Question

I have a total of $Rs.300$ in coins of denominations $Rs.1,Rs.2$ and $Rs.5$. The number of $Rs.2$ coins is $3$ times the number of $Rs.5$ coins. The total number of coins is $160$. How many coins of each denomination are with me?

Answer 1

Step 1: List the given conditions:

Let the number of $Rs.1$ coins be $x$, the number of $Rs.2$ coins be $y$ and the number of $Rs.5$ coins be $z$.

The given conditions are,

$y=3z$ . . .$\left(1\right)$

$x+y+z=160$ . . .$\left(2\right)$

$x+2y+5z=300$ . . .$\left(3\right)$

Step 2: Establish the pair of linear equations:

Substituting equation $\left(1\right)$ in equation $\left(2\right)$ we get;

$x+3z+z=160$

$\Rightarrow$ $x+4z=160$ . . .$\left(4\right)$

Substituting equation $\left(1\right)$ in equation $\left(3\right)$ we get,

$$\begin{gathered} x+2\left(3z\right)+5z=300 \\ \Rightarrow x+6z+5z=300 \end{gathered}$$

$\Rightarrow$ $x+11z=300$ . . .$\left(5\right)$

Subtracting equation $\left(4\right)$ from equation $\left(5\right)$ we get,

$x+11z-x-4z=300-160$

$\Rightarrow$ $7z=140$

$\Rightarrow$ $z=20$

Step 3: Evaluate the number of coins

Substituting the value of $z$ in equation $\left(1\right)$ we get,

$y=3\times 20=60$

Substituting the value of $z$ in equation $\left(4\right)$ we get,

$x+4\left(20\right)=160$

$\Rightarrow$ $x=160-80$

$\Rightarrow$ $x=80$

Hence, there are $80$ $Rs.1$ coin, $60$ $Rs.2$ and $20$ $Rs.5$.