A collection of dimes and quarters is worth $19.85. There are 34 more dime than quarters. Write a system of equations that can be used to find how many of each type of coin there are in the collection.
I'll work this one to give you an idea of how to do the others.
Let D = # of dimes and Q = # of quarters.
D = Q + 34
10D + 25Q = 1985
47 Quarters and 81 Dimes
Let's say the number of quarters in the collection is represented by 'q' and the number of dimes is represented by 'd'. We can write the following system of equations:
1. The value of the coins:
0.25q + 0.10d = 19.85
2. The number of dimes is 34 more than the number of quarters:
d = q + 34
So, the system of equations is:
0.25q + 0.10d = 19.85
d = q + 34
To write a system of equations, let's first assign variables to represent the number of dimes and quarters in the collection.
Let:
D = the number of dimes
Q = the number of quarters
Now, let's translate the problem information into equations:
1. The value of the collection:
The total value in cents can be expressed as:
Value of dimes (in cents) + Value of quarters (in cents) = Total value (in cents)
Since dimes are worth 10 cents and quarters are worth 25 cents, the equation becomes:
10D + 25Q = 1985 (where 1985 represents $19.85 in cents)
2. The number of dimes is 34 more than the number of quarters:
This can be expressed as:
D = Q + 34
Now we have a system of two equations:
10D + 25Q = 1985
D = Q + 34
These equations can be used to find the values of D and Q, representing the number of dimes and quarters, respectively, in the collection.