A jar contains 40 coins consisting of dimes and quarters and having a total value of $4.90. How many of each kind if coins are there?
number of dimes --- x
number of quarters --- 40-x
10x + 25(40-x) = 490
solve for x, then sub into my definitions.
34 dimes
Let's assume the number of dimes is represented by 'd' and the number of quarters is represented by 'q'.
We can write two equations based on the given information:
1. The total number of coins equation: d + q = 40
2. The total value equation: 0.10d + 0.25q = 4.90
To solve this system of equations, we can use the substitution method.
First, solve the first equation for 'd' in terms of 'q':
d = 40 - q
Now substitute this value of 'd' into the second equation:
0.10(40 - q) + 0.25q = 4.90
Distribute the 0.10:
4 - 0.10q + 0.25q = 4.90
Combine like terms:
0.15q + 4 = 4.90
Subtract 4 from both sides:
0.15q = 0.90
Divide both sides by 0.15:
q = 6
Now substitute this value of 'q' back into the first equation to find 'd':
d + 6 = 40
d = 40 - 6
d = 34
Therefore, there are 34 dimes and 6 quarters in the jar.
To solve this problem, we need to set up a system of equations based on the given information. Let's use the following variables:
Let x be the number of dimes.
Let y be the number of quarters.
We have two pieces of information from the problem:
1. There are a total of 40 coins:
x + y = 40
2. The total value of all the coins is $4.90:
0.10x + 0.25y = 4.90
Now, we can solve this system of equations using substitution or elimination method.
Let's use the elimination method:
Multiply the first equation by 0.10 to get rid of decimals:
0.10x + 0.10y = 4
Now, subtract the new equation from the second equation to eliminate x:
0.10x + 0.25y - (0.10x + 0.10y) = 4.90 - 4
0.15y - 0.10y = 0.90
0.05y = 0.90
y = 0.90 / 0.05
y = 18
Now, substitute the value of y back into the first equation to solve for x:
x + y = 40
x + 18 = 40
x = 40 - 18
x = 22
So, there are 22 dimes and 18 quarters.