Word Problems 7

Problem 7: The percentage of water in 20 litters of adulterated milk is 10. The quantity of water to be added to it to increase the percentage of water to 25 is ;

a) 4 litters                          b) 5 litters                  c) 7 litters                   d) 8 litters




We are given that in 20 liters mixture of milk and water there is  10 percent water.

i-e  Water = 10 % of 20 liter

or water = \frac{10}{100} \times20

\Rightarrow water = 2 liters

So water in 20 liter mixture is 2 liters .

Now suppose if we add ‘y’ liters of more water { then total water will become y+2} and water percentage becomes 25 % of mixture {which will be y+20 now}

In other words we supposing that ;

\left ( y+2 \right ) water = \frac{25}{100} of \left ( y+20 \right ) mixture


\left ( y+2 \right ) = \frac{25}{100} \times \left ( y+20 \right )

Simplifying and shifting ‘y’ terms on one side;

\Rightarrow\left ( y+2 \right ) = \frac{1}{4} \times \left ( y+20 \right )

\Rightarrow 4 \left ( y+2 \right ) = \left ( y+20 \right )

\Rightarrow4y+8 = y+20

\Rightarrow4y-y= 20-8


\Rightarrow y = 4

So, we will have to add 4 more liters to make water concentration 25 % of 20+4=24 liters of mixture




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