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average of A and B is 71; of B and C is 76; of A and C is 79. How to find the average of A,B,C. I know the way of solving this problem using equations.But it takes time.So wanted to know if any other logic can be applied to it to solve faster.

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Wow, Latex-slinging. – copper.hat May 10 '12 at 5:31
An even faster method is: $\frac{2}{3} \begin{bmatrix} 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1\end{bmatrix}^{-1} \begin{bmatrix} 71 \\ 76 \\ 79 \end{bmatrix}$. – copper.hat May 10 '12 at 5:44
up vote 1 down vote accepted

What could be faster than:

$\frac{A+B+C}{3} = \frac{\frac{A+B}{2}+\frac{B+C}{2}+\frac{A+C}{2}}{3} = \frac{71+76+79}{3}$

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Just add the averages and divide by $3$: $$\frac{A+B}2+\frac{A+C}2+\frac{B+C}2=A+B+C\;\;.$$

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Add the three averages, you get $A+B+C$ (because first average is $\frac{A}{2}+\frac{B}{2}$, etc); then divide by $3$. So $$\frac{1}{3}(71+76+79) = \frac{226}{3}.$$

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A trick you might use (if you're doing this in your head) is to "center" the numbers on a guess near the average, and then compute the average offset and add back the guess.

For example, for your three numbers $71, 76, 79,$ I would guess the average is around $75$. So now I take the offsets from my guess: $-4, 1,$ and $4$. Next, take the average of these 3 numbers, which is clearly $1/3$. And last, add back $75$, so the answer becomes $75 \frac{1}{3}$.

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