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by sanju09 » Thu Jan 14, 2010 5:25 am
A set of consecutive positive integers beginning with 1 is written on the blackboard. A student came along and erased one number. The average of the remaining number is 35 7/17 (read thirty five whole number seven upon seventeen). What was the number erased?
(A) 7
(B) 8
(C) 9
(D) 10
(E) None of these
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by rohan_vus » Thu Jan 14, 2010 6:41 am
The avg with erased integer is 35 7/17..which means the number of remaining integers got be multiple of 17 ..as sum of integers is always an integer

Let consider orig no of integers be n..sum of consecutive integers = n*(n+1)/2 ---(1)
Sum of integers with erased integer = (n-1) *(35 + 7/17)-----(2)

Diff of (1) and (2) is the erased integer -
(1) should be greater than (2) and that can only happen if (n-1) is multiple of 17 and (n-1)/2 should be close to 35..--(3)

With this consideration , lets pick n-1 = 68 as its fairly close to our condition (3)

Sum of orig integers = 69*35
Sum with erased integers = 68*35 + 28
So erased integer is 69*35 - 68*35 - 28 ==> 35-28= 7...

A is the ans
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by ace_gre » Thu Jan 14, 2010 1:14 pm
(1) should be greater than (2) and that can only happen if (n-1) is multiple of 17 and (n-1)/2 should be close to 35..--(3)
Hi Rohan, I am having difficulty in understanding the reasoning here. Can you help explain more on how you pick n-1=68??

Thanks!
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