Question

Let A, B and C be three positive integers such that the sum of A and the mean of B and C is 5. In addition, the sum of B and the mean of A and C is 7. Then the sum of A and B is

• A. 6
• B. 5
• C. 7
• D. 4

Answer 1

The Correct Option is A

The correct answer is (A): \(6\)

Given \(A, B\) and \(C\) are positive integers such that \(A+\frac{B+C}2 = 5\)...(1)

\(B+\frac{A+C}{2} = 7\)...(2)

\((2)-(1) ⇒ \frac{B}{2}-\frac{A}{2} = 2 ⇒ B-A = 4\)

The least value for \(A=1\) in which case \(B=5\). 

Hence \(A+B = 6\)

Answer 2

From the question it is clear that \(A+\frac{B+C}2 = 5\)
\(⇒ 2A+B+C=10.......(1)\)
\(B+\frac{A+C}{2} = 7\)
\(⇒ A+2B+C=14.....(2)\)
\((1)-(2) \)
\(⇒ B=4+A\)
Given positive integers A, B, and C.
In the case that \(A=1, B=5\), and \(C=3.\)
 If \(B=6\) and \(A=2\), then \(C=0\). however, as C is positive, this is untrue.
In the same way, cases are invalid if \(A>2\) and C is negative. \(A+B=6\) as a result.