Mother has P1 coin, P5 coins and P10 bill in her purse. In all, she has P112. She has two more P1 coin than P5 coins and one
More P10 bills than P1 coins. How many of each kind does she have?
# of P5 --- x
# of P1 --- x+2
# of P10 -- x+2 + 1 = x+3
equation dealing with "value"
5x + x+2 + 10(x+3) = 112
finish it up.
X = 5
To find out how many of each kind of coin and bill the mother has, we can set up a system of equations based on the given information.
Let's denote the number of P1 coins as x.
The number of P5 coins is x - 2 (two more P1 coins than P5 coins).
The number of P10 bills is x + 1 (one more P10 bill than P1 coins).
Now, let's set up the equations based on the total amount of money:
1x + 5(x - 2) + 10(x + 1) = 112
Simplifying the equation:
x + 5x - 10 + 10x + 10 = 112
16x = 112 + 10 - 10
16x = 112
Divide both sides by 16 to solve for x:
x = 112 / 16
x = 7
Therefore, the mother has 7 P1 coins, 7 - 2 = 5 P5 coins, and 7 + 1 = 8 P10 bills.