# NIELIT 2019 Feb Scientist D - Section C: 18

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A dishonest milkman professes to sell his milk at $C.P$. but he mixes it with water and thereby gains $25\%$. The percentage of water in the mixture is :

1. $25\%$
2. $20\%$
3. $4\%$
4. None of these

Ans is option (B)

Let the milkman sells $1$L of milk originally at Rs $100$

Let him mix $’x’$ L of water in the mixture. Then, total volume of mixture is now  $’x+1’$ L

Now, he would have gained Rs $100$ on selling  pure milk. But now, with $’x+1’$ L of impure milk, he got Rs $125$ i.e. Rs $25$ extra.

$\therefore$   $1$ L  $\rightarrow$  Rs $100$

$’x+1’$ L  $\rightarrow$   Rs $125$

$\therefore$  $100(x+1)=125$   $\Rightarrow$  $x=\frac{1}{4}$ L

$\therefore$ Percentage of water in the current mixture  $=$   $\frac{\frac{1}{4}}{\frac{1}{4}+1}\times100=20$%

Suppose, 100 litres of milk costs Rs.100, 100 litres of milk is sold at 25% profit, so he would have sold it for Rs.125.

As water doesn't have any cost, water added by him will be 125 – 100 = 25 litres [1 litre milk cost Rs.1].

100 litres of milk and 25 litres of water added, so mixture now became 125 litres.

The percentage of water in the mixture is $\frac{25}{125} * 100 = 20$%

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