In a collection of nickels, dimes and quarters, there are twice as many dimes as nickels, and 4 fewer quarters than dimes. The total value of the coins is $5.75.
How many nickels are there in the collection ?
Dealing with NUMBER of coins:
nickels ---- x
dimes ----- 2x
quarters ---- 2x - 4
dealing with VALUE of our coins:
5x + 10(2x) + 25(2x-4) = 575
solve for x and back-substitute
To solve this problem, we can use a system of equations. Let's start by assigning variables to the unknowns:
Let N represent the number of nickels.
Let D represent the number of dimes.
Let Q represent the number of quarters.
According to the problem:
1) There are twice as many dimes as nickels: D = 2N.
2) There are 4 fewer quarters than dimes: Q = D - 4.
3) The total value of the coins is $5.75: 0.05N + 0.10D + 0.25Q = 5.75.
Now we can substitute the variables in equation 3 with the given information from equations 1 and 2:
0.05N + 0.10(2N) + 0.25(D - 4) = 5.75.
Simplifying this equation, we get:
0.05N + 0.20N + 0.25D - 1 = 5.75.
Combining like terms, we have:
0.25N + 0.25D = 6.75.
Dividing both sides of the equation by 0.25, we get:
N + D = 27.
Now we can use equation 1 to express D in terms of N:
D = 2N.
Substituting this expression into the equation N + D = 27, we have:
N + 2N = 27.
Combining like terms, we get:
3N = 27.
Dividing both sides of the equation by 3, we get:
N = 9.
Therefore, there are 9 nickels in the collection.