 Problem of the Week

Problem C and Solution

Sibling Rivalry

Problem

Akira and Hideo are twins with different jobs. Akira earns five-eighths of what Hideo earns, but Akira’s expenses are half of Hideo’s. Akira ends up saving 40% of his income. What percentage of his income does Hideo save? Solution

Solution 1: Using only one variable
Let $$h$$ represent Hideo’s income. Then Akira’s income is $$\frac{5}{8}h$$.
Since Akira saves 40% of his income, his expenses are $$100\%-40\%=60\%$$ of his income. Therefore, Akira’s expenses are $$60\%\times \frac{5}{8}h=\frac{60}{100}\times \frac{5}{8}h=\frac{3}{8}h$$.
Akira’s expenses are one-half of Hideo’s expenses so Hideo’s expenses are twice Akira’s expenses. Therefore, Hideo’s expenses are $$2\times \frac{3}{8}h=\frac{3}{4}h=0.75h=75\% \mbox{ of } h$$. Since Hideo’s expenses are 75% of his income, he saves $$100\%-75\%=25\%$$ of his income.
Therefore, Hideo saves 25% of his income.
Solution 2: Using two variables
Let $$x$$ represent Hideo’s income and $$y$$ represent Hideo’s expenses.
Then Akira’s income is $$\frac{5}{8}x$$ and his expenses are $$\frac{1}{2}y$$.
Since Akira saves 40% of his income, his expenses are 60% of his income.
\begin{aligned} \frac{1}{2}y&=0.60\times \frac{5}{8}x\\ \frac{1}{2}y&=\frac{6}{10}\times \frac{5}{8}x\\ \frac{1}{2}y&=\frac{3}{8}x\\ y&=\frac{3}{4}x\\[-4mm]\end{aligned} Hideo saves whatever is left of his income after expenses. Therefore Hideo saves $x-y=x-\frac{3}{4}x=\frac{1}{4}x=0.25x=25\% \mbox{ of } x.$ Therefore, Hideo saves 25% of his income.
Solution 3: Using two variables a bit differently
Let $$8x$$ represent Hideo’s income and $$2y$$ represent Hideo’s expenses.
Then Akira’s income is $$\frac{5}{8}(8x)=5x$$ and his expenses are $$\frac{1}{2}(2y)=y$$.
Since Akira saves 40% of his income, his expenses are 60% of his income.
\begin{aligned} y&=0.60\times 5x\\ y&=\frac{6}{10}\times 5x\\ y&=3x\\[-4mm]\end{aligned} Hideo earns $$8x$$ and his expenses are $$2y$$ so his savings are $$8x-2y$$. We want the ratio of his savings to his income, $$\dfrac{8x-2y}{8x}=\dfrac{8x-2(3x)}{8x}=\dfrac{2x}{8x}=\dfrac{1}{4}$$ or 25% .
Therefore, Hideo saves 25% of his income.