There are 60 coins in a box containing only nickels, dimes, and quarters. If there is one more than twice as many dimes as quarters, and the total value of all dimes is five cents less than the total value of all nickels, how many of each denomination are in the box?

n+d+q = 60

d = 2q+1
10d = 5n-5

Now go for it.

Got it. 8 quarters, 17 dimes, and 35 nickels. Thanks for the equations, that was what I had trouble setting up.

Let's solve this problem step by step.

Step 1: Let's assume the number of quarters in the box is "q".

Step 2: We are given that there are twice as many dimes as quarters, plus one more dime. So the number of dimes, "d", can be written as 2q + 1.

Step 3: We know there are 60 coins in total, so the number of nickels, "n", can be calculated as follows: n = 60 - (q + (2q + 1)) = 60 - 3q - 1 = 59 - 3q.

Step 4: The value of all the nickels can be calculated by multiplying the number of nickels by 5 cents: 5n = 5(59 - 3q) = 295 - 15q.

Step 5: We are also given that the total value of all dimes is five cents less than the total value of all nickels. So we can write the equation as follows: 10d = 5n - 5 = 295 - 15q - 5 = 290 - 15q.

Now, we can solve the equation.

Step 6: Substitute the value of "d" from Step 2 into the equation in Step 5:

10(2q + 1) = 290 - 15q.

Step 7: Simplify the equation:

20q + 10 = 290 - 15q.

Step 8: Combine like terms:

20q + 15q = 290 - 10.

Step 9: Simplify further:

35q = 280.

Step 10: Divide both sides of the equation by 35:

q = 280 / 35 = 8.

So, there are 8 quarters in the box.

Step 11: Substitute the value of "q" into the equation in Step 2 to find the number of dimes:

d = 2(8) + 1 = 17.

So, there are 17 dimes in the box.

Step 12: Substitute the value of "q" into the equation in Step 3 to find the number of nickels:

n = 59 - 3(8) = 59 - 24 = 35.

So, there are 35 nickels in the box.

In conclusion, there are 8 quarters, 17 dimes, and 35 nickels in the box.

To solve this problem, we need to set up a system of equations based on the given information.

Let's assign variables to the number of nickels, dimes, and quarters in the box. Let N represent the number of nickels, D represent the number of dimes, and Q represent the number of quarters.

Based on the given information, we can set up the following equations:

Equation 1: N + D + Q = 60 (since there are a total of 60 coins)

Equation 2: D = 2Q + 1 (since there is one more than twice as many dimes as quarters)

Equation 3: 10D = 5N - 5 (since the total value of all dimes is 5 cents less than the total value of all nickels)

Now we can solve the system of equations:

First, let's substitute Equation 2 into Equation 1:

N + (2Q + 1) + Q = 60

N + 3Q + 1 = 60

Next, let's solve Equation 3 for N:

10D = 5N - 5

Divide both sides by 5:

2D = N - 1

N = 2D + 1

Now let's substitute this new expression for N into Equation 1:

(2D + 1) + D + Q = 60

3D + 1 + Q = 60

Finally, let's substitute Equation 2 into this equation as well:

3(2Q + 1) + 1 + Q = 60

Simplify the equation:

6Q + 3 + 1 + Q = 60

Combine like terms:

7Q + 4 = 60

Subtract 4 from both sides:

7Q = 56

Divide both sides by 7:

Q = 8

Now we can substitute Q = 8 into Equation 2:

D = 2Q + 1

D = 2(8) + 1

D = 17

Finally, we can substitute Q = 8 and D = 17 into Equation 1 to find N:

N + D + Q = 60

N + 17 + 8 = 60

N + 25 = 60

N = 35

Therefore, there are 35 nickels, 17 dimes, and 8 quarters in the box.