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

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Sajjad1994
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Set X contains 10 consecutive integers. If the sum of the 5 smallest m [#permalink] New post   13 Mar 2017, 10:47
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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
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A
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BrentGMATPrepNow
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Re: Set X contains 10 consecutive integers. If the sum of the 5 smallest m [#permalink] New post   13 Mar 2017, 11:24
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SajjadAhmad 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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Re: Set X contains 10 consecutive integers. If the sum of the 5 smallest m [#permalink] New post   14 Mar 2017, 13:54
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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
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Re: Set X contains 10 consecutive integers. If the sum of the 5 smallest m [#permalink] New post   14 Mar 2017, 17:41
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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
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Re: Set X contains 10 consecutive integers. If the sum of the 5 smallest m [#permalink] New post   03 Apr 2017, 04:10
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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
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Re: Set X contains 10 consecutive integers. If the sum of the 5 smallest m [#permalink] New post   03 Apr 2017, 08:44
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"
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Re: Set X contains 10 consecutive integers. If the sum of the 5 smallest m [#permalink] New post   12 Jun 2018, 11:16
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SajjadAhmad 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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Re: Set X contains 10 consecutive integers. If the sum of the 5 smallest m [#permalink] New post   01 Feb 2024, 07:41
Sajjad1994 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

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

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

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) \(290\)
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Re: Set X contains 10 consecutive integers. If the sum of the 5 smallest m [#permalink]
01 Feb 2024, 07:41
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