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Roll Number of Jessica

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by pannalal » Sun Sep 24, 2017 9:40 am
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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by [email protected] » 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

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by Matt@VeritasPrep » 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.
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