A container holds a liquid mixture where water constitutes \frac{3}{8} of the total volume… — JAMB UTME

A container holds a liquid mixture where water constitutes $\frac{3}{8}$ of the total volume. When $15\text{ liters}$ of pure water is added to the mixture, water then accounts for $50\%$ of the new total volume. What was the initial total volume of the mixture in liters?

Answer: 60 liters

Answer

The initial total volume of the mixture was 60 liters.

The correct answer is 60 liters. By modeling the initial volume of water as

\frac{3}{8}V

, adding 15 liters yields a new water volume of

\frac{3}{8}V + 15

out of a total volume of

V + 15

. Setting

\frac{3}{8}V + 15 = 0.5(V + 15)

and solving gives

V = 60

Step-by-Step Solution

Define the variable and write an algebraic expression for the initial amount of water.

V

is the initial total volume in liters, initial water volume =

\frac{3}{8}V

Water makes up

\frac{3}{8}

of the total initial volume.

Account for the addition of 15 liters of pure water to both water volume and total volume.

New water volume =

\frac{3}{8}V + 15

; New total volume =

V + 15

Adding pure water increases both the specific water volume and the total mixture volume by 15 liters.

Formulate an equation relating new water volume to 50% of the new total volume.

\frac{3}{8}V + 15 = 0.5(V + 15)

Water constitutes 50% (or

\frac{1}{2}

) of the updated mixture.

Solve the linear equation for

V

7.5 = \frac{1}{8}V \implies V = 60

Subtracting

\frac{3}{8}V

and

7.5

from both sides isolates

\frac{1}{8}V

on one side.

Key Concept

Solving multi-step fraction and percentage mixture problems

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