Question

Milk and water in two vessels A and B are in the ratio 4 : 3 and 2 : 3 respectively. In what ratio the liquids in both the plate vessels should be mixed to obtain a new mixture in vessel C, only five maintaining half milk and half water?

• A. 1:1
• B. 7:5
• C. 2:4
• D. 1:3

Answer 1

The Correct Option is B

The problem requires determining the ratio in which liquids from vessels A and B are mixed to achieve a 1:1 mixture of milk and water in vessel C.

First, calculate the fractions of milk and water in each vessel: 
Vessel A has a milk to water ratio of 4:3. This means out of a total of 7 parts, there are 4 parts milk and 3 parts water. Therefore, the fraction of milk in vessel A is:

\[\text{Milk in A} = \frac{4}{7}\]

And the fraction of water in vessel A is:

\[\text{Water in A} = \frac{3}{7}\]

Vessel B has a milk to water ratio of 2:3. Out of 5 parts, there are 2 parts milk and 3 parts water. So, the fraction of milk in vessel B is:

\[\text{Milk in B} = \frac{2}{5}\]

And the fraction of water in vessel B is:

\[\text{Water in B} = \frac{3}{5}\]

To find the ratio in which to mix A and B to have equal parts of milk and water in vessel C, set up the equation for milk and water:

If \( x \) is the part of mixture A and \( y \) is the part of mixture B, the equation for milk becomes:

\[\frac{4}{7}x + \frac{2}{5}y = \frac{1}{2}(x+y)\]

Similarly, the equation for water becomes:

\[\frac{3}{7}x + \frac{3}{5}y = \frac{1}{2}(x+y)\]

Simplifying the equations using a common denominator and algebraic steps:

\( \frac{4}{7}x + \frac{2}{5}y = \frac{1}{2}(x + y) \)

\( \Rightarrow 40x + 28y = 35x + 35y \)

\( \Rightarrow 5x = 7y \)

The simplest form of ratio \( x:y \) is 7:5.

Thus, the ratio in which liquids A and B should be mixed is 7:5.