QUESTION

# A 15 litre mixture contains 20% alcohol and rest water, If 3 litre of water is added in it, find the % alcohol in the new mixture.$(a){\text{ 16}}{\text{.66% }} \\ (b){\text{ 12}}{\text{.26% }} \\ (c){\text{ 18}}{\text{.88% }} \\ (d){\text{ 21}}{\text{.27% }} \\$

Hint – In this question let the quantity of alcohol be x litres. The quantity of water will eventually be (15-x) litres as the total mixture is 15 litres. Use the question’s constraint that the mixture contains 20% of alcohol. When 3 litres of water is added this now changes the total amount of mixture to 18 litres.

Given data
15 litres of mixtures contains 20 percent alcohol and the rest water.
Let alcohol be x litres.
So, water will be (15 - x) litres.
Now, according to the given condition, x is the 20 percent of 15.
$\Rightarrow x = \dfrac{{20}}{{100}} \times 15 = \dfrac{{15}}{5} = 3$ litres.
So water will be (15 - 3) = 12 litres.
Now it is given that 3 litres of water is mixed with the mixture.
So, total litres in the new mixture will be $\left( {15 + 3} \right) = 18$ litres.
Now we have to find out percentage of alcohol in new mixture,
Let, the percentage of alcohol in the new mixture be y.
So, y is equal to litres of alcohol divided by total litres in the new mixture multiplied by 100.
$y = \dfrac{3}{{18}} \times 100 = \dfrac{{100}}{6} = \dfrac{{50}}{3} = 16.66$ %
So, 16.66% is the required percentage of alcohol in the new mixture.
Hence, option (a) is correct.

Note – These types of questions are pure general mathematics based problems and there is no predetermined formula. The information of the questions helps forming relations between the variables, so it is advised to pay special attention to even the smallest of constraints of the question.