Question

Robert has $\$200$ of which he spent $\frac{1}{5}$th on lunch and $\frac{1}{10}$th on notebooks. How much more must he spend on lunch so that he saves exactly $\$100?$

• A. $\$\frac{45}{2}$
• B. $\$40$
• C. $\$50$
• D. $\$\frac{200}{6}$

Answer 1

The correct option is B $\$40$

Let Robert spend $\$p$ more for lunch.
$200-\frac{1}{5}\times 200-\frac{1}{10}\times 200-p=100$
$200-\frac{1}{5}\times 200-\frac{1}{10}\times 200-100=p$
$(200-100)-[200\times \frac{1}{5}+200\times \frac{1}{10}]=p$ [Associative property]
$100-200\times (\frac{1}{5}+\frac{1}{10})=p$ [Distributive property]
$100-200\times (\frac{2}{10}+\frac{1}{10})=p$
$100-200\times \left(\frac{2+1}{10}\right)=p$
$100-200\times \frac{3}{10}=p$
$100-20\times 3=p$
$100-60=p$
$p=\$40$
Robert must spend $\$40$ more on lunch so that he saves exactly $\$100$.