Question

A $200$-litre container initially has $x$ litres of milk (only milk). $6$ L of milk is removed and $5$ L water is added. Then $6$ L of the mixture is replaced with $6$ L water. Finally, milk and water are in the ratio $9:16$. Find $x$.

• A. 6 litres
• B. 9 litres
• C. 15 litres
• D. 16 litres

Hint:

In replacement problems, track fractions just before the draw. Write milk and water after each step explicitly and use the final ratio to solve for the initial quantity.

Answer 1

The Correct Option is C

After the first operation: milk $=x-6$, water $=5$, total $=x-1$.
Before the second replacement, the milk fraction is $\dfrac{x-6}{x-1}$ and the water fraction is $\dfrac{5}{x-1}$. Removing $6$ L of mixture reduces milk by $6\cdot \dfrac{x-6}{x-1}$ and water by $6\cdot \dfrac{5}{x-1}$. Then $6$ L water is added back.
[2mm] Final amounts: \[ \text{Milk}=(x-6)-\frac{6(x-6)}{x-1}=(x-6)\Big(1-\frac{6}{x-1}\Big)=\frac{(x-6)(x-7)}{x-1}, \] \[ \text{Water}=5-\frac{30}{x-1}+6=11-\frac{30}{x-1}. \] Given the final ratio $9:16$, \[ \frac{\frac{(x-6)(x-7)}{x-1}}{\,11-\frac{30}{x-1}\,}=\frac{9}{16}. \] Solving yields $x\approx 14.79$ L, i.e.\ $\boxed{15\ \text{litres}}$ to the nearest litre, matching the option.