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OG Negative Exponents Q

This topic has 4 expert replies and 0 member replies
AbeNeedsAnswers Master | Next Rank: 500 Posts Default Avatar
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OG Negative Exponents Q

Post Wed Jul 19, 2017 10:04 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    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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    Post 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

    x = 3.

    The correct answer: C

    Hope this helps!

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    GMAT/MBA Expert

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

    (2⁻¹⁴ + 2⁻¹⁵ + 2⁻¹⁶ + 2⁻¹⁷)/5 = x * 2⁻¹⁷

    Let's multiply both sides by 5:

    (2⁻¹⁴ + 2⁻¹⁵ + 2⁻¹⁶ + 2⁻¹⁷) = 5x * 2⁻¹⁷

    Then divide both sides by 2⁻¹⁷:

    2³ + 2² + 2 + 1 = 5x

    which simplifies to

    3 = x

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    Post 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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    Post Tue Jul 25, 2017 11:04 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
    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)

    We pull out the common factor 2^-17 from each term on the left side of the equation:

    2^-17[2^3 + 2^2 + 2^1 + 1] = (k)(2^-17)(5)

    We eliminate 2^-17 from each side of the equation:

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

    15 = 5k

    3 = k

    Answer: C

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