Question

There are 3 containers of equal capacity which are equally filled where 1/4 part of liquid from the first container was shifted into the second container and 1/2 part of initial quantity from the second container was shifted into the third container then what will be the ratio of final quantity of liquid in all the containers after the shifting has been stopped?

• A. 1 : 1 : 2
• B. 1 : 2 : 1
• C. 1 : 2 : 2
• D. 1 : 1 : 1

Answer 1

Answer:

A. 1 : 1 : 2

Given:

3 equal containers of equal volume.

$\frac{1}{4}$ part was shifted from 1^st to 2^nd container.

$\frac{1}{2}$ part was shifted from 2^nd to 3^rd container.

Calculation:

According to question,

Since, all containers are equally filled and possess same volume.

So, let the amouint filled in all the containers be unity.

Initially,

1^st container = 1

2^nd container = 1

3^rd container = 1

During shifting,

1^st container = 1 - $\frac{1}{4}=\frac{3}{4}$

2^nd container = 1 + $\frac{1}{4}=\frac{5}{4}$

Since $\frac{1}{2}$ part of initial quantity is taken out from the 2^nd container 

∴ $\frac{5}{4}-\frac{1}{2}=\frac{3}{4}$

 

3^rd container = 1 + $\frac{1}{2}=\frac{3}{2}$

After shifting,

1^st container = $\frac{3}{4}$

2^nd container = $\frac{3}{4}$

3^rd container = $\frac{3}{2}$

Ratio = $\frac{3}{4}$ : $\frac{3}{4}$ : $\frac{3}{2}$

⇒ 3 : 3 : 6

⇒ 1 : 1 : 2

∴The correct answer is 1 : 1 : 2.