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by talaangoshtari » Mon Oct 05, 2015 3:25 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 numbers is 35 7/17. What was the number erased?

A. 7
B. 8
C. 9
D. 10
E. 11
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Source: — Problem Solving |

by GMATGuruNY » Mon Oct 05, 2015 4:25 am
talaangoshtari wrote: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 numbers is 35 7/17. What was the number erased?

A. 7
B. 8
C. 9
D. 10
E. 11


After one value is removed:
Since all of the values are INTEGERS, the sum here must be an integer.
Sum = (number)(average).
Since the average = 35 7/17, and the sum must be an integer, the number of integers must be a MULTIPLE OF 17.

For any evenly spaced set, average = median.
After one of the consecutive integers is removed, most of the remaining set will still be evenly spaced.
As a result, the average of the remaining set -- 35 7/17 -- will still be close to the median.
Implication:
The number of integers = 4*17 = 68, with the result that 35 7/17 will be close to the median of the 68 mostly consecutive integers.
Thus:
Sum = (number)(average) = (68)(35 7/17) = 2408.

Original set:
Since 68 integers remain after one of the integers is removed, the original set contains 69 integers.
Sum of the first n positive integers = (n)(n+1)/2.
Thus:
Sum = (69)(70)/2 = 2415.

Removed integer = (original sum) - (sum after one integer is removed) = 2415-2408 = 7.

The correct answer is A.
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by [email protected] » Mon Oct 05, 2015 9:24 am
Hi talaangoshtari,

This question is layered with subtle "clues" as to how you can find the correct answer:

1) We're asked which number was removed from the list and the answers ARE numbers. We can use them against the prompt.
2) The average of the new group of numbers is 35 7/17. The denominator of THAT fraction tells us that the total number of terms in the new list MUST be a multiple of 17:

17, 34, 51, 68, 85, etc.

3) Since we're removing 1 number from a list of consecutive positive integers, we could quickly limit down the possible number of terms in the ORIGINAL list:

If new = 17, old = 18, but the numbers from 1 to 18 would NOT have an average in the mid-30s (it would be much smaller). ELIMINATE this option.
If new = 34, old = 35, but we run into the same problem. The average won't match here either. ELIMINATE this option.
If new = 51, old = 52, same problem here. ELIMINATE this option.

If new = 68, old = 69....HERE we have a group of numbers that might just be what we're looking for.

In the "old group", the average is the "middle term" = 35, which is REALLY close to the new average once we remove a number.

Now, we just have to figure out which of the 5 answers was removed (and that changed the new average to 35 7/17). There are a couple of ways to do this math, but I'm going to use the answers to save myself some steps:

Sum of 1 to 69, inclusive = 69(35) = 2415

Remove Answer B....2415 - 8 = 2407

New average = 2407/68 = 35 27/68

Now, we can compare 27/68 to 7/17

27/68 and 28/68

This is REALLY close, but is not the answer. Eliminate B. The answer is either A or C.

Since 28/68 is a BIGGER fraction than 27/68, we need the sum (of the 68 integers) to be BIGGER...which means we need to remove a SMALLER number.

Final Answer: A

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by vishalwin » Mon Oct 05, 2015 11:28 am
Hello Mitch,

Can you please explain how get this "The number of integers = 4*17 = 68, with the result that 35 7/17 will be close to the median of the 68 mostly consecutive integers."


Thanks & Regards,
Vishal
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