Question

A container has 60 liters of 40% alcohol solution. How many liters of a 50% alcohol solution should be added to this solution so that the resultant solution will have 42% alcohol?

Answer 1

Answer:

The volume of $50\%$ alcohol solution added to $40\%$ alcohol solution calculated is $15L$.

Explanation:

Given data,

The volume of $40\%$ alcohol solution in a container = $60L$

Let the volume of $50\%$ alcohol solution in the container = $x L$

The resultant volume of $42\%$ alcohol solution in the container = $(x+60) L$

To find:

The volume of $50\%$ alcohol solution added to $40\%$ alcohol solution = $x L$ =?

Now, according to the question, we get an equation:

• $40\%$ of $60L$ + $50\%$ of $xL$ = $42\%$ of $(x+60) L$

Therefore, the equation becomes:

• $(\frac{40}{100}\times 60 )+ (\frac{50}{100}\times x) = \frac{42}{100}\times (x+60)$
• $24+ 0.5 x= 0.42\times (x+60)$
• $24+ 0.5 x= 0.42x+25.2$
• $0.5x-0.42x = 25.2-24$
• $0.08x = 1.2$
• $x = 15L$

Hence, the volume of $50\%$ alcohol solution added to $40\%$ alcohol solution = $15L$.