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If a + 2c = 8, b + 5c = 13, and 3a + 4b= 50, what is the average (arit

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If a + 2c = 8, b + 5c = 13, and 3a + 4b= 50, what is the average (arit [#permalink]

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Re: If a + 2c = 8, b + 5c = 13, and 3a + 4b= 50, what is the average (arit [#permalink]

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New post 02 Jan 2018, 04:51
Bunuel wrote:
If a + 2c = 8, b + 5c = 13, and 3a + 4b= 50, what is the average (arithmetic mean) of a, b, and c?

A. 3
B. 4
C. 5
D. 6
E. 7


a + 2c = 8 ----> a = 8 - 2c ----->(I)
b + 5c = 13 -----> b = 13 - 5c ------> (II)

Use the values of a & b in 3a + 4b = 50

3 (8 - 2c) + 4 (13 - 5c) = 50

24 - 6c + 52 - 20c = 50

-26c = 50 - 76

c = 1

We can calculate b = 8 & a = 6

Average = \(\frac{(6 + 8 + 1)}{3}\) = 5

(C)
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Re: If a + 2c = 8, b + 5c = 13, and 3a + 4b= 50, what is the average (arit   [#permalink] 02 Jan 2018, 04:51
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