Question

In one pot the ratio of milk and water is 3:5 and in another pot it is 6:1 in what ratio should the content of the two pots be mixed so as to make the ratio between milk and water is to be 7:3

Answer 1

Step-by-step explanation:

the above picture is the answer to that question

Answer 2

Answer:

The content of the two pots should be mixed in ratio 44 : 91 as to make the ratio between milk and water is to be 7:3.

Step-by-step explanation:

Consider that 'a' is the amount taken from mixture pot 1,

In First pot, milk and water in ratio 3:5

The quantity of the milk  $=\frac{\big 3}{\big 8}$

The quantity of water $=\frac{\big 5}{\big 8}$

In 2nd pot, milk and water in ratio 6 : 1,

The quantity of the milk  $=\frac{\big 6}{\big 7}$

The quantity of water $=\frac{\big 5}{\big 7}$

After mixing the content of both pots in ratio 7 : 3,

The quantity of the milk  $=\frac{\big 7}{\big 10}(a+b)$

The quantity of water $=\frac{\big 5}{\big 8} (a+b)$

To achieve the ratio 7 :3, with respect to milk,

$\frac{7}{10} (a+b)=a(\frac{3}{8}) +b(\frac{6}{7} )$

$\frac{7}{10} a+\frac{7}{10}b =\frac{3}{8}a +\frac{6}{7}b$

$\frac{7}{10} a-\frac{3}{8}a =\frac{6}{7}b-\frac{7}{10}b$

$\frac{(28-15)a}{40} =\frac{(60-49)}{70}b$

$\frac{13}{40}a =\frac{11}{70}b$

$\frac{a}{b}=\frac{44}{91}$

Similarly, with respect to water

$\frac{3}{10} (a+b)=a(\frac{5}{8}) +b(\frac{1}{7} )$

$\frac{3}{10} a-\frac{5}{8}a =\frac{1}{7}b-\frac{3}{10}b$

$\frac{(12-25)a}{40} =\frac{(10-21)}{70}b$

$\frac{a}{b}=\frac{44}{91}$

Therefore, the content of the two pots should be mixed in ratio 44: 91.