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1. 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 $\frac{1}{3}$ of first, $\frac{1}{2}$ of second and $\frac{1}{7}$ 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%

Solution:

Capacities of vessels

3 : 2 : 1

\begin{aligned} M : W T Mixture \\ V  - 1 \to (5 : 2 = 7)_{\times 5} \\ V  - 2 \to (4 : 1 = 5)_{\times 7} \\ V  - 3 \to (4 : 1 = 5)_{\times 7} \end{aligned}

Equate the mixture

\begin{aligned} M : W T Mixture \\ (V  - 1) \to 25 : 10 = 35 \\ (V  - 2) \to 28 : 7 = 35 \\ (V  - 3) \to 28 : 7 = 35 \end{aligned}

\begin{aligned} Capacities M : W = Total Mix . \\ (V - 1) \times 3 \to 75 : 30 = 105 \\ (V - 2) \times 2 \to 56 : 14 = 70 \\ (V - 1) \times 1 \to 28 : 7 = 35 \end{aligned}

Water taken out

\Rightarrow \frac{1}{3} of water in (V  - 1)

+ \frac{1}{2} of water in (V  - 2)

+ \frac{1}{7} of water in (V  - 3)

\begin{aligned} \Rightarrow \frac{1}{3} \times 30 + \frac{1}{2} \times 14 + \frac{1}{7} \times 7 \\ \Rightarrow 10 + 7 + 1 \\ \Rightarrow 18 \end{aligned}

Similarly mixture will be

\Rightarrow \frac{1}{3} \times 105 + \frac{1}{2} \times 70 + \frac{1}{7}

× 35
⇒ 75

\begin{aligned} ∴ % of water  = \frac{18}{75} \times 100 \\ = 24 % \end{aligned}

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