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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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New post 02 Jan 2018, 00:31
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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, 03:51
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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]

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New post 04 Jan 2019, 20:46
Answer C

1. a+2c=8
2. b+5c=13
3. 3a+4b=50

Multiply equation 1. by 5 and equation 2. by 2 to cancel out c. After multiplying, equation 1. becomes 5a+10c=40
and equation 2. becomes 2b+10c=26
Subtracting these 2 equations: 5a-2b=14 .........(equation A)
Now we have 2 equations to solve:
A) 5a-2b=14
3. 3a+4b=50

Multiply equation A by 2 to equalize b and then add the two equations to give the value of a. Substitute the value of a in earlier equations to find value of b and c. The values are:
a= 6, b=8, c=1

Mean = (a+b+c)/3 = 15/3= 5
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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 03 Jun 2020, 21:53
a+2c = 8
b+5c = 13
3a + 4b = 50
Inserting each equation in one another

3a + 4(13-5c) = 50
3a + 52 - 20c = 50
3a-20c = -2
3(8-2c) - 20c = -2
24 - 6c - 20c = -2
26 = 26c
c=1
a= 6
b = 8

Mean = 15/3 = 5

IMO(C)
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Re: If a + 2c = 8, b + 5c = 13, and 3a + 4b= 50, what is the average (arit   [#permalink] 03 Jun 2020, 21:53

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

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