Question

If $m-n=16$ and <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $m^2+n^2=400$, then find $mn$.

• A. 72
• B. 78
• C. 84
• D. 88

Answer 1

The correct option is A 72

It is given that, $m-n=16$ ------(i)
And, $m^2$ + <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $n^2=400$ -----(ii)
Now, squaring (i) we get,
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> ${(m-n)}^2$ = ${16}^2$
⇒ $m^2$ + <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $n^2-2mn=256$
⇒ $400-2mn=256$
⇒ $-2mn=256-400$
⇒ $-2mn=-144$
⇒ $mn=72$
Hence, the value of $mn$ is $72$.