Finding the error in A=15-B, C=B+9, D=B+21

[Ilikebugs]

View attachment 6392 I know that its well above average.

So I got A= 15-B, C=B+9 and D=B+21, but I think I made a mistake somewhere

 

[Evgeny.Makarov]

Let's denote the four numbers by $a$, $b$, $c$ and $d$. Then we have the following system.
\[
\left\{
\begin{aligned}
a+\frac13b+\frac13c+\frac13d=25\\
\frac13a+b+\frac13c+\frac13d=37\\
\frac13a+\frac13b+c+\frac13d=43\\
\frac13a+\frac13b+\frac13c+d=51\\
\end{aligned}
\right.
\]
Multiply each equation by 3 and add them. Recall that you need to find $\dfrac{a+b+c+d}{4}$.

 

[Ilikebugs]

6(A+B+C+D)=468

(A+B+C+D)=78

Average is 19.5?

 

[Evgeny.Makarov]

You are right.