Problem D and Solution

Sale Boats

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 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.