The average of m, n and p is q and the average of q and r is s. Which

\(\frac{m+n+p}{3}=q \implies m+n+p=3q \quad \quad (1)\)
\(\frac{q+r}{2}=s \implies q+r=2s \quad \quad \quad (2)\)

From (2) we have \(r=2s-q\)
From (1) we have \(r=2s-q=2s-\frac{m+n+p}{3}=\frac{6s-m-n-p}{3}\)

The answer is B
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We are given that the average of m, n and p is q and the average of q and r is s. Thus:

q = (m + n + p)/3

AND

s = (q + r)/2

2s = q + r

r = 2s - q

We can substitute (m + n + p)/3 for q into the second equation and we have:

r = 2s - (m + n + p)/3 = 6s/3 - (m + n + p)/3 = (6s - m - n -p)/3

Answer: B
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B is the answer.

(m+n+p)/3 = q and (q+r)/2=s

r=2s-q

now substitute the value of q in r= 2s-q

and we will get B as the answer

Another approach is the INPUT-OUTPUT approach.

First find some values of m, n, p, q, r and s that satisfy all of the given information.

For example, the following values satisfy all of the given information: m = 1, n = 1, p = 1, q = 1, r = 1 and s = 1
In other words, when m = 1, n = 1, p = 1, q = 1 and s = 1, the value of r is 1

QUESTION: Which of the following is r in terms of m, n, p and s?
Since r = 1, we're looking for an answer choice that yields an OUTPUT of 1, when we INPUT m = 1, n = 1, p = 1, q = 1 and s = 1

A. (6s - m + n + p)/3 = (6 - 1 + 1 + 1)/3 = 7/3. No good - we need an OUTPUT of 1. ELIMINATE A.

B. (6s - m - n - p)/3 = (6 - 1 - 1 - 1)/3 = 1. Perfect. KEEP B.

C. (2s - m - n - p)/3 = (2 - 1 - 1 - 1)/3 = -1/3. No good - we need an OUTPUT of 1. ELIMINATE C.

D. (s + m + n + p)/2 = (1 + 1 + 1 + 1)/2 = 2. No good - we need an OUTPUT of 1. ELIMINATE D.

E. 2s - m + n + p = 2 - 1 + 1 + 1 = 3. No good - we need an OUTPUT of 1. ELIMINATE E.

Answer: B
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