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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
in Quantitative Aptitude retagged by
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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  $=$   $\frac{\frac{1}{4}}{\frac{1}{4}+1}\times100=20$%
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