A container has liquids X, Y, Z in ratio 4 : 6 : 5. 20 liters removed, then 10L of X and 15L of Y added. New Y is 25L more than new X. What was initial total quantity?

Question

Accepted Answer

1. Let total be $T$. Initial amounts: X = $\frac{4}{15}T$, Y = $\frac{6}{15}T=\frac{2}{5}T$.

2. After removing 20 L, scale factor is $(T-20)/T$. So X' = $\frac{4}{15}(T-20)$, Y' = $\frac{2}{5}(T-20)$.

3. Add 10 L to X and 15 L to Y. Then X_new = $\frac{4}{15}(T-20)+10$, Y_new = $\frac{2}{5}(T-20)+15$.

4. Use Y_new = X_new + 25. Solve:
$\frac{2}{5}(T-20)+15=\frac{4}{15}(T-20)+35$. This gives $(2/15)(T-20)=20$, so $T-20=150$ and $T=170$.

FINAL_ANSWER: c

Answer

Answer

Answer