d Word Problem 20 | Job Exam Rare Mathematics (JEMrare) - Solved MCQs

# Word Problem 20

Problem 20: A jar contains three parts of pure milk and one part of water. How much of the mixture must be drawn and water substituted, in order that the resulting mixture may be half milk and half water?

a) 1/4th              b) 1/2nd                  c) 1/3rd                  d) 1/5th

Solution:

We are given following information;

Milk = 3 parts
Water = 1 part

Mixture to be withdrawn = ?

so that,

Milk = water

We can infer that at present;

Mixture = water + milk = 1+3 = 4 parts

Now we assume that to make water equal to milk ;

The mixture we withdraw = y

and

Water then added be y  ( to bring the level of liquid in the jar  as it was before)

The data can be arranged as;

 Mixture Milk Water 4 3 1 After withdrawing y parts of mixture 4-y 3 / 4 of mixture =3(4-y) / 4 1 / 4  of mixture  =  (4-y) / 4 After adding y parts of water later 4 3 / 4 of mixture =3(4-y) / 4 {(4-y) / 4}  + y  = 4+3y / 4

So now finally water and milk are equal !

i-e

$\frac{3(4-y)}{4} = \frac{4+3y}{4}$

$3(4-y)=4+3y$

$12-3y= 4+3y$

$8 =6y$

$y= \frac{4}{3}$

So, we have to withdraw 4/3 of mixture !

Finally,

We use relation;

$Water Parts To Be Added = \frac{Mixture Withdrawn}{Original Mixture} = \frac{y}{4}$

Putting value of y from above, we get;

$Water To Be Added = \frac{\frac{4}{3}}{4}= \frac{1}{3}$

So if 1/3rd of mixture be withdrawn and replaced by same quantity of water – then milk and water will be equal in the mixture