Question

Mukesh bought some oranges. He sold one-third of them at 10% profit, one-sixth of them at 40% profit, and the remaining at a certain profit percentage. If he made an overall profit of 25%, find the profit percent made in selling the remaining oranges.​

Answer 1

Therefore the profit made in selling remaining oranges = 30%.

Step-by-step explanation:

Let Mukesh bought x oranges.

Let the cost price of each orange be y.

Total cost price of the oranges is = xy

Total profit = $\frac{25}{100} xy$

He sold one - third of them 10% profit.

One third of x

$=\frac{1}{3} x$

Cost price of one - third oranges is $\frac{1}{3} xy$

profit =$\frac{10}{100} \times \frac{1}{3} xy$

He sold one sixth of them 40% profit.

One-sixth of x

$=\frac{1}{6} x$

Cost price of one sixth oranges is $\frac{1}{6} xy$

Profit $=\frac{40}{100} \times \frac{1}{6} xy$

Remaining oranges are $=[1-(\frac{1}{3} +\frac{1}{6} )]x$ $=\frac{1}{2} x$

Let he sold remaining oranges at a%profit.

The cost price of one-half of them = $\frac{1}{2} xy$

then profit =$\frac{a}{100} \times \frac{1}{2} xy$

Total profit = $\frac{10}{100} \times \frac{1}{3} xy+\frac{40}{100} \times \frac{1}{6} xy+\frac{a}{100} \times \frac{1}{2} xy$

According to the problem,

$(\frac{10}{100} \times \frac{1}{3} xy)+(\frac{40}{100} \times \frac{1}{6} xy)+(\frac{a}{100} \times \frac{1}{2} xy)=\frac{25}{100} xy$

$\Rightarrow \frac{10}{3}+\frac{40}{6} +\frac{a}{2} =25$

$\Rightarrow \frac{a}{2} =25-(\frac{10}{3}+\frac{40}{6} )$

$\Rightarrow \frac{a}{2} =25-10$

$\Rightarrow \frac{a}{2} =15$

$\Rightarrow a=30$

Therefore the profit made in selling remaining oranges = 30%.