# Problem of the Week Problem D and Solution Sale Boats

## Problem

Harold, a marina manager, purchased two boats. He then sold the boats, the first at a profit of $$40\%$$ and the second at a profit of $$60\%$$. The total profit on the sale of the two boats was $$54\%$$ and $$\88\,704$$ was the total selling price of the two boats.

What did Harold originally pay for each of the two boats?

## Solution

Solution 1

Let $$a$$ represent what Harold paid for the first boat, in dollars, and $$b$$ represent what he paid for the second boat, in dollars.

The profit on the sale of the first boat was $$40\%$$ or $$0.4a$$ dollars. Thus, the first boat sold for $$a+0.4a=1.4a$$ dollars. The profit on the sale of the second boat was $$60\%$$ or $$0.6b$$ dollars. Thus, the second boat sold for $$b+0.6b=1.6b$$ dollars. The total selling price of the two boats was $$\88\,704$$, so we have \begin{align*} 1.4a+1.6b&= 88\,704 \tag{1}\end{align*} Harold bought both boats for a total of $$(a+b)$$ dollars. The profit on the sale of the two boats was $$54\%$$ or $$0.54(a+b)$$ dollars. The two boats sold for $$(a+b)+0.54(a+b)=1.54(a+b)$$ dollars. But the total selling price was $$\88\,704$$, so \begin{aligned} 1.54(a+b)&=88\,704\\ a+b&=88\,704\div 1.54\\ a+b&=57\,600\\ a & = 57\,600 - b \end{aligned} Substituting $$a = 57\,600 - b$$ into equation $$(1)$$ gives \begin{aligned} 1.4(57\,600 - b)+1.6b&= 88\,704 \\ 80\,640 - 1.4b+1.6b&= 88\,704 \\ 0.2b&= 8064 \end{aligned}

Dividing by $$0.2$$, we get $$b= 40\,320$$. Since $$b=40\,320$$ and $$a+b=57\,600$$, then $$a=17\,280$$ follows.

Therefore, Harold paid $$\17\,280$$ for the first boat and $$\40\,320$$ for the second boat.

Solution 2

Let $$a$$ represent what Harold paid for the first boat, in dollars, and $$b$$ represent what he paid for the second boat, in dollars.

The profit on the sale of the first boat was 40% or $$0.4a$$ dollars. The first boat sold for $$a+0.4a=1.4a$$ dollars. The profit on the sale of the second boat was 60% or $$0.6b$$ dollars. The second boat sold for $$b+0.6b=1.6b$$ dollars. The total selling price of the two boats was $$\88\,704$$ so we have \begin{aligned} 1.4a+1.6b&= 88\,704\end{aligned} Multiplying by $$5$$, we get \begin{align*} 7a+8b&=443\,520 \tag{1} \end{align*}

Harold bought both boats for a total of $$(a+b)$$ dollars. The profit on the sale of the two boats was $$54\%$$ or $$0.54(a+b)$$ dollars. The total profit is the sum of the profit from the sale of each boat, so \begin{aligned} 0.54(a+b)&=0.4a+0.6b\\ 0.54a+0.54b&=0.4a+0.6b \\ 0.14a&=0.06b \end{aligned} Multiplying by $$50$$, we get \begin{align*} 7a&=3b \tag{2}\end{align*}

Substituting $$3b$$ for $$7a$$ into equation $$(1)$$, we get $$3b+8b=443\,520$$ or $$11b=443\,520$$, and $$b=40\,320$$ follows.

Substituting $$b=40\,320$$ into equation $$(2)$$, we get $$7a=120\,960$$, and $$a=17\,280$$ follows.

Therefore, Harold paid $$\17\,280$$ for the first boat and $$\40\,320$$ for the second boat.