in a group of quarters and nickels, there are four more nickels than quarters. how many nickels and quarters are there if the coins are worth $2.30?
To solve this problem, we need to set up a system of equations.
Let's assume the number of quarters in the group is represented by 'q', and the number of nickels is represented by 'n'.
According to the problem, there are four more nickels than quarters. So, we can write the equation:
n = q + 4 ---(Equation 1)
The value of each quarter is $0.25 and the value of each nickel is $0.05. Since the total value of all the coins is $2.30, we can write another equation:
0.25q + 0.05n = 2.30 ---(Equation 2)
Now we have a system of equations:
n = q + 4 ---(Equation 1)
0.25q + 0.05n = 2.30 ---(Equation 2)
To solve this system of equations, we can use the substitution method or the elimination method.
Let's use the substitution method.
From Equation 1, we can rewrite it as:
q = n - 4
Now substitute this value of 'q' in Equation 2:
0.25(n - 4) + 0.05n = 2.30
Simplify the equation:
0.25n - 1 + 0.05n = 2.30
Combine like terms:
0.30n - 1 = 2.30
Add 1 to both sides:
0.30n = 3.30
Divide both sides by 0.30:
n = 11
Now substitute this value of 'n' back into Equation 1 to find 'q':
q = 11 - 4
q = 7
Therefore, there are 7 quarters and 11 nickels in the group.
quarters ---- x
nickels ----- x + 4
25x + 5(x+4) = 230
solve for x