Question

A sum of Rs.$3.75$ was paid in $25$ paisa,$10$ paisa and $5$ paisa coins.The number of $10$ paisa coins was four times the number of $25$ paisa coins and twice of the number of $5$ paisa coins.How many were there of each type?

Answer 1

Let the number of $10$ paisa coins be $x$, number of $25$ paisa coin$=\frac{x}{4}$

Number of $5$ paisa coin$=\frac{x}{2}$

According to the given problem,

$\Rightarrow \frac{10x}{100}+\frac{x}{4}\times \frac{25}{100}+\frac{x}{2}\times \frac{5}{100}=3.75$

$\Rightarrow \frac{10x}{100}+\frac{25x}{400}+\frac{5x}{200}=\frac{375}{100}$

$\Rightarrow \frac{40x}{400}+\frac{25x}{400}+\frac{10x}{400}=\frac{375}{100}$

$\Rightarrow \frac{40x+25x+10x}{400}=\frac{375}{100}$

$\Rightarrow \frac{75x}{4}=\frac{375}{1}$

$\therefore x=\frac{1500}{75}=20$

Number of $10$ paisa coins$=20$

Number of $25$ paisa coins$=5$

Number of $5$ paisa coins$=10$