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Roll Number of Jessica

This topic has 2 expert replies and 0 member replies

Top Member

Roll Number of Jessica

Post Sun Sep 24, 2017 9:40 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    There are 'n' students in a grade 6 of a school. Each student is given unique roll number from 1 to 'n'. On a particular day, each student was present except Jessica. The sum of roll numbers of students present is 458. The roll number of Jessica is

    A) 7
    B) 8
    C) 9
    D) 10
    E) None of the above

    OA will be given later.

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    Post Sun Sep 24, 2017 10:22 am
    Hi pannalal,

    Since we're dealing with a sum from 1 to N, we can use 'bunching' to save some time. To start, we need to have a sum that's relatively close to 458 and we can do a couple of 'brute force' calculations to hone in on what N could be:

    IF... N = 20
    the numbers 1 to 20 can be 'bunched' into 10 groups of 21...
    (10)(21) = 210. That's not close enough to 458 though, so we need N to be HIGHER...

    IF... N = 30
    the numbers 1 to 30 can be 'bunched' into 15 groups of 31...
    (15)(31) = 465. That's fairly close to 458, so 30 is almost certainly the total number of students.
    465 - 458 = 7, and that result 'fits' the restrictions of the prompt, so Jessica's assigned number must be 7.

    Final Answer: A

    GMAT assassins aren't born, they're made,
    Rich

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    Thanked by: pannalal

    GMAT/MBA Expert

    Post Tue Sep 26, 2017 11:08 pm
    Here's another approach that doesn't involve guesstimation:

    The sum from 1 to n = n * (n + 1) / 2, so n * (n + 1)/2 - j = 458.

    Since n * (n + 1)/2 > 458, n ≥ 30. But n ≥ j, and if n > 30, then n * (n + 1)/2 - n > 458, meaning there is no solution j. So 30 ≥ n.

    Since n ≥ 30 and 30 ≥ n, n must be 30, and j must be 7.

    Thanked by: pannalal
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