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Three vessels whose capacities are in the ratio of 3 : 2 : 1 are completely filled with milk mixed with water. The ratio of milk and water in the mixture of vessels are 5 : 2, 4 : 1 and 4 : 1 respectively. Taking $\large\frac{1}{3}$ of first, $\large\frac{1}{2}$ of second and $\large\frac{1}{7}$ of third mixture, a new mixture kept in a new vessel is prepared. The percentage of water in the new mixture is.

A. 32% B. 28% C. 30% D. 24% Answer: Option D
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Solution(By Apex Team)

$\begin{array}{l}3:2:1\\ \text{ M:W }\quad\text{ T Mixture }\\ \mathrm{V}-1\rightarrow(5:2=7)_{\times5}\\ \mathrm{V}-2\rightarrow(4:1=5)_{\times7}\\ \mathrm{V}-3\rightarrow(4:1=5)_{\times7}\\ \text{Equate the mixture}\\ \text{ M:W T Mixture }\\ (\mathrm{V}-1)\rightarrow25:10\quad=35\\ (\mathrm{V}-2)\rightarrow28:7=35\\ (\mathrm{V}-3)\rightarrow28:7=35\\ \text{ Capacities }\quad\mathrm{M}:\mathrm{W}=\text{ Total Mix. }\\ (\mathrm{V}-1)\times3\rightarrow75:30=105\\ (\mathrm{V}-2)\times2\rightarrow56:14=70\\ (\mathrm{V}-1)\times1\rightarrow28:7=35\\ \end{array}$ $\begin{aligned}&\text{Water taken out}\\ &\Rightarrow\frac{1}{3}\text{ of water in }(\mathrm{V}-1)+\frac{1}{2}\text{ of water in }(\mathrm{V}-2)+\frac{1}{7}\text{ of water in }(\mathrm{V}-3)\\ &\Rightarrow\frac{1}{3}\times30+\frac{1}{2}\times14+\frac{1}{7}\times7\\ &\Rightarrow10+7+1\\ &\Rightarrow18\\ &\text{Similarly mixture will be}\\ &\Rightarrow\frac{1}{3}\times105+\frac{1}{2}\times70+\frac{1}{7}\times35\\ &\Rightarrow75\\ &\therefore\%\text{ of water }=\frac{18}{75}\times100\\ &=24\%\end{aligned}$

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