Problem leading simple equations 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
 A: 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 $P_5 = 124-53 = 71$
A: 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}
$
A: 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?
