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If a + b = 9c, and the average (arithmetic mean) of a, b, c, and d in

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Joined: 02 Sep 2009
Posts: 50730
If a + b = 9c, and the average (arithmetic mean) of a, b, c, and d in  [#permalink]

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New post 21 Dec 2017, 03:14
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

88% (00:55) correct 13% (01:28) wrong based on 28 sessions

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Re: If a + b = 9c, and the average (arithmetic mean) of a, b, c, and d in  [#permalink]

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New post 21 Dec 2017, 05:00
Bunuel wrote:
If a + b = 9c, and the average (arithmetic mean) of a, b, c, and d in terms of c is 4c, then what is d in terms of c?

A. c/3
B. 2c
C. 5c/2
D. 5c
E. 6c


a + b = 9c,

Also, (a+b+c+d)/4 = 4c

i.e. (a+b+c+d) = 16c

i.e. (9c+c+d) = 16c

i.e. 10c+d = 16c

i.e. d = 6c

Answer: option E
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Re: If a + b = 9c, and the average (arithmetic mean) of a, b, c, and d in  [#permalink]

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New post 21 Dec 2017, 09:55
Bunuel wrote:
If a + b = 9c, and the average (arithmetic mean) of a, b, c, and d in terms of c is 4c, then what is d in terms of c?

A. c/3
B. 2c
C. 5c/2
D. 5c
E. 6c


Since the equations in the question look simple (no powers or roots), we'll go for an exact (Precise) approach.
Writing down the details of our question as equations gives:
a + b = 9c
(a + b + c + d) / 4 = 4c
Substituting equation (1) into (2) and multiplying by 4 gives 9c + c + d = 16c
Then d = 6c.

(E) is our answer.
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Re: If a + b = 9c, and the average (arithmetic mean) of a, b, c, and d in  [#permalink]

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New post 21 Dec 2017, 10:16
(a+b+c+d)/4=4c
9c+c+d=16c
So d=6c
E answer


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Re: If a + b = 9c, and the average (arithmetic mean) of a, b, c, and d in &nbs [#permalink] 21 Dec 2017, 10:16
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