# Problem leading simple equations [closed]

A sum of Rs. 8.85 is made up of 124 coins which are either 10 paisa coins or 5 paisa coins ; how many coins are there each Note : Rs. 1 = 100 paisas

• Have you worked with simultaneous equations? Can you set up two equations for finding the solution? Commented Jun 3, 2015 at 13:05

1 RS = 100 paisas, so 8.85 RS = 885 paisas

885 paisas is made up of 124 coins, either 10 paisa coins or 5 paisa coins

Let x = the number of 10 paisa coins and y = the number of 5 paisa coins

then $124 = x + y$, so $x = 124 - y$

885 paisas = the number of 10 paisa coins * 10 (what they are worth) + the number of 5 paisa coins * 5 (what they are worth)

\begin{align} 885 &= 10x + 5y \\ &= 10 (124 - y) + 5y \quad\quad(\text{ substituting in }x = 124 - y )\\ &= 1240 - 10y + 5y \\ &= 1240 - 5y \\ -355 &= -5y \\ 71 &= y \end{align}

Since: $124 - y = x$

\begin{align} 124 - 71 &= x \\ 53 &= x \end{align}

So there are 53 10-paisa coins and 71 5-paisa coins.

Check:

\begin{align} 10x + 5y &= 10(53) + 5(71) \\ &= 530 + 355 \\ &= 885 \quad \checkmark \end{align}

Let $P_{10}$ represent a 10 paisa coin and $P_5$ represent a 5 paisa coin.

Then the two equations you want to work with are:

\begin{align} P_5+P_{10} &=124 \\ 0.05P_5+0.10P_{10} &=8.85 \end{align}

To solve by substitution:

\begin{align} P_5 &=124-P_{10} \\ 0.05(124-P_{10})+0.10P_{10} &=8.85 \\ 6.2-0.05P_{10}+0.10P_{10} &=8.85 \\ 0.05P_{10} &=2.65 \\ P_{10} &=53 \\ \end{align}

Substitute to solve for $P_5 = 124-53 = 71$

If $x$ is the number of $5$ paisa coins than the number of $10$ paisa coins is $(124-x)$ and the total value of the coins is: $$5\times x+10 \times (124-x)=885$$ You can solve this?