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# Mixture and Alligation #18

 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%

Explanation:

Solution 1

If a trader professes to sell his goods at cost price, but uses false weights, then
Gain% $=\left[\dfrac{\text{Error}}{(\text{True Value- Error})} × 100 \right]\%$

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 $=\dfrac{1}{5}×100=20\%$