Joe has a collection of nickels and dimes that is worth $6.20. If the number of dimes was doubled and the number of nickels was decreased by 37, the value of the coins would be $8.85. How many nickels and dimes does he have?
original number of nickels : x
original number of dimes : y
5x + 10y = 620 or
x + 2y = 124 (#1)
new case:
number of nickels : 2y
number of dimes : x-37
5(x-37) + 10(2y) = 885
5x + 20y = 1070
x + 4y = 214 (#2)
subtract: #2 - #1
2y = 90
y = 45
sub into #1
x + 2(45) = 124
x = 34
He had 34 nickels and 45 dimes
check: 34nickels + 45 dimes = 170+450 = 620
case 2:
90 dimes + (-3nickels) = 900 - 15 = 885
Even though there is a mathematical solution , and the answer can be verified arithmetically, the answer
makes no real sense
So I would say there is no solution, and the problem as stated is flawed.
post
To solve this problem, we can set up a system of equations using the given information. Let's use the variables n for the number of nickels and d for the number of dimes.
From the problem statement, we can create two equations:
Equation 1: 0.05n + 0.10d = 6.20 (the value of the nickels and dimes is $6.20)
Equation 2: 0.05(n - 37) + 0.10(2d) = 8.85 (after doubling the number of dimes and decreasing the number of nickels by 37, the value is $8.85)
Now, we can solve this system of equations to find the values of n (number of nickels) and d (number of dimes).
From Equation 1:
0.05n + 0.10d = 6.20
Multiply both sides by 100 to get rid of decimals:
5n + 10d = 620
From Equation 2:
0.05(n - 37) + 0.10(2d) = 8.85
Multiply both sides by 100 to get rid of decimals:
5(n - 37) + 10(2d) = 885
5n - 185 + 20d = 885
5n + 20d = 1070
Now we have a system of equations:
5n + 10d = 620
5n + 20d = 1070
We can subtract the first equation from the second to eliminate n:
5n + 20d - (5n + 10d) = 1070 - 620
10d = 450
d = 45
Now, substitute the value of d (45) back into one of the original equations to solve for n:
5n + 10d = 620
5n + 10(45) = 620
5n + 450 = 620
5n = 620 - 450
5n = 170
n = 34
Therefore, Joe has 34 nickels and 45 dimes in his collection.