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# There is a set of 10 consecutive integers. If the sum of the largest 5

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Manager
Joined: 07 Jun 2018
Posts: 52
Location: United States
There is a set of 10 consecutive integers. If the sum of the largest 5  [#permalink]

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25 Dec 2018, 14:14
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Difficulty:

25% (medium)

Question Stats:

74% (01:49) correct 26% (01:54) wrong based on 40 sessions

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There is a set of 10 consecutive integers. If the sum of the largest 5 integers in the set is 355, what is the sum of the smallest 5 integers?

a. 315
b. 320
c. 325
d. 330
e. 355
VP
Joined: 07 Dec 2014
Posts: 1197
There is a set of 10 consecutive integers. If the sum of the largest 5  [#permalink]

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25 Dec 2018, 15:54
Jazzmin wrote:
There is a set of 10 consecutive integers. If the sum of the largest 5 integers in the set is 355, what is the sum of the smallest 5 integers?

a. 315
b. 320
c. 325
d. 330
e. 355

terms 1-5 are each 5 less than terms 6-10 respectively
355-5*5=330
D
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 2888
Re: There is a set of 10 consecutive integers. If the sum of the largest 5  [#permalink]

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26 Dec 2018, 10:53

Solution

Given:
• There is a set of 10 consecutive integers
• The sum of the largest 5 integers in the set is 355

To find:
• The sum of the smallest 5 integers

Approach and Working:
As all the 10 numbers are consecutive integers, we can say:
• 6th number – 1st number = 5
• 7th number – 2nd number = 5
• 8th number – 3rd number = 5
• 9th number – 4th number = 5
• 10th number – 5th number = 5

Therefore, sum of last 5 numbers – sum of first 5 numbers = 5 x 5 = 25
Or, sum of first 5 numbers = sum of smallest 5 numbers = 355 – 25 = 330

Hence, the correct answer is option D.

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VP
Joined: 31 Oct 2013
Posts: 1375
Concentration: Accounting, Finance
GPA: 3.68
WE: Analyst (Accounting)
Re: There is a set of 10 consecutive integers. If the sum of the largest 5  [#permalink]

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26 Dec 2018, 13:40
Jazzmin wrote:
There is a set of 10 consecutive integers. If the sum of the largest 5 integers in the set is 355, what is the sum of the smallest 5 integers?

a. 315
b. 320
c. 325
d. 330
e. 355

Consecutive integers***

Sum of largest 5 integers = 335.

If we divide 335 by 5 we will get the 8th one:

335/5 = 71

8th = 71.

So, 5th =68.

4th = 67
3rd=66
2nd= 65
1st = 64.

68 + 67 + 66 + 65 + 64 = 330.

CEO
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Posts: 3879
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Concentration: Sustainability, Marketing
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Re: There is a set of 10 consecutive integers. If the sum of the largest 5  [#permalink]

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26 Dec 2018, 17:00
Jazzmin wrote:
There is a set of 10 consecutive integers. If the sum of the largest 5 integers in the set is 355, what is the sum of the smallest 5 integers?

a. 315
b. 320
c. 325
d. 330
e. 355

let 1st term be x
since all are consective so
2nd=x+1, 3rd x+2 ... so on
so 6th to 10th term sum = 355
or say
5x+35=355
x=64 1st term

so sum of first 5 terms would be
sn = n/2 ( 2a+(n-1)d)
n=5 , a = 64 d= 1
solve we would get
sn= 330 IMO D
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Manager
Joined: 24 Nov 2018
Posts: 109
Location: India
GPA: 3.27
WE: General Management (Retail Banking)
Re: There is a set of 10 consecutive integers. If the sum of the largest 5  [#permalink]

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26 Dec 2018, 18:50
Jazzmin wrote:
There is a set of 10 consecutive integers. If the sum of the largest 5 integers in the set is 355, what is the sum of the smallest 5 integers?

a. 315
b. 320
c. 325
d. 330
e. 355

Let the no. be x-5, x-4, x-3, x-2, x-1, x, x+1, x+2, x+3, x+4,

Sm of largest 5 integeres =5x+(1+2+3+4)=5x+10=355--> 5x=355-10=345

Sum of smallest 5 integers= 5x-(1+2+3+4+5)=5x-15=345-15=330.

IMO, Option D.
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Re: There is a set of 10 consecutive integers. If the sum of the largest 5   [#permalink] 26 Dec 2018, 18:50
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