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# A mixture of $40$liters of milk and water $10\%$water. How much water must be added to make $20\%$water in the mixture?(A)$3$liters (B)$4$ liters (C) $5$ liters (D) $6$ liters

Last updated date: 14th Sep 2024
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Hint: In these types of questions there is no need for particular formula. Only the knowledge of percentage in deep is sufficient. The percentage is nothing but a number or ratio that represents a fraction of $100$. For example, if $50\%$ of the total number of students in the class are male, that means 50 out of every 100 students are male. If there are 500 students, then 250 of them are male.

Total quantity of mixture = $40$ liters
Quantity of water in the mixture in percentage = $10\%$
$\therefore$ Quantity of water in the mixture in liter = $40 \times \dfrac{{10}}{{100}}$liters = $4$ liters
Since the remaining part is milk,
Quantity of milk in the mixture in percentage = $90\%$
$\therefore$ Quantity of milk in the mixture in liter = $40 \times \dfrac{{90}}{{100}}$liters = $4 \times 9$ liters = $36$ liters
Suppose $x$ quantity of water is added to the mixture to make 20% of water in the mixture.
Now in the new mixture,
Total quantity of mixture is = $40 + x$ liters
Quantity of water in the mixture in liter = $4 + x$ liters
Quantity of milk in the mixture in liter = $36 + x$ liters
It is given that,
Quantity of water in the new mixture in percentage = $20\%$
$\therefore$ Quantity of water in the new mixture in liter = $(40 + x) \times \dfrac{{20}}{{100}} = 4 + x$
$\Rightarrow (40 + x)\dfrac{1}{5} = 4 + x$
$\Rightarrow 40 + x = 5(4 + x)$
$\Rightarrow 40 + x = 20 + (5 \times x)$
$\Rightarrow 40 - 20 = 5 \times x - x$
$\Rightarrow 20 = 4x$
$\Rightarrow x = 5$liters
Hence 5 liters water must be added to make 20% water in the mixture.

Note:
This question can also be solved by equating milk quantity in the new mixture to the milk quantity in the old mixture since the quantity of milk in both the mixture is same. That means
$\therefore$ Quantity of milk in the new mixture in liter = Quantity of milk in the old mixture in liter
$(40 + x) \times \dfrac{{80}}{{100}} = 36$
$\Rightarrow (40 + x)\dfrac{4}{5} = 36$
$\Rightarrow 40 + x = \dfrac{{36 \times 5}}{4}$
$\Rightarrow 40 + x = 45$
$\Rightarrow x = 45 - 40$
$\Rightarrow x = 5$liters