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

- $6$
- $5$
- $7$
- $4$

1 vote

Given that, $ A>0, B>0, C>0 $

And, $ A + \frac{B+C}{2} = 5 $

$ \Rightarrow 2A + B + C = 10 \quad \longrightarrow (1) $

And, $ B + \frac{A+C}{2} = 7 $

$ \Rightarrow 2B + A + C = 14 \quad \longrightarrow (2) $

Subtract equation $(2)$ from $(1),$ we get

$\require{cancel} \begin{array} {cccc} 2A + B + \cancel{C} = 10 \\ A + 2B + \cancel{C} = 14 \\\hline \boxed{A – B = -4} \end{array}$

$\textbf{Case 1:}$ The least value of $B$ is $1.$

Then, $ A – 1 =\; – 4 $

$ \Rightarrow \boxed{ A = 5} $

So, $ A + B = 5 + 1 = 6.$

$\textbf{Case 2:}$ The least value for $A$ is $1.$

Then, $ 1 – B = \;– 4 $

$ \Rightarrow \boxed{ B = 5} $

So, $ A + B = 1 + 5 = 6.$

$\therefore$ The sum of $A$ and $B$ is $6.$

Correct Answer $: \text{A}$

And, $ A + \frac{B+C}{2} = 5 $

$ \Rightarrow 2A + B + C = 10 \quad \longrightarrow (1) $

And, $ B + \frac{A+C}{2} = 7 $

$ \Rightarrow 2B + A + C = 14 \quad \longrightarrow (2) $

Subtract equation $(2)$ from $(1),$ we get

$\require{cancel} \begin{array} {cccc} 2A + B + \cancel{C} = 10 \\ A + 2B + \cancel{C} = 14 \\\hline \boxed{A – B = -4} \end{array}$

$\textbf{Case 1:}$ The least value of $B$ is $1.$

Then, $ A – 1 =\; – 4 $

$ \Rightarrow \boxed{ A = 5} $

So, $ A + B = 5 + 1 = 6.$

$\textbf{Case 2:}$ The least value for $A$ is $1.$

Then, $ 1 – B = \;– 4 $

$ \Rightarrow \boxed{ B = 5} $

So, $ A + B = 1 + 5 = 6.$

$\therefore$ The sum of $A$ and $B$ is $6.$

Correct Answer $: \text{A}$