Let x = number of 5 peso coins
Then 50-x = number of 1 peso coins
5x + 1(50-x) = 130
5x + 50 - x = 130
4x + 50 = 130
4x = 80
x = 20
He has 20 5-peso coins
and
50-x =
50-20 =
30 1-peso coins
Edwin
Answer by 12345qwerty(1) (Show Source): You can put this solution on YOUR website! Can you answer this
A sum of money amounting to P37 consists of 1-peso bills, 2-peso bills and 5-peso bills. The number of 1-peso bills exceeds seven times the number of 5 peso bills by one, and the number of 2-peso bills is nine less than the number of 1-peso bills. How many of each kind of bills are there?
x+y=50 the number of coins is 50
5x+1y=130 the value of the coins is 130 pesos
The difference between the two equations is 4x=80, leading to x=20; that makes y=50-20=30
ANSWER: 20 5-peso coins and 30 1-peso coins
CHECK: 20(5)+30(1)=100+30=130
Informally (if an algebraic solution is not required, and if the speed of getting the answer is important)....
(1) imagine starting with all 50 coins being 1-peso coins; that would make the total 50 pesos, which is 80 pesos short of the actual total.
(2) exchanging a 1-peso coin for a 5-peso coin keeps the total number of coins the same but increases the total value by 5-1=4 pesos.
(3) 80/4 means there have to be 20 5-peso coins; and that means 30 1-peso coins
The words of explanation make this informal solution longer than the formal algebraic solution. But without all the words, here are the complete calculations for this method: