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Saul paid a bill of 4.60 with nickels, dimes, and quarters. The number of nickels was 3 less than the number of dimes. The number of dimes was 5 more then the number of quarters. How many coins of each type did he use?

User Ola
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1 Answer

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Let's solve this problem step by step.

Let's denote the number of nickels as N, the number of dimes as D, and the number of quarters as Q.

From the given information, we know that:

1. Saul paid a total bill of $4.60, which can be represented as 460 cents.

2. The number of nickels was 3 less than the number of dimes, so N = D - 3.

3. The number of dimes was 5 more than the number of quarters, so D = Q + 5.

Next, let's convert the values of nickels, dimes, and quarters into cents:

1 nickel = 5 cents

1 dime = 10 cents

1 quarter = 25 cents

Given that Saul paid $4.60 in total, we can write the equation:

5N + 10D + 25Q = 460

Now, substitute the values of N and D based on the given information:

5(D - 3) + 10D + 25Q = 460

5D - 15 + 10D + 25Q = 460

15D + 25Q = 475

Since we have two variables (D and Q), we need another equation to solve the system of equations. Using the information that D = Q + 5, we can substitute D in the equation above:

15(Q + 5) + 25Q = 475

15Q + 75 + 25Q = 475

40Q = 400

Q = 10

Plugging the value of Q back into the equation D = Q + 5:

D = 10 + 5

D = 15

Now, we can substitute the values of D and Q back into the equation N = D - 3:

N = 15 - 3

N = 12

So, Saul used 12 nickels, 15 dimes, and 10 quarters to pay the bill.

NICE

User OppfinnarJocke
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