Question

A vessel contains liquids A and B in the ratio of 3:1. If 8 litres of the mixture is removed and the same quantity of
liquid B is added, the ratio becomes 1:3. What quantity does the vessel hold?

Answer 1

Given:

A vessel contains liquids A and B in the ratio of 3:1. If 8 litres of the mixture is removed and the same quantity of liquid B is added, the ratio becomes 1:3. What quantity does the vessel hold?

To find:

What quantity does the vessel hold?

Solution:

Let "3x" and "x" be the quantity of liquids in vessels A and B respectively.

After the removal of 8 litres of the mixture,  

The quantity of liquid in A becomes = $3x - \frac{3}{4} \times 8 = 3x - 6$

and

The quantity of liquid in B becomes = $x - \frac{1}{4} \times 8 = x - 2$

Now, as given in the question we can form the equation as,

$\frac{3x - 6}{x - 2 + 8} = \frac{1}{3}$

$\implies \frac{3x - 6}{x + 6} = \frac{1}{3}$

$\implies 3(3x - 6) = (x + 6)$

$\implies 9x - 18 = (x + 6)$

$\implies 9x - x = 18 + 6$

$\implies 8x = 24$

$\implies x = 3$

∴ The total quantity of the vessel = $3x + x = 4x = 4 \times 3 = \bold{12\:litres }$

Thus, the vessel holds 12 litres of liquid.

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Also View:

A vessel contains liquid a and b in ratio 5:3. if 16 litres of the mixture are removed and the same quantity of liquid b is added, the ratio becomes 3:5.what quantity does the vessel hold?

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