Question

Two bottles A and B contain diluted acid. In bottle A, the amount of water is double the amount of acid while in bottle B, the amount of acid is 3 times that of water. How much mixture(in litres) should be taken from each bottle A and B respectively in order to prepare 5 liters diluted acid containing an equal amount of acid and water?

• A. 1,4
• B. 4,1
• C. 2,3
• D. 3,2

Answer 1

Answer:

D. 3,2

Solution:

Given:

Two bottles A and B contain diluted acid.

In bottle A, the amount of water is double the amount of acid.

In bottle B, the amount of acid is 3 times that of water.

Resulted solution quantity = 5 liter and ratio of acid and water in solution is 1 : 1.

Formula used:

Individual share in ratio = (individual ratio/sum of ratio) × total quantity

Calculation:

Let the amount of diluted acid be taken from bottle A and B is X and Y respectively.

Ratio between Acid and Water in bottle A = 1 : 2

Ratio between Acid and Water in bottle B = 3 : 1

Acid quantity in resulted solution $= \frac{X}{3}+\frac{3Y}{4}$

Water quantity in resulted solution $= \frac{2X}{3}+\frac{Y}{4}$

According to question, acid quantity and water quantity is equal in resulted solution.

$\frac{X}{3}+\frac{3Y}{4}=\frac{2X}{3}+\frac{Y}{4}$

$\frac{3Y-Y}{4}=\frac{2X-X}{3}$

$\frac{2Y}{4}=\frac{X}{3}$

$\frac{X}{Y}=\frac{3}{2}$

Quantity of resulted solution = 5 litres

∴ Mixture(in litres) should be taken from bottle A is 3 litres and from bottle B is 2 litres.

Alternate Method

Acid : Water in bottle A = 1 : 2

Acid in bottle $A=\frac{1}{3}$

Acid : Water in bottle B = 3 : 1

Acid in bottle B = ¾

Acid : water in required solution = 1 : 1

Acid in required solution = ½

By using allegation method

  1/3                         3/4

 

                     1/2

 

$(\frac{3}{4}-\frac{1}{2})=\frac{1}{4}$    $(\frac{1}{2}-\frac{1}{3})=\frac{1}{6}$

⇒ $ratio=\frac{1}{4} : \frac{1}{6}$

⇒ 3 : 2    

Quantity of resulted solution = 5 litres

∴ Mixture(in litres) should be taken from bottle A is 3 litres and from bottle B is 2 litres.