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?

Answers

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:

[tex]0.05N + 0.10D + 0.25Q = 5.25[/tex]

The number of coin implies:

[tex]N + D + Q = 48[/tex]

and

[tex]N = 7 + Q[/tex]

Required

Determine the number of Dimes

[tex]0.05N + 0.10D + 0.25Q = 5.25[/tex]

[tex]N + D + Q = 48[/tex]

[tex]N = 7 + Q[/tex]

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

[tex]0.05N + 0.10D + 0.25Q = 5.25[/tex]

[tex]0.05(7 + Q) + 0.10D + 0.25Q = 5.25[/tex]

[tex]0.35 + 0.05Q+ 0.10D + 0.25Q = 5.25[/tex]

Collect Like Terms

[tex]0.10D + 0.25Q+0.05Q = 5.25 - 0.35[/tex]

[tex]0.10D + 0.30Q = 4.90[/tex] ----- (1)

[tex]N + D + Q = 48[/tex]

[tex]7 + Q + D + Q = 48[/tex]

Collect Like Terms

[tex]D + Q + Q= 48 - 7[/tex]

[tex]D + 2Q= 41[/tex]

Make Q the subject

[tex]2Q = 41 - D[/tex]

[tex]Q = \frac{41}{2} - \frac{D}{2}[/tex]

[tex]Q = 20.5 - 0.50D[/tex]

Substitute 20.5 - 0.50D for Q in  (1)

[tex]0.10D + 0.30Q = 4.90[/tex]

[tex]0.10D + 0.30(20.5 - 0.50D) = 4.90[/tex]

[tex]0.10D + 0.30*20.5 - 0.30*0.50D = 4.90[/tex]

[tex]0.10D + 6.15 - 0.15D = 4.90[/tex]

[tex]0.10D - 0.15D = 4.90 - 6.15[/tex]

[tex]-0.05D = -1.25[/tex]

Solve for D

[tex]D = \frac{-1.25}{-0.05}[/tex]

[tex]D = 25[/tex]

Hence, there are 25 Dimes


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Answers

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Step-by-step explanation:

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Answers

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Answers

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Step-by-step explanation:

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Step-by-step explanation:

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Answers

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Answer:

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