AAPL wrote:Set X contains 10 consecutive integers. If the sum of the 5 smallest members of set X is 265, then what is the sum of the 5 largest members of set X?

A. 290
B. 285
C. 280
D. 275
E. 270

-------ASIDE-------------------------------------------
Notice that we can write the first 10 consecutive integers as follows:
1, 2, 3, 4, 5, (1 + 5), (2 + 5), (3 + 5), (4 + 5), (5 + 5)

So, the sum of the first 5 integers = 1 + 2 + 3 + 4 + 5

The sum of the last 5 integers = (1 + 5) + (2 + 5) + (3 + 5) + (4 + 5) + (5 + 5)
= 1 + 2 + 3 + 4 + 5 + 5 + 5 + 5 + 5 + 5
= 1 + 2 + 3 + 4 + 5 + 25

In other words, the sum of the last 5 numbers = (the sum of the first 5 numbers) + 25

--------NOW ONTO THE QUESTION-----------------
The same approach can be applied to the given information.
The sum of the last 5 numbers = (the sum of the first 5 numbers) + 25
= 265 + 25
= 290

Answer: A

AAPL wrote:Set X contains 10 consecutive integers. If the sum of the 5 smallest members of set X is 265, then what is the sum of the 5 largest members of set X?

A. 290
B. 285
C. 280
D. 275
E. 270

We can see that the 6th, 7th, 8th, 9th and 10th integers are each 5 more than the 1st, 2nd, 3rd, 4th and 5th integers, respectively. Thus, the sum of the 5 largest integers will be 25 more than the sum of the 5 smallest integers. Since the sum of the 5 smallest integers is 265, the sum of the 5 largest integers will be 265 + 25 = 290.

Alternate Solution:

Letting the smallest integer be x, we see that the sum of the first five integers will be:

x + x + 1 + x + 2 + x + 3 + x + 4 = 265

5x + 10 = 265

5x = 255

x = 51

The sum of the 6th through 10th integers is:

x + 5 + x + 6 + x + 7 + x + 8 + x + 9 = 5x + 35 = 5 * 51 + 35 = 255 + 35 = 290.

Answer: A