Question

Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is?​

Answer 1

$let \: the \: third \: number \: be \: x \\ \therefore \: first \: number \: = \: x + 20\% \: of \: x \\ = x + \frac{20x}{100} \\ = \frac{100x + 20x}{100} \\ = \frac{120x}{100} \\ = \frac{6x}{5} \\ and \: second \: number \: = x + 50\% \: of \: x \\ = x + \frac{50x}{100} \\ = \frac{100x + 50x}{100} \\ = \frac{150x}{100} \\ = \frac{3x}{2} \\ so \: required \: ratio = \frac{ \frac{6x}{5} }{ \frac{3x}{2} } \\ = \frac{6x}{5} \times \frac{2}{3x} \\ = \frac{2}{5} \times \frac{2}{1} \\ = \frac{4}{5} \\ = 4 \ratio \: 5$

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