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R = ?

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AbeNeedsAnswers Master | Next Rank: 500 Posts Default Avatar
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R = ?

Post Mon Jul 03, 2017 8:39 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    If p ≠ 0 and p - (1-p^2)/p = r/p, then r =

    A) p + 1

    B) 2p - 1

    C) p^2+ 1

    D) 2p^2 - 1

    E) p^2 + p - 1

    D

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    Post Mon Jul 03, 2017 9:18 pm
    AbeNeedsAnswers wrote:
    If p ≠ 0 and p - (1-p^2)/p = r/p, then r =

    A) p + 1

    B) 2p - 1

    C) p^2+ 1

    D) 2p^2 - 1

    E) p^2 + p - 1
    Let p=-1.

    Substituting p=-1 into p - (1-p²)/p = r/p, we get:
    -1 - [1 - (-1)²]/-1 = r/(-1)
    -1 - 0 = -r
    -1 = -r
    r = 1.

    Since the question stem asks for the value of r, our target value is r=1.
    Now plug p=-1 into the answers to see which yields our target value of 1.
    Only D works:
    2p² - 1 = 2(-1)² - 1 = 1.

    The correct answer is D.

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    Post Mon Jul 03, 2017 9:26 pm
    AbeNeedsAnswers wrote:
    If p ≠ 0 and p - (1-p^2)/p = r/p, then r =

    A) p + 1

    B) 2p - 1

    C) p^2+ 1

    D) 2p^2 - 1

    E) p^2 + p - 1

    D
    We have p - (1-p^2)/p = r/p

    => [p^2 -(1-p^2)]/p = r/p; taking LCM of LHS

    Since p ≠ 0, we can cancel p from both the sides.

    => p^2 -(1-p^2) = r

    => r = p^2 -1 + p^2

    r = 2p^2 -1

    The correct answer: D

    Hope this helps!

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    Post Tue Jul 04, 2017 8:12 am
    p - (1 - p²)/p = r/p

    Start by multiplying both sides by p:
    p[p - (1 - p²)/p = r/p]
    p² - (1 - p²) = r
    p² - 1 + p² = r
    2p² - 1 = r

    The answer is D.

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    Post Wed Jul 05, 2017 1:01 am
    p - (1 - p²)/p = r/p

    multiply both sides by p:

    p² - (1 - p²) = r

    do the subtraction on the left side:

    2p² - 1 = r

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    Post Wed Jul 12, 2017 4:27 pm
    AbeNeedsAnswers wrote:
    If p ≠ 0 and p - (1-p^2)/p = r/p, then r =

    A) p + 1

    B) 2p - 1

    C) p^2+ 1

    D) 2p^2 - 1

    E) p^2 + p - 1
    Let’s simplify the given equation p - (1-p^2)/p = r/p by multiplying both sides by p:

    p^2 - (1-p^2) = r

    p^2 - 1 + p^2 = r

    2p^2 - 1 = r

    Answer: D

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