¹� + 2�¹�)/5 = x * 2�¹�

Let's multiply both sides by 5:

(2�¹� + 2�¹� + 2�¹� + 2�¹�) = 5x * 2�¹�

Then divide both sides by 2�¹�:

2³ + 2² + 2 + 1 = 5x

which simplifies to

3 = x
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by Brent@GMATPrepNow » Mon Jul 24, 2017 11:20 am
AbeNeedsAnswers wrote:The value of [2^(−14) + 2^(−15) + 2^(−16) + 2^(−17)]/ 5 is how many times the value of 2^(−17)?

A. 3/2
B. 5/2
C. 3
D. 4
E. 5

C
Here's my solution

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Brent Hanneson - Creator of GMATPrepNow.com
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Re: OG Negative Exponents Q

by Scott@TargetTestPrep » Mon Mar 02, 2020 5:46 am
AbeNeedsAnswers wrote:
Wed Jul 19, 2017 10:04 pm
The value of [2^(−14) + 2^(−15) + 2^(−16) + 2^(−17)]/ 5 is how many times the value of 2^(−17)?

A. 3/2
B. 5/2
C. 3
D. 4
E. 5

C
We can create the following equation in which k is an integer:

[2^(-14) + 2^(-15) + 2^(-16) + 2^(-17)] / 5 = (k)(2^-17)

[2^(-14) + 2^(-15) + 2^(-16) + 2^(-17)] = (k)(2^-17)(5)

Let’s multiply each side of the equation by 2^17:

2^17[2^(-14) + 2^(-15) + 2^(-16) + 2^(-17)] = (2^17)(2^-17)(k)(5)

2^3 + 2^2 + 2^1 + 1 = 5k

(8 + 4 + 2 + 1) = 5k

15 = 5k

3 = k

Answer: C

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