Question

Product $M$ is produced by mixing chemical $X$ and chemical $Y$ in the ratio $5:4$. Chemical $X$ is prepared by mixing two raw materials, $A$ and $B$, in the ratio $1:3$. Chemical $Y$ is prepared by mixing raw materials, $B$ and $C$, in the ratio $2:1$. Then the final mixture is prepared by mixing $864$ units of product $M$ with water. If the concentration of the raw material $B$ in the final mixture is $50%$, how much water had been added to product $M$?

• A. 328 units
• B. 368 units
• C. 392 units
• D. 616 units
• E. None of the above

Hint:

In nested mixtures, first compute the target component’s fraction at each level, then weight by the outer mixing ratio. Finally apply the given percentage condition on the total mixture.

Answer 1

The Correct Option is B

Step 1: Composition of $X$ and $Y$ in terms of $A,B,C$.
For $X$ (ratio $A:B=1:3$) $\Rightarrow$ fraction of $B$ in $X=\dfrac{3}{1+3}=\dfrac{3}{4}$.
For $Y$ (ratio $B:C=2:1$) $\Rightarrow$ fraction of $B$ in $Y=\dfrac{2}{2+1}=\dfrac{2}{3}$. Step 2: Fraction of $B$ in product $M$ (mix of $X:Y=5:4$).
\[ \text{Fraction of }B\text{ in }M =\frac{5\cdot \frac{3}{4}+4\cdot \frac{2}{3}}{5+4} =\frac{\frac{15}{4}+\frac{8}{3}}{9} =\frac{\frac{45+32}{12}}{9} =\frac{\frac{77}{12}}{9} =\frac{77}{108}. \] Step 3: Amount of $B$ present in $864$ units of $M$.
\[ B\text{-amount from }864\text{ units of }M =864\times \frac{77}{108}=8\times 77=616\ \text{units}. \] Step 4: Mix with water and use the $50%$ condition.
Let water added $=w$ units. Final mixture $=(864+w)$ units and $B$-concentration is $50%$: \[ \frac{616}{864+w}=0.5\ \Rightarrow\ 616=\frac{864+w}{2}\ \Rightarrow\ 1232=864+w\ \Rightarrow\ w=368. \] \[ \boxed{368\ \text{units}} \]