Ike's piggy bank contained only quarters and nickles. When he counted the money there was $21.65. If there were 121 more nickles than quarters, how many of each type of coin did Ike have in the bank?
number of quarters ---- x
number of nickels ------ x + 121
25x + 5(x+121) = 2165
solve for x
To solve this problem, we can set up a system of equations. Let's represent the number of quarters as "q" and the number of nickels as "n".
From the information given in the problem, we know that the value of quarters is $0.25 (25 cents) each, and the value of nickels is $0.05 (5 cents) each. We can use this information to create our first equation:
0.25q + 0.05n = 21.65
Now, let's incorporate the second piece of information given, which states that there are 121 more nickels than quarters:
n = q + 121
We now have a system of two equations:
Equation 1: 0.25q + 0.05n = 21.65
Equation 2: n = q + 121
To solve this system, we can use substitution or elimination.
Let's start by substituting Equation 2 into Equation 1:
0.25q + 0.05(q + 121) = 21.65
Simplifying the equation, we get:
0.25q + 0.05q + 6.05 = 21.65
Combining like terms:
0.30q + 6.05 = 21.65
Subtracting 6.05 from both sides of the equation:
0.30q = 15.60
Dividing both sides by 0.30:
q = 52
Now, we can substitute the value of q into Equation 2 to find the value of n:
n = 52 + 121
n = 173
Therefore, Ike has 52 quarters and 173 nickels in his piggy bank.