A trader bought a quantity of palm oil. He sold \frac{1}{3} of the total quantity at a pro… — JAMB UTME

A trader bought a quantity of palm oil. He sold $\frac{1}{3}$ of the total quantity at a profit of $20\%$ , $\frac{1}{2}$ of the remaining palm oil at a loss of $10\%$ , and the rest of the palm oil at a profit of $x\%$ . If his overall profit on the entire transaction was $15\%$ , what is the value of $x$ ?

Answer: 35 %

Answer

The value of

x

35

The palm oil is divided into three equal portions of

\frac{1}{3}

each. The first portion contributes

\frac{20}{3}\%

profit, the second portion contributes a loss of

-\frac{10}{3}\%

, and the final portion contributes

\frac{x}{3}\%

profit. Summing these contributions yields

\frac{10 + x}{3}\%

. Equating this to the target overall profit of

15\%

gives

10 + x = 45

, which solves to

x = 35

Step-by-Step Solution

Determine the fractional portion of palm oil sold at each stage.

First portion =

\frac{1}{3}

. Remaining after first sale =

1 - \frac{1}{3} = \frac{2}{3}

. Second portion =

\frac{1}{2} \times \frac{2}{3} = \frac{1}{3}

. Final portion =

1 - \left(\frac{1}{3} + \frac{1}{3}\right) = \frac{1}{3}

Finding the exact fraction of the total quantity sold at each step is necessary to weight their respective profit/loss rates correctly.

Formulate the net percentage profit equation.

\text{Net Profit } \% = \left(\frac{1}{3} \times 20\%\right) + \left(\frac{1}{3} \times (-10\%)\right) + \left(\frac{1}{3} \times x\%\right) = \frac{20 - 10 + x}{3}\% = \frac{10 + x}{3}\%

The overall profit percentage is the sum of individual fractional contributions, treating loss as a negative percentage.

Set the net profit equal to

15\%

and solve for

x

\frac{10 + x}{3} = 15 \implies 10 + x = 45 \implies x = 35

Solving this linear equation gives the required profit percentage for the final portion.

Key Concept

Weighted Average Profit and Loss across fractional components of an asset

Estimated Time:2m 0s

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