Question

In one glass, milk and water are mixed in the ratio of $3:5$ and in another glass they are mixed in the ratio $6:1$. In what ratio should the content of the two glasses be mixed together so that the next mixture contains milk and water in the ratio of $1:1$?

Answer 1

Step 1: Find the quantity of milk and water in the first mixture

It is given that, in the first mixture, milk and water are mixed in the ratio of $3:5$.

Let, the amount of the first mixture $=x$

Then, the quantity of milk in the first mixture $=\frac{3}{3+5}ofx$

$\Rightarrow$ the quantity of milk in the first mixture $=\frac{3}{8}x$ $\left[\because a\mathrm{of}b=a\times b\right]$

And, the quantity of water in the first mixture $=\frac{5}{3+5}ofx$

$\Rightarrow$ the quantity of water in the first mixture $=\frac{5}{8}x$ $\left[\because a\mathrm{of}b=a\times b\right]$

Step 2: Find the quantity of milk and water in the second mixture

It is given that, in the second mixture, milk and water are mixed in the ratio of $6:1$.

Let, the total amount of the second mixture $=y$

Then, the quantity of milk in the second mixture $=\frac{6}{6+1}ofy$

$\Rightarrow$ the quantity of milk in the second mixture $=\frac{6}{7}y$ $\left[\because a\mathrm{of}b=a\times b\right]$

And, the quantity of water in the second mixture $=\frac{1}{6+1}ofy$

$\Rightarrow$ the quantity of water in the second mixture $=\frac{1}{7}y$ $\left[\because a\mathrm{of}b=a\times b\right]$

Step 3: Calculate the ratio of the first mixture and the third mixture

Now, if we say that $x$ amount of the first mixture and $y$ amount of the second mixture is added to make the third mixture in which the ratio of milk and water as $1:1$. Then,

$\frac{\mathrm{total}\mathrm{quantity}\mathrm{of}\mathrm{milk}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{and}\mathrm{second}\mathrm{mixture}}{\mathrm{total}\mathrm{quantity}\mathrm{of}\mathrm{water}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{and}\mathrm{second}\mathrm{mixture}}=1:1$

$\Rightarrow$ $\frac{{\displaystyle \frac{3}{8}}x+{\displaystyle \frac{6}{7}}y}{{\displaystyle \frac{5}{8}}x+{\displaystyle \frac{1}{7}}y}=\frac{1}{1}$

$\Rightarrow$ $\frac{21x+48y}{56}÷\frac{35x+8y}{56}=\frac{1}{1}$

$\Rightarrow$ $\frac{21x+48y}{56}\times \frac{56}{35x+8y}=\frac{1}{1}$

$\Rightarrow$ $\frac{21x+48y}{35x+8y}=\frac{1}{1}$

$\Rightarrow$ $21x+48y=35x+8y$ [using cross-multiplication]

$\Rightarrow$ $14x=40y$

$\Rightarrow$ $\frac{x}{y}=\frac{40}{14}$

$\Rightarrow$ $\frac{x}{y}=\frac{20}{7}$

$\Rightarrow$ $x:y=20:7$

i.e., the amount of the first mixture $:$ the amount of the second mixture $=20:7$

Hence, the required ratio is $20:7$.