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Set X contains 10 consecutive integers. If the sum of the

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by AAPL » Sat Jun 02, 2018 6:00 am

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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

The OA is A.

We are given: n, (n+1), (n+2), (n+3), (n+4), (n+5), (n+6), (n+7), (n+8), (n+9) and n+, ...., + (n+4) = 265

n, (n+1), (n+2), (n+3), (n+4) = 5n+10
5n+10=265
5n=255
n=51

(n+5), (n+6), (n+7), (n+8), (n+9) = 5n+35
255+25=290

Answer A.

Has anyone another strategic approach to solve this PS question? Regards!
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by Brent@GMATPrepNow » Sat Jun 02, 2018 6:17 am
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
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by GMATGuruNY » Sat Jun 02, 2018 6:34 am
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
For any set of consecutive integers:
average = median.
sum = (count)(median).

Since the sum of the 5 smallest integers = 265, we get:
265 = (5)(median)
median = 265/5 = 53.

Thus, the 5 smallest integers are 51, 52, 53, 54, and 55, implying that the 5 largest integers are 56, 57, 58, 59, and 60.
Since the median of the 5 largest integers = 58, we get:
sum = (count)(median) = 5*58 = 290.

The correct answer is A.
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by Scott@TargetTestPrep » Tue Jun 12, 2018 10:18 am
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

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