Question

If x is 30% more than y; find:
(i) $\frac{x}{y}$
(ii) $\frac{y+x}{x}$
(iii) $\frac{y}{y-x}$

Answer 1

Consider y = a
We know that
$x=a\times \frac{(100+30)}{100}$
By further calculation
= $a\times \frac{130}{100}$
= $\frac{13}{10}a$
(i) $\frac{x}{y}=\frac{\frac{13a}{10}}{a}$
We can write it as
$\frac{13a}{10a}=\frac{13}{10}$
(ii) $\frac{y+x}{x}$ = $\frac{a+\frac{13a}{10}}{\frac{13a}{10}}$
By further calculation
= $\frac{23a}{10}\times \frac{10}{13a}$
= $\frac{23}{13}$
(iii) $\frac{y}{y-x}$ = $\frac{a}{a-\frac{13a}{10}}$

We can write it as
= $\frac{a}{\frac{-3a}{10}}$
So we get=
= $\frac{10a}{-3a}$
= $\frac{-10}{3}$