Gail has a collection of nickels, dimes, and quarters that is worth $7.95 altogether. If she has three more nickels than dimes, and 38 coins altogether, then how many of each kind of coin does she have?
dimes -- x
nickels -- x+3
quarters --- 38-x-(x+3) = 35-2x
10x + 5(x+3) + 25(35-2x) = 795
10x + 5x + 15 + 875 - 50x = 795
-35x = -80
no integer solution, so the question is flawed.
check my arithmetic, and check your typing
To solve this problem, we can use a system of equations.
Let's assign variables to represent the number of nickels, dimes, and quarters that Gail has. Let:
- N be the number of nickels
- D be the number of dimes
- Q be the number of quarters
According to the problem, Gail has three more nickels than dimes, so we can write the equation: N = D + 3 --- Equation 1
The second piece of information states that the total number of coins Gail has is 38, so the equation is: N + D + Q = 38 --- Equation 2
Next, we need to convert the values of the coins into cents since $7.95 is in cents. A nickel is 5 cents, a dime is 10 cents, and a quarter is 25 cents.
The value of N nickels in cents is 5N.
The value of D dimes in cents is 10D.
The value of Q quarters in cents is 25Q.
According to the problem, the total value of the coins is $7.95, which is equivalent to 795 cents. So, the equation is: 5N + 10D + 25Q = 795 --- Equation 3
Now we have a system of three equations:
N = D + 3 --- Equation 1
N + D + Q = 38 --- Equation 2
5N + 10D + 25Q = 795 --- Equation 3
To solve this system of equations, we can substitute the value of N from Equation 1 into Equations 2 and 3.
Substituting N = D + 3 in Equation 2, we get: (D + 3) + D + Q = 38
Simplifying, we have: 2D + Q = 35 --- Equation 4 (after rearranging)
Substituting N = D + 3 in Equation 3, we get: 5(D + 3) + 10D + 25Q = 795
Simplifying, we have: 15D + 25Q = 780 --- Equation 5 (after simplifying and rearranging)
Now we have a new system of equations:
2D + Q = 35 --- Equation 4
15D + 25Q = 780 --- Equation 5
Solving this system will give us the values for D (dimes) and Q (quarters). Let's solve it:
From Equation 4, we can express Q in terms of D: Q = 35 - 2D.
Substituting this into Equation 5, we get: 15D + 25(35 - 2D) = 780
Simplifying, we have: 15D + 875 - 50D = 780
Combining like terms, we get: -35D = -95
Dividing both sides by -35, we find: D = 3
Now that we have the value for D, we can substitute it back into Equation 4 to find Q:
2(3) + Q = 35
6 + Q = 35
Subtracting 6 from both sides, we get: Q = 29
Finally, substituting the values of D (3) and Q (29) into Equation 1, we can solve for N:
N = D + 3 = 3 + 3 = 6
So, Gail has 6 nickels, 3 dimes, and 29 quarters.