Question

There are three containers of equal capacity The ratio of sulphuric acid to water in the first container is 3 : 2 that in the second container is 7 : 3 and in the third container it is 11 : 4 If all the liquids are mixed together then the ratio of sulphuric acid to water in the mixture will be

• A. 61 : 29
• B. 61 : 28
• C. 60 : 29
• D. 59 : 29

Answer 1

The correct option is A 61 : 29

Let the quantity of mixture in each container be x. Then,
Sulphuric acid in 1 st container $=\frac{3x}{5}$ Water in 1 st container $=\frac{2x}{5}$
Sulphuric acid in 2 nd container $=\frac{7x}{10}$ Water is 2nd container $=\frac{3x}{10}$
Sulphuric acid in 3rd container $=\frac{11x}{15}$
Water in 3rd container $=\frac{4x}{15}\therefore$ Required ratio $=(\frac{3x}{5}+\frac{7x}{10}+\frac{11x}{15}):(\frac{2x}{5}+\frac{3x}{10}+\frac{4x}{15})=\left(\frac{36+42+44}{60}\right):\left(\frac{24+18+16}{60}\right)=122:58=61:29.$