Question

Niall 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 $\$100$?
[Hint: Assume Niall spends $ $p$ on lunch and then solve the obtained equation to find $p$.]

Answer 1

Let Niall spend $\$p$ more for lunch.

$200-\frac{1}{5}\times 200-\frac{1}{10}\times 200-100=p$

$(200-100)-[200\times \frac{1}{5}+200\times \frac{1}{10}]=p\text{[Associative property]}$

$100-200\times (\frac{1}{5}+\frac{1}{10})=p\text{[Distributive property]}$

$100-200\times (\frac{2}{10}+\frac{1}{10})=p$

$100-200\times \left(\frac{2+1}{10}\right)=p$

$100-200\times \left(\frac{3}{10}\right)=p$

$100-20\times 3=p$

$100-60=p$

$p=\$40$

Niall must spend $\$40$ more on lunch so that he saves exactly $\$100$.