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If a, b, c, d, e are consecutive positive integers such that a<b<c<d<e  [#permalink]

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If a, b, c, d, e are consecutive positive integers such that a<b<c<d<e, then what is the positive difference between averages of Set (b, c, d, e) and (a, b, c, d)

A) 0.25
B) 0.5
C) 0.75
D) 1
E) 1.5

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Re: If a, b, c, d, e are consecutive positive integers such that a<b<c<d<e  [#permalink]

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GMATinsight wrote:
If a, b, c, d, e are consecutive positive integers such that a<b<c<d<e, then what is the positive difference between averages of Set (b, c, d, e) and (a, b, c, d)

A) 0.25
B) 0.5
C) 0.75
D) 1
E) 1.5

Source: http://www.GMATinsight.com

since a,b,c,d,e are consecutive positive integers

they can also be written as
a,a+1,a+2,a+3,a+4,

average of Set (b, c, d, e) = $$\frac{(4a+10)}{4}$$
average of Set (a, b, c, d) = $$\frac{(4a+6)}{4}$$
difference = $$\frac{(4a+10)}{4}$$ - $$\frac{(4a+6)}{4}$$ =1
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Re: If a, b, c, d, e are consecutive positive integers such that a<b<c<d<e  [#permalink]

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1
Put $$1,2,3,4,5$$ for $$a,b,c,d,e$$ respectively and find the average for each set and then the difference of average.

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Re: If a, b, c, d, e are consecutive positive integers such that a<b<c<d<e  [#permalink]

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GMATinsight wrote:
If a, b, c, d, e are consecutive positive integers such that a<b<c<d<e, then what is the positive difference between averages of Set (b, c, d, e) and (a, b, c, d)

A) 0.25
B) 0.5
C) 0.75
D) 1
E) 1.5

We see that b = a + 1, c = a + 2, d = a + 3 and e = a + 4. The difference between the averages of the two given sets is:

(b + c + d + e)/4 - (a+ b + c + d)/4

(e - a)/4

(a + 4 - a)/4

4/4 = 1

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GMAT 1: 680 Q49 V34 Re: If a, b, c, d, e are consecutive positive integers such that a<b<c<d<e  [#permalink]

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GMATinsight wrote:
If a, b, c, d, e are consecutive positive integers such that a<b<c<d<e, then what is the positive difference between averages of Set (b, c, d, e) and (a, b, c, d)

A) 0.25
B) 0.5
C) 0.75
D) 1
E) 1.5

Source: http://www.GMATinsight.com

since they are consecutive, let a=a
=> b=a+1,c=a+2,d=a+3 and e = a+4

(b+c+d+e)/4 - (a+b+c+d)/4

=> e-a/4
=>a+4-a/4

= 1.

Option D is the answer.
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Re: If a, b, c, d, e are consecutive positive integers such that a<b<c<d<e  [#permalink]

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a<b<c<d<e

They are in ascending order. So plug in values - {1<2<3<4<5)

b+c+d+e/4 = 2+3+4+5/4 = 14/4=7/2
a+b+c+d/4 = 1+2+3+4/4 = 10/4=5/2

Difference = 7/2-5/2 = 2/2 = 1 Re: If a, b, c, d, e are consecutive positive integers such that a<b<c<d<e   [#permalink] 15 Oct 2018, 03:46
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