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

Question

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

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sujoykrdatta · Answer

a+b+c+d = 4(4c) => a+b+d = 16c-c = 15c
Since a+b = 9c => d = 6c

Option E

007Phoenix007 · Answer

Given a+b=9c
(a+b+c+d)/4=4c
9c+c+d=16c
10c+d=16c
d=6c

So, Option E is correct

Dereno · Answer

Given that (a+b+c+d)/4 = 4c

Then a+b+c+d = 4*4c = 16c

and a+b = 9c

9c + c + d = 16c

d =  6c

Option E

findingmyself · Answer

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?

Mean= (Sum of items)/(Number of Items)

Mean, 4c= (a+b+c+d)/4 
 Cross multiply and you will get, 16c= a+b+c+d 
 Now substitute the value of a+b in the above equation 
 16c=9c+c+d 
 d=6c

Souradeep17 · Answer

a+b=9c 
average of a+b+c+d=4c
sum of a+b+c+d=16c
a+b+c=9c+c=10c
d=16c-10c=6c (option E)

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