A PRE-RMO 2014 problem Q14

Let the total number of members be $k$ , total amount of milk $=$ $x$ ounce, and total amount of coffee $=$ $y$ ounce.

We know that $\frac{x}{7}$ $+$ $\frac{2y}{17}$ $=$ $8$

Therefore, $17x$ $+$ $14y$ $=$ $8*7*17$

$56*17$ $=$ $14x$ $+$ $3x$ $+$ $14y$

$56*17$ $=$ $14$ $(x+y)$ $+$ $3x$

$56*17$ $-$ $112k$ $=$ $3x$

$56$ ( $17-2k)$ $=$ $3x$

$\frac{x}{7}$ $=$ $\frac{8}{3}$ $(17-2k)$

It is given $0$ $<$ $\frac{x}{7}$ $<$ $8$

$0$ $<$ $17-2k$ $<$ $3$

$0$ $>$ $k$ $-$ $\frac{17}{2}$ $>$ $\frac{-3}{2}$

$\frac{17}{2}$ $>$ $k$ $>$ $7$

$7$ $<$ $k$ $<$ $8.5$

The only integral value of $k$ satisfying this is $k=8$

Therefore, there are $8$ members in Manjul's family.