CEMC Cayley, 2023

Algebra Difficulty 1.7 Junior Find the answer Canada

The average of $a$ , $b$ and $c$ is 16. The average of $c$ , $d$
and $e$ is 26. The average of $a$ , $b$ , $c$ , $d$ , and $e$ is 20. The value of $c$ is

Pick one

Solution

Since the average of $a$ ,
$b$ and $c$ is 16, then $\dfrac{a+b+c}{3} = 16$ and so $a+b+c=3 \times 16 = 48$ .

Since the average of $c$ , $d$ and $e$ is 26, then $\dfrac{c+d+e}{3} = 26$ and so $c+d+e=3 \times 26 = 78$ .

Since the average of $a$ , $b$ , $c$ , $d$ , and $e$ is 20, then $\dfrac{a+b+c+d+e}{5} = 20$ .

Thus, $a+b+c+d+e = 5 \times 20 = 100$ .

We note that $(a+b+c)+(c+d+e) = (a+b+c+d+e) + c$ and so $48 + 78 = 100 + c$ which gives $c = 126 - 100 = 26$ .

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