alligation

Question

There are two glasses having mixtures of wine and water. In glass 1 the ratio of wine and water is 1:2 and in glass 2 the ratio of wine and water is 4:1. Find the amount of mixture that should be taken from glass 1 in order to make 280 ml of mixture containing equal amount of water and wine.

(a) 180 ml  (b) 140 ml

(c) 200 ml  (d) None of these

Answer 1

Let $a$ be the amount of mixture taken from glass 1 and $b$ be the same taken from glass 2.

$$\begin{equation} \frac{1} {1+2} \times a + \frac{4 }{4+1} \times b = \frac{2}  {1+2} \times a + \frac{1} {4+1} \times b  \\\implies \frac{a} {3} = \frac{3b} {5} \label{628eq1} \end{equation}$$

$$\begin{equation} a+b = 280 \label{628eq2} \end{equation}$$

From $\ref{628eq1}$ and $\ref{628eq2}$,

$$\frac{a}{3} = \frac{3(280-a)}{5} \\ \implies 5a = 9\times280 - 9a \\\implies a = 180.$$