If the average of a, b and c is m more than the average of a and b

Question

If the average of a, b and c is m more than the average of a and b, how much greater is c than the average of a and b?

A. m
B. 3m/2
C. 3m
D. -m
E. -3m

Bunuel · Accepted Answer

If the average of a, b and c is m more than the average of a and b, how much greater is c than the average of a and b?

A. m
B. 3m/2
C. 3m
D. -m
E. -3m

The average of a, b and c is  \frac{a + b + c}{3}

The average of a and b is  \frac{a + b}{2}

We are told that  \frac{a + b + c}{3}  is m more than  \frac{a + b}{2} . Hence:

\frac{a + b + c}{3} - \frac{a + b}{2} = m

2a + 2b + 2c - 3a -3b = 6m

2c - a - b = 6m

The question asks: how much greater is c than the average of a and b. Thus, we need to find the value of c - (a + b)/2:

c - \frac{a + b}{2} = \frac{2c - a - b}{2} = \frac{ 6m}{2} = 3m

Answer: C.

saynchalk · Answer

I got till 2c > a+b

Lost here, 'm' is throwing me off

honouredone · Answer

I plugged in values: a=2, b=4 and c=6.

So, avg of a + b + c = (2 + 4 + 6)/3 = 4
avg of a + b = 3.

So avg of a + b + c is 1 greater than avg of a + b, so m = 1.

Now C is 6 and avg of a + b is 3. And 6 is 3m ( 3x 1 ) greater. SO, ans is C

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