Question

A company makes a mixture which contains 2% alcohol. If 10 Liters of alcohol is added to the mixture, then concentration increases to 5%. What is the approx. quantity of the mixture?

Answer 1

In this problem we have to deal with percentage quantity. The relationship between Percentage and amount of the substance in the mixture can be given as:
$Percentage = \dfrac{{Volume{\text{ }}of{\text{ }}dissolved{\text{ }}subs\tan ce}}{{Total{\text{ }}volume{\text{ }}of{\text{ }}liquid}} \times 100$

Complete answer:
Let us assume the quantity of the given mixture to be ‘x’ litres. We are given that in the initial mixture the alcohol quantity is 2% of the entire concentration of the mixture.
Mathematically amount of alcohol using the above formula will be $= \dfrac{2}{{100}} \times x = \dfrac{x}{{50}}L$ --(1)
Second case given to us is when 10L of alcohol is added to the overall mixture, the concentration becomes 5%.
When 10L of alcohol is added to the initial solution, the total volume of the mixture becomes $= (x + 10)L$
Therefore, the amount of alcohol referring to equation (1) after the addition of 10L of alcohol can be given as $= \dfrac{{(x + 10)}}{{50}}L$
We are given that after the addition of alcohol the concentration becomes 5%. So, the percentage of alcohol in new mixture $= \dfrac{{\dfrac{{10 + x}}{{50}}}}{{x + 10}} = \dfrac{5}{{100}}$
Cross Multiplying and simplifying the equation, to find the value of x
$100(500 + x) = 250(x + 10)$
$2(500 + x) = 5(x + 10)$
$x = \dfrac{{950}}{3} = 316.66L$
Hence the approx. The quantity of mixture will be $\approx 317L$. This is the required answer.

Note:
For this problem we don’t require any specific formula. We just need to understand the concept of percentage and apply simple logic to convert percentage into volume. Remember to maintain the units same throughout the problem (in here we have maintained the unit Litres). If approx. quantity is asked, we can round-off the answer to the nearest whole number, as in this case we have rounded it off to 317 L.