grant's change rack contained $8.80 in quarters, dimes, and nickels. There were two more than five times as many nickels as quarters and four less than twice as many dimes as quarters. How many of each kind of coin was there in the change rack?
please help
translate the words to symbols:
25q + 10d + 5n = 880
n = 5q+2
d = 2q-4
n = 67
d = 22
q = 13
Let's solve this problem step by step.
Let's assume the number of quarters as x.
According to the problem statement, there were two more than five times as many nickels as quarters. This can be represented as:
Number of nickels = 5x + 2
Also, it states that there were four less than twice as many dimes as quarters. This can be represented as:
Number of dimes = 2x - 4
Now, let's calculate the total value of the coins.
The value of each quarter is $0.25, so the total value of quarters is 0.25x.
The value of each dime is $0.10, so the total value of dimes is 0.10(2x - 4).
The value of each nickel is $0.05, so the total value of nickels is 0.05(5x + 2).
The total value of all coins is given as $8.80. So, we can write the equation as:
0.25x + 0.10(2x - 4) + 0.05(5x + 2) = 8.80
Now, let's solve this equation step by step to find the value of x, which represents the number of quarters.
Step 1: Distribute and simplify the equation:
0.25x + 0.20x - 0.40 + 0.25x + 0.10 = 8.80
Step 2: Combine like terms:
0.70x - 0.30 = 8.80
Step 3: Add 0.30 to both sides:
0.70x = 9.10
Step 4: Divide both sides by 0.70:
x = 13
Therefore, there are 13 quarters. To find the number of nickels and dimes, substitute the value of x into the formulas we derived earlier.
Number of nickels = 5x + 2 = 5(13) + 2 = 65 + 2 = 67
Number of dimes = 2x - 4 = 2(13) - 4 = 26 - 4 = 22
Therefore, there are 13 quarters, 67 nickels, and 22 dimes in the change rack.
To solve this problem, we need to use algebraic equations to represent the given information and find the solution. Let's break down the information into equations step by step.
Let's assume the number of quarters as "q," the number of dimes as "d," and the number of nickels as "n."
1) "There were two more than five times as many nickels as quarters."
This can be expressed as: n = 5q + 2
2) "Four less than twice as many dimes as quarters."
This can be expressed as: d = 2q - 4
Now, let's move on to the values of the coins:
The value of one quarter is $0.25, the value of one dime is $0.10, and the value of one nickel is $0.05.
3) "Grant's change rack contained $8.80 in quarters, dimes, and nickels."
The total value of the quarters, dimes, and nickels can be calculated as:
0.25q + 0.10d + 0.05n = 8.80
Now, we have a system of three equations:
n = 5q + 2 (Equation 1)
d = 2q - 4 (Equation 2)
0.25q + 0.10d + 0.05n = 8.80 (Equation 3)
To solve this system of equations, we can use the substitution method or the elimination method.
Let's use the substitution method:
From Equation 2, we can write:
d = 2q - 4
Substitute this value of "d" into Equation 3:
0.25q + 0.10(2q - 4) + 0.05n = 8.80
Simplify the equation:
0.25q + 0.20q - 0.40 + 0.05n = 8.80
0.45q + 0.05n = 9.20 (Equation 4)
Now we have two equations:
n = 5q + 2 (Equation 1)
0.45q + 0.05n = 9.20 (Equation 4)
To solve this system of equations, we can substitute Equation 1 into Equation 4:
0.45q + 0.05(5q + 2) = 9.20
Simplify the equation:
0.45q + 0.25q + 0.10 = 9.20
0.70q + 0.10 = 9.20
0.70q = 9.10
q = 9.10 / 0.70
q ≈ 13
Now, we have the value of "q." Substitute this value into Equation 1 to find "n":
n = 5(13) + 2
n = 65 + 2
n = 67
Finally, substitute the values of "q" and "n" into Equation 2 to find "d":
d = 2(13) - 4
d = 26 - 4
d = 22
Therefore, there are approximately 13 quarters, 22 dimes, and 67 nickels in Grant's change rack.