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The four integers a, b, c, and d are such that a < b < c < d, d = 7a

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The four integers a, b, c, and d are such that a < b < c < d, d = 7a [#permalink]

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New post 23 Nov 2017, 05:57
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The four integers a, b, c, and d are such that a < b < c < d, d = 7a, and b - a = c - b = d - c. What is the average (arithmetic mean) of the four integers?

(1) c = 25

(2) c/a = 5
[Reveal] Spoiler: OA

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Re: The four integers a, b, c, and d are such that a < b < c < d, d = 7a [#permalink]

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New post 23 Nov 2017, 06:29
Bunuel wrote:
The four integers a, b, c, and d are such that a < b < c < d, d = 7a, and b - a = c - b = d - c. What is the average (arithmetic mean) of the four integers?

(1) c = 25

(2) c/a = 5



b - a = c - b MEANS b is average of a and c..
c - b = d - c MEANS c is average of b and d..
this means a, b,c and d are equally spaced..

since d=7a, terms are \(a,3a,5a,7a\)......
if we know any term, we can get all terms..
average = 4a

(1) c = 25
5a = 25....a=5
average = 4*5=20
sufficient

(2) c/a = 5
nothing new we kcow this info from statement itself
c/a = 5a/a = 5
insuff

A
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Re: The four integers a, b, c, and d are such that a < b < c < d, d = 7a [#permalink]

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New post 23 Nov 2017, 08:19
a < b < c < d, d = 7a, and b - a = c - b = d - c.. This means that the numbers a, b, c, d are in AP.

Let us assume the numbers to be a, a+x, a+2x, a+3x respectively (x is the common difference here). We are given that a+3x = 7a. This gives us x=2a. So the four numbers a, b, c, d become:- a, 3a, 5a and 7a respectively. Average of these four numbers = 16a/4 = 4a.

So, we need the value of a to find the answer.

(1) c=25. OR 5a=25. This will give us the value of a. Sufficient.

(2) c/a = 5. we have c=5a. So c/a will be 5 only, no matter what the value of a. This doesnt give us any value. Insufficient.

Hence A answer

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Re: The four integers a, b, c, and d are such that a < b < c < d, d = 7a   [#permalink] 23 Nov 2017, 08:19
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