Set X contains 10 consecutive integers. If the sum of the 5 smallest m

Question

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

BrentGMATPrepNow · Accepted Answer

-------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

matthewsmith_89 · Answer

You could even do this way

x + (x +1) + (x+2) + (x+3) + (x +4) = 265
x = 51

Therefore

56 + 57 + 58 + 59 + 60 = 290

Answer is A

Funsho84 · Answer

I thought about this question using the average...

Knowing there are 10 numbers in the set which are all consecutive, I took the average of 265.

265/5 = 53

Knowing 53 is the average of 265. I counted 2 numbers below and 2 numbers above: 51-52-53-54-55
I then counted 5 numbers above 55 as the next 5 digits to complete the set of 10: 56-57-58-59-60

Adding those numbers up, you get 290.

Answer A

Shiv2016 · Answer

The 10 numbers are:
a,(a+1),(a+2)..................(a+9)

Given:
a+a+1+a+2+a+3+a+4= 265
5a+10= 265
a= 51

The last 5 integers in the set are: 56,57,58,59, and 60

Their sum= 290

emockus · Answer

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"

ScottTargetTestPrep · Answer

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

Abhishek009 · Answer

X = a , (a + 1), (a + 2), (a + 3)................. (a + 8), (a + 9)

a + (a + 1) + (a + 2) + (a + 3) + (a + 4) = 265

Or,  5a + 10 = 265

So,  a = 51

Thus sum of 5 largest no's is

(a + 5) + (a + 6) + (a + 7) + (a + 8) + (a + 9)

Or,  5a + 35

Thus, sum will be  5*51 + 35 = 290 , Answer must be (A)  290

Set X contains 10 consecutive integers. If the sum of the 5 smallest m

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Set X contains 10 consecutive integers. If the sum of the 5 smallest m

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

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My Rewards