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18. A dishonest milkman sells his milk at cost price but he mixes it with water and thereby gains 25%. What is the percentage of water in the mixture? | |

A. 24% | B. 22% |

C. 25% | D. 20% |

Answer: Option D

Explanation:

**Solution 1**

If a trader professes to sell his goods at cost price, but uses false weights, then

Gain% $=\left[\dfrac{\text{Error}}{(\text{True Value- Error})} × 100 \right]\%$

Gain% $=\left[\dfrac{\text{Error}}{(\text{True Value- Error})} × 100 \right]\%$

Here Gain= 25%

error = quantity of water he mixes in the milk $=x$

true value = true quantity of milk = T

So the formula becomes, $25=\dfrac{x}{(T - x)} × 100$

$\Rightarrow 1= \dfrac{x}{(T - x)}× 4\\\Rightarrow T - x = 4x\\\Rightarrow T = 5x$

Percentage of water in the mixture

$=\dfrac{x}{T}× 100 = \dfrac{x}{5x}× 100 \\= \dfrac{1}{5}× 100 = 20\%$

**Solution 2**

Let CP of 1 litre milk = Rs.1

SP of 1 litre mixture = CP of 1 litre milk = Rs.1

Gain = 25%

Hence CP of 1 litre mixture

$=\dfrac{100} {\left(100 + \text{Gain} \% \right)}× \text{SP}\\= \dfrac{100}{(100+25)}× 1 = \dfrac{100}{125}= \dfrac{4}{5}$

By rule of alligation, | ||||||||||

CP of 1 litre milk | CP of 1 litre water | |||||||||

1 | 0 | |||||||||

CP of 1 litre mixture $\dfrac{4}{5}$ | ||||||||||

$\dfrac{4}{5}-0=\dfrac{4}{5}$ | $1-\dfrac{4}{5}=\dfrac{1}{5}$ |

=> Quantity of milk : Quantity of water $=\dfrac{4}{5}:\dfrac{1}{5}=4:1$

Hence percentage of water in the mixture $=\dfrac{1}{5}×100=20\%$

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