I have 50 coins, some 10c and the rest 20c. Altogether they total $8.20. How many of
each coin do I have?
Set it up as two equations
10c = P and 20c=Q
so P+Q=50 or P=50-Q
10P+20Q=820
10(50-Q)+20Q=820
500-10Q+20Q=820
10Q=320 so Q=32
hence P=50-32=18
check
(18x10)+(32x20)=180+640=820
To solve this problem, you can use a system of equations. Let's represent the number of 10 cent coins as x and the number of 20 cent coins as y.
We are given two pieces of information:
1. You have a total of 50 coins: x + y = 50.
2. The total value of the coins is $8.20: 0.10x + 0.20y = 8.20.
To solve the system of equations, you can use either the substitution method or the elimination method. Let's use the elimination method:
Since we have an equation for x in terms of y from the first equation (x = 50 - y), we can substitute this value into the second equation:
0.10(50 - y) + 0.20y = 8.20.
Now, let's simplify and solve for y:
5 - 0.10y + 0.20y = 8.20,
0.10y = 8.20 - 5,
0.10y = 3.20,
y = 3.20 / 0.10,
y = 32.
So, you have 32 coins of 20 cents each.
Now substitute this value back into one of the original equations to solve for x:
x + 32 = 50,
x = 50 - 32,
x = 18.
Therefore, you have 18 coins of 10 cents each and 32 coins of 20 cents each.