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The average (arithmetic mean) of the weights of a father and
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03 Jan 2020, 06:35
03 Jan 2020, 06:35
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The average (arithmetic mean) of the weights of a father and mother is twice the average of the weights of their 4 children. If the average weight of the 6 family members is 128 pounds, what is the average weight, in pounds, of the 4 children?
A. 64
B. 91
C. 92
D. 94
E. 96
Kudos for the right answer and explanation
A. 64
B. 91
C. 92
D. 94
E. 96
Kudos for the right answer and explanation
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Re: The average (arithmetic mean) of the weights of a father and
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03 Jan 2020, 09:00
03 Jan 2020, 09:00
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Let the Sum of Weights of 4 Children be C and Weight of Father be F and Mother be M
The average (arithmetic mean) of the weights of a father and mother is twice the average of the weights of their 4 children
=> (F+M)/2 = 2 * C/4 (As there are 4 children so average weight = sum/4)
=> F+M = C ...(1)
average weight of the 6 family members is 128 pounds
(F+M+C)/6 = 128
=> F+M+C = 768
Putting F+M = C from (1) we get
C+C = 768
=> C=768/2 = 384
Average weight of 4 children = Sum/4 = C/4 = 384/4 = 96
So, Answer will be E
Hope it helps!
The average (arithmetic mean) of the weights of a father and mother is twice the average of the weights of their 4 children
=> (F+M)/2 = 2 * C/4 (As there are 4 children so average weight = sum/4)
=> F+M = C ...(1)
average weight of the 6 family members is 128 pounds
(F+M+C)/6 = 128
=> F+M+C = 768
Putting F+M = C from (1) we get
C+C = 768
=> C=768/2 = 384
Average weight of 4 children = Sum/4 = C/4 = 384/4 = 96
So, Answer will be E
Hope it helps!
Re: The average (arithmetic mean) of the weights of a father and
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03 Jan 2020, 09:08
03 Jan 2020, 09:08
1
Let age of mother = M and age of father = F
We are given
\(\frac{M+F}{2} = 2 \times (\frac{W_1+W_2+W_3+W_4}{4})\)
\(M + F = 4 \times (\frac{W_1+W_2+W_3+W_4}{4})\)
\(M + F = W_1+W_2+W_3+W_4 \)
Average weight of the 6 family members = 128
\(\frac{M + F + W_1+W_2+W_3+W_4}{6} = 128\)
\(\frac{2 \times (W_1+W_2+W_3+W_4)}{6} = 128\)
\(W_1+W_2+W_3+W_4 = 6 \times 64\)
We need to find the average weight of the 4 children that is \(\frac{W_1+W_2+W_3+W_4}{4} = \frac{6 \times 64}{4}\)
\(= 6 \times 16\)
\(= 96\)
OA, E
We are given
\(\frac{M+F}{2} = 2 \times (\frac{W_1+W_2+W_3+W_4}{4})\)
\(M + F = 4 \times (\frac{W_1+W_2+W_3+W_4}{4})\)
\(M + F = W_1+W_2+W_3+W_4 \)
Average weight of the 6 family members = 128
\(\frac{M + F + W_1+W_2+W_3+W_4}{6} = 128\)
\(\frac{2 \times (W_1+W_2+W_3+W_4)}{6} = 128\)
\(W_1+W_2+W_3+W_4 = 6 \times 64\)
We need to find the average weight of the 4 children that is \(\frac{W_1+W_2+W_3+W_4}{4} = \frac{6 \times 64}{4}\)
\(= 6 \times 16\)
\(= 96\)
OA, E
Carcass wrote:
The average (arithmetic mean) of the weights of a father and mother is twice the average of the weights of their 4 children. If the average weight of the 6 family members is 128 pounds, what is the average weight, in pounds, of the 4 children?
A. 64
B. 91
C. 92
D. 94
E. 96
Kudos for the right answer and explanation
A. 64
B. 91
C. 92
D. 94
E. 96
Kudos for the right answer and explanation
