Question

Shelley has a collection of dimes, nickels and quarters that is worth $5.25. There are a total of 48 coins. If there are 7 more nickels than quarters, how many dimes are there?

Answer 1

25 dimes

Explanation:

Given

Coins: Nickels (N), Quarters (Q) , and Dimes (D)

$0.05 = 1 nickel, $0.10 = 1 dime and $0.25 = 1 quarter

So, the amount implies:

`0.05N + 0.10D + 0.25Q = 5.25`

The number of coin implies:

`N + D + Q = 48`

and

`N = 7 + Q`

Required

Determine the number of Dimes

`0.05N + 0.10D + 0.25Q = 5.25`

`N + D + Q = 48`

`N = 7 + Q`

Substitute 7 + Q for N in the first and second equation

`0.05N + 0.10D + 0.25Q = 5.25`

`0.05(7 + Q) + 0.10D + 0.25Q = 5.25`

`0.35 + 0.05Q+ 0.10D + 0.25Q = 5.25`

Collect Like Terms

`0.10D + 0.25Q+0.05Q = 5.25 - 0.35`

`0.10D + 0.30Q = 4.90` ----- (1)

`N + D + Q = 48`

`7 + Q + D + Q = 48`

Collect Like Terms

`D + Q + Q= 48 - 7`

`D + 2Q= 41`

Make Q the subject

`2Q = 41 - D`

`Q = (41)/(2) - (D)/(2)`

`Q = 20.5 - 0.50D`

Substitute 20.5 - 0.50D for Q in (1)

`0.10D + 0.30Q = 4.90`

`0.10D + 0.30(20.5 - 0.50D) = 4.90`

`0.10D + 0.30*20.5 - 0.30*0.50D = 4.90`

`0.10D + 6.15 - 0.15D = 4.90`

`0.10D - 0.15D = 4.90 - 6.15`

`-0.05D = -1.25`

Solve for D

`D = (-1.25)/(-0.05)`

`D = 25`

Hence, there are 25 Dimes