# Profit/Loss percentage question

Tom buys $$18$$ lobsters at the same cost. He sells $$16$$ lobsters at a $$60\%$$ profit. Each of the remaining lobsters are sold at the same loss. It is given that the overall percentage profit is $$50\%$$.

A) Find the percentage of loss of selling the remaining lobsters.

B) Suppose that the total selling price of the lobster $$\1755$$, find the loss for each remaining lobster.

This is what I tried: let $$x$$ be the cost of $$1$$ lobster. Therefore, the profit will be $$9.6x$$ for selling $$16$$ lobsters. Also, the selling price for $$16$$ lobsters would then be $$25.6x$$ That’s all I was able to get so far!

Let $$c$$ be each lobster's cost price. Then the total revenue (= total/net selling price) from selling the $$18$$ lobsters is $$1.5$$ times their total cost price, which is $$1.5(\18c) = \27c.$$ This total revenue is made up of two parts: (i) The total revenue from selling $$16$$ lobsters each at a selling price of $$\1.6c,$$ which is $$16(\1.6c) = \25.6c.$$ (ii) The total revenue from selling $$2$$ lobsters each at a selling price of $$2\left(1 - \frac{r}{100}\right)(\c),$$ where $$r\%$$ is the percentage loss for selling each of these $$2$$ lobsters. Therefore, we have

$$\27c \;\; = \;\; \25.6c \; + \; \2\left(1 - \frac{r}{100}\right)c$$

Dividing both sides by $$c$$ and rearranging a bit (and working with pure numbers) gives

$$27 - 25.6 \; = \; 2 - \frac{r}{50}$$

$$-0.6 \; = \; -\frac{r}{50}$$

$$r \; = \; 30$$

Therefore, the answer to (A) is $$30\%.$$

For (B), the total selling price is equal to $$\1755.$$ Therefore, $$\27c = \1755,$$ or $$c = \65,$$ and hence the loss amount for each of the $$2$$ remaining lobsters is $$30\%$$ of $$\65,$$ which equals $$(0.3)(\65) = \19.50.$$

• Heyy thanks I used the same method but was confused as a (now deleted) comment had different ideas of solving the question. Thanks for the effort :) Commented May 6, 2021 at 7:13