Joe has a collection of nickels and dimes that is worth $6.05 if the number of dimes was double and the number of nickels was decreased by 10 the value of the coins would be $9.85 how many nickels and dimes does he have?
case 1:
let the number of nickels be n
let the number of dimes be d
5n + 10d = 605
n + 2d = 121
case 2
number of nickels ---> n - 10
number of dimes ---> 2d
5(n-10) + 10(2d) = 985
5n - 50 + 20d = 985
5n + 20d = 1035
n + 4d = 207
subtract them
2d = 86
d = 43 then n = 35
He has 43 dimes and 35 nickels
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To solve this problem, we can use a system of equations. Let's assign variables for the number of nickels and dimes Joe has.
Let's say that Joe has "n" nickels and "d" dimes.
From the given information, we can establish the following equations:
Equation 1: The value of Joe's collection is $6.05.
0.05n + 0.10d = 6.05
Equation 2: If the number of dimes was double and the number of nickels was decreased by 10, the value of the coins would be $9.85.
0.05(n - 10) + 0.10(2d) = 9.85
Now let's solve this system of equations:
First, simplify Equation 2:
0.05n - 0.50 + 0.20d = 9.85
0.05n + 0.20d = 10.35
Next, multiply both sides of Equation 1 by 100 to eliminate the decimals:
5n + 10d = 605
Now we have two equations:
5n + 10d = 605 (Equation 3)
0.05n + 0.20d = 10.35 (Equation 4)
We can multiply Equation 4 by 20 to eliminate the decimals:
n + 4d = 207 (Equation 5)
Now we have a system of two equations:
5n + 10d = 605 (Equation 3)
n + 4d = 207 (Equation 5)
We can solve this system of equations to find the values of "n" and "d."
Multiplying Equation 5 by -5, we get:
-5n - 20d = -1035 (Equation 6)
Adding Equation 3 and Equation 6, we cancel out the "n" terms:
(5n + 10d) + (-5n - 20d) = (605 - 1035)
Simplifying, we have:
-10d = -430
Dividing both sides by -10, we get:
d = 43
Plugging the value of "d" back into Equation 5:
n + 4 * 43 = 207
n + 172 = 207
n = 207 - 172
n = 35
Therefore, Joe has 35 nickels and 43 dimes.