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tricky VIAC (Variables in the answer choices) question

This topic has 3 expert replies and 1 member reply

tricky VIAC (Variables in the answer choices) question

Post Mon May 08, 2017 6:07 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + 97 + 99, then 1² - 2² + 3² - 4² + 5² - 6² + . . . . . + 97² - 98² + 99² - 100² =

    A) J² - K²
    B) -50(J² - K²)
    C) -K - J
    D) K² - J²
    E) (-J - K)²

    Answer: C

    Source: www.gmatprepnow.com
    Difficulty level: 700

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    Post Wed May 10, 2017 1:05 pm
    Brent@GMATPrepNow wrote:
    If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + 97 + 99, then 1² - 2² + 3² - 4² + 5² - 6² + . . . . . + 97² - 98² + 99² - 100² =

    A) J² - K²
    B) -50(J² - K²)
    C) -K - J
    D) K² - J²
    E) (-J - K)²

    We have several differences of squares hiding in the expression 1² - 2² + 3² - 4² + 5² - 6² + . . . . . + 97² - 98² + 99² - 100²

    1² - 2² + 3² - 4² + 5² - 6² + . . . . . + 97² - 98² + 99² - 100² = 1² - 2² + 3² - 4² + 5² - 6² + . . . . . + 97² - 98² + 99² - 100²
    = (1 - 2)(1 + 2) + (3 - 4)(3 + 4) + (5 - 6)(5 + 6) + . . . . . + (97 - 98)(97 + 98) + (99 - 100)(99 + 100)
    = (-1)(1 + 2) + (-1)(3 + 4) + (-1)(5 + 6) + . . . . . + (-1)(97 + 98) + (-1)(99 + 100)
    = (-1)[(1 + 2) + (3 + 4) + (5 + 6) + . . . . . + (97 + 98) + (99 + 100)]
    = (-1)(1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100)

    IMPORTANT: within the sum, 1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100, we have all of the ODD integers from 1 to 99 inclusive, and we have all of the EVEN integers from 2 to 100 inclusive.

    So, we can say that 1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100 = K + J

    So, we're replace 1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100 with K + J.
    We get: (-1)(1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100) = (-1)(K + J)
    = -K - J

    Answer: C

    Cheers,
    Brent

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    GMAT/MBA Expert

    Post Thu May 11, 2017 8:34 pm
    I'm thinking ...

    1² - 2² + 3² - 4² ± ... + 99² - 100² =>

    (1² + 3² + 5² + ... + 99²) - (2² + 4² + ... + 100²) =>

    (1² + 3² + 5² + ... + 99²) - ((1+1)² + (3+1)² + ... + (99+1)²) =>

    (1² + 3² + 5² + ... + 99²) - (1² + 3² + ... + 99² + 2*1 + 2*3 + ... + 2*99 + 1 + 1 + ... + 1) =>

    (1² + 3² + 5² + ... + 99²) - ((1² + 3² + 5² + ... + 99²) + 2*(1 + 3 + ... + 99) + 50) =>

    -2*(1 + 3 + ... + 99) - 50 =>

    -2*K - 50 =>

    -K - K - 50

    And since J = K + 50, we can write this as

    -K - (K + 50)

    or

    -K - J

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    GMAT/MBA Expert

    Post Thu May 11, 2017 8:37 pm
    We could also cheat with a pattern:

    n² - (n + 1)² =>

    n² - (n² + 2n + 1) =>

    -(2n + 1) =>

    -(n + n + 1)

    for any value of n.

    Since we've got 1² - 2² + 3² - 4² ..., we've really got -(1 + 2) -(3 + 4) .... -(99 + 100), or -1 -2 -3 -4 .... - 99 - 100, or -(1 + 2 + 3 + ... + 100), or -(K + J), or -K - J.

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    hazelnut01 Master | Next Rank: 500 Posts Default Avatar
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    Post Thu May 11, 2017 9:44 pm
    Matt@VeritasPrep wrote:
    I'm thinking ...

    1² - 2² + 3² - 4² ± ... + 99² - 100² =>

    (1² + 3² + 5² + ... + 99²) - (2² + 4² + ... + 100²) =>

    (1² + 3² + 5² + ... + 99²) - ((1+1)² + (3+1)² + ... + (99+1)²) =>

    (1² + 3² + 5² + ... + 99²) - (1² + 3² + ... + 99² + 2*1 + 2*3 + ... + 2*99 + 1 + 1 + ... + 1) =>

    (1² + 3² + 5² + ... + 99²) - ((1² + 3² + 5² + ... + 99²) + 2*(1 + 3 + ... + 99) + 50) =>

    -2*(1 + 3 + ... + 99) - 50 =>

    -2*K - 50 =>

    -K - K - 50

    And since J = K + 50, we can write this as

    -K - (K + 50)

    or

    -K - J
    Hi Matt, How do we know J = K + 50?

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