Question

From the following figure, we can say:

$\frac{2x}{3}+\frac{3y}{2}=8\frac{1}{3};\frac{3x}{2}+\frac{2y}{3}=13\frac{1}{3}$

• A. $x=1,y=4$
• B. $x=8,y=3$
• C. $x=6,y=5$
• D. None of these

Answer 1

The correct option is D None of these

$\frac{2x}{3}+\frac{3y}{2}=8\frac{1}{3}\Rightarrow \frac{4x+9y}{6}=\frac{25}{3}\Rightarrow 4x+9y=50$ .......(i)
$\frac{3x}{2}+\frac{2y}{3}=13\frac{1}{3}\Rightarrow \frac{9x+4y}{6}=\frac{40}{3}\Rightarrow 9x+4y=80$ ......(ii)
Adding (i) and (ii). we get,
$13x+13y=130$
$\therefore x+y=10$ ......(iii)
On subtracting (i) from (ii). we get,
$5x-5y=30$
$\therefore x-y=6$ ......(iv)
Again adding (iii) and (iv). we get,
$2x=16\Rightarrow x=8$
On substituting $x=8$ in (iv). we get,
$8-y=6$
$\therefore y=2$
$\therefore x=8,y=2$