Solve
A collection of dimes and quarters is worth $7.65.
There are 45 coins in all . How many of each are there?
d+q = 45
10d+25q = 765
24 dimes, 21 quarters
To solve this problem, you can use a system of equations. Let's represent the number of dimes as 'd' and the number of quarters as 'q'.
We know that the value of dimes is $0.10 each and the value of quarters is $0.25 each.
The total value of the dimes can be calculated as 0.10d and the total value of the quarters as 0.25q.
We're given two pieces of information:
1. The total value of the collection is $7.65, which can be written as the equation 0.10d + 0.25q = 7.65.
2. There are a total of 45 coins, so the number of dimes plus the number of quarters is 45. This gives us the equation d + q = 45.
To solve the system of equations, we can use substitution or elimination. Let's use the elimination method.
We can multiply the second equation by 0.10 to match the coefficients of 'd' in both equations:
0.10d + 0.10q = 4.50
Now we can subtract this new equation from the first equation:
(0.10d + 0.25q) - (0.10d + 0.10q) = 7.65 - 4.50
0.10d + 0.25q - 0.10d - 0.10q = 3.15
0.15q = 3.15
To isolate 'q', divide both sides of the equation by 0.15:
q = 3.15 / 0.15
q ≈ 21
Now that we know the value of 'q', we can substitute it back into one of the original equations to find the value of 'd'. Let's use the equation d + q = 45:
d + 21 = 45
Subtract 21 from both sides of the equation:
d = 45 - 21
d = 24
Therefore, there are 24 dimes and 21 quarters in the collection.