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02 Jan 2018, 01:31
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
78% (02:35) correct
22%
(02:59)
wrong
based on 262
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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. 3
B. 4
C. 5
D. 6
E. 7
02 Jan 2018, 04:51
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Bunuel
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)
04 Jan 2019, 21:46
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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
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
