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

Answer:

25 dimes

Step-by-step 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

$2Q = 41 - D$

$Q = \frac{41}{2} - \frac{D}{2}$

$Q = 20.5 - 0.50D$

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

$D = \frac{-1.25}{-0.05}$

$D = 25$

Hence, there are 25 Dimes