There are 3 containers of equal capacity. The ratio of sulphuric acid

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

There are 3 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»

(a) 122 : 58 = 61 : 29

Let the quantity of mixture in each container be x.

Then, Sulphuric acid in 1st container = \(\frac{3x}{5}\)

Water in 1st container = \(\frac{2x}{5}\)

Sulphuric acid in 2nd container = \(\frac{7x}{10}\)

Water is 2nd container = \(\frac{3x}{10}\)

Sulphuric acid in 3rd container = \(\frac{11x}{15}\)

Water in 3rd container = \(\frac{4x}{15}\)

∴ Required ratio

= \(\bigg(\frac{3x}{5}+\frac{7x}{10}+\frac{11x}{15}\bigg):\) \(\bigg(\frac{2x}{5}+\frac{3x}{10}+\frac{4x}{15}\bigg)\)

= \(\bigg(\frac{36+42+44}{60}\bigg):\)\(\bigg(\frac{24+18+16}{60}\bigg)\)

= 122 : 58 = 61 : 29.