July 11, 2010

Quantitative Ability-Question of the Day

QUESTION:-
When one number was erased from the set of natural numbers till N, average of the remaining numbers was 881/29. Which number was erased?

OPTIONS
1) 6
2) 7
3) 8
4) 9
5) 10


SOLUTION:-






Now we consider the average after deletion, 881/29

Since the denominator of the average after deleting one number is 29, value of N can be 30.
If the above fraction has been arrived at after removing the common multiple of 2 in both the denominator and numerator, value of N can be 29 × 2 + 1 = 59.
Similarly it can be 29 × 3 + 1 = 88 or 29 × 4 + 1 = 117 and so on.
Suppose N = 30.
Then sum of all the natural numbers till 30 is 30 × 31/2 = 465, which is less than 881.
So, N = 30 is not possible.
Suppose N = 59
Then sum of all the natural numbers till 59 is 59 × 60/2 = 1770, which is greater than 881 and hence possible.
Suppose the number deleted is x.



∴1770-x = 1762
∴ x = 8
Since we get an integer value of x, our assumption that N = 59 is true.
For higher values of N, x > N, which is not possible.
Hence, option 3.

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