A cashier has 25 coins consisting of nickels, dimes, and quarters with a value of $4.10. If the number of dimes is 1 more than twice the number of nickels, how many of each type of coin does he have?
Ok, your turn.
For the first two, I have given you the equation
What have you done for these last two?
Show me your preliminary steps.
n+d+q=25
n+d(1+2n)+q=4.10
I'm not sure if I am doing this right
no, you are just writing stuff down.
based on what it said:
number of nickels --- x
number of dimes---- 2x+1
number of quarters = 25-(x + 2x+1) = 24 - 3x
now form a "value" equation:
5x + 10(2x+1) + 25(24-3x) = 410
To solve this problem, we can set up a system of equations.
Let's first assign variables to represent the number of nickels, dimes, and quarters. Let's say:
- N = number of nickels
- D = number of dimes
- Q = number of quarters
We can now translate the given information into equations:
1) The cashier has a total of 25 coins: N + D + Q = 25
2) The total value of the coins is $4.10: 0.05N + 0.10D + 0.25Q = 4.10
3) The number of dimes is 1 more than twice the number of nickels: D = 2N + 1
Now we have a system of three equations. We can solve this system to find the values of N, D, and Q.
First, let's rearrange equation 3 to isolate N:
2N = D - 1
N = (D - 1) / 2
Now we can substitute this expression for N in equations 1 and 2, resulting in:
(D - 1) / 2 + D + Q = 25 (equation 1)
0.05((D - 1) / 2) + 0.10D + 0.25Q = 4.10 (equation 2)
Now we have a system of two equations with two variables (D and Q). We can solve this system using algebra or another method like substitution or elimination.
Simplifying equation 1:
D/2 - 1/2 + D + Q = 25
2D + Q = 24 + 1/2
Now simplifying equation 2:
0.05(D - 1)/2 + 0.10D + 0.25Q = 4.10
0.05(D - 1) + 0.20D + 0.25Q = 4.10
0.05D - 0.05 + 0.20D + 0.25Q = 4.10
0.25D + 0.25Q = 4.10 + 0.05
Now we have a system of two equations:
2D + Q = 24.5 (equation 3)
0.25D + 0.25Q = 4.15 (equation 4)
Now we can solve this system using substitution or elimination to find the values of D and Q.