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Mixture to be withdrawn = ?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\[ Water Parts To Be Added = \frac{Mixture Withdrawn}{Original Mixture} = \frac{y}{4} \]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So if 1/3rd of mixture be withdrawn and replaced by same quantity of water – then milk and water will be equal in the mixture
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