Question

15 liters of mixtures contains 20 percent alcohol and the rest water. If 3 liters of water be mixed with it, the percentage of alcohol in the new mixture would be
$
  {\text{a}}{\text{. 15 percent}} \\
  {\text{b}}{\text{. 16}}\dfrac{2}{3}{\text{ percent}} \\
  {\text{c}}{\text{. 17 percent}} \\
  {\text{d}}{\text{. 18}}\dfrac{1}{2}{\text{ percent}}{\text{.}} \\
$

Answer 1

Hint – Assume any variable be the liters of alcohol then first try to find out liters of alcohol as variable is the 20 percent of 15 so, use this concept to reach the solution of the problem.

Given data
15 liters of mixtures contains 20 percent alcohol and the rest water.
Let alcohol be x liters.
So, water will be (15 - x) liters.
Now, according to the given condition, x is the 20 percent of 15.
$ \Rightarrow x = \dfrac{{20}}{{100}} \times 15 = \dfrac{{15}}{5} = 3$ Liters.
So water will be (15 - 3) = 12 liters.
Now it is given that 3 liters of water is mixed with the mixture.
So, total liters in the new mixture will be $\left( {15 + 3} \right) = 18$liters.
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 liters of alcohol divided by total liters in the new mixture multiplied by 100.
$y = \dfrac{3}{{18}} \times 100 = \dfrac{{100}}{6} = \dfrac{{50}}{3} = 16\dfrac{2}{3}$ Percent.
So, $16\dfrac{2}{3}$ is the required percentage of alcohol in the new mixture.
Hence, option (b) is correct.

Note – In such types of questions first find out the liters of alcohol from the given mixture as above, then calculate the value of total liters in the new mixture, then divide liters of alcohol to the total liters in the new mixture and multiply by 100, which is the required percentage of alcohol in the new mixture.