Problem: solve word problem using polynomials
Joe has a collection of nickles and dimes worth $2.15 If the number of dimes was doubled and the number of nickles was increased by 28, the value of the coins would be $4.65. How many nickles and dimes does he have?
Let's say he has n nickles and d dimes.
0.05n + 0.10d = 2.15
The problem proposes to increase d to 2d and increase n to n + 28.
0.05(n + 28) + 0.10(2d) = 4.65
We have two equation to work with. Let's multiply both equations
by 100 to eliminate the decimals.
100(0.05n + 0.10d) = 2.15 * 100
5n + 10d = 215
100(0.05(n + 28) + 0.10(2d)) = 4.65 * 100
5(n + 28) + 10(2d) = 465
We need to expand this second equation.
5(n + 28) + 10(2d) = 465
5n + 140 + 20d = 465
5n + 140 + 20d - 140 = 465 - 140
5n + 20d = 325
Here's what we have:
1) 5n + 10d = 215
2) 5n + 20d = 325
We can immediately eliminate the n by subtracting equatioin 1 from equation 2.
5n + 20d = 325
-(5n + 10d = 215)
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10d = 110
10d = 110
10d/10 = 110/10
d = 11
This tells us that Joe currently has 11 dimes. Let's use equation 1 directly above
to solve for n.
5n + 10d = 215
5n + 10(11) = 215
5n + 110 = 215
5n + 110 - 110 = 215 - 110
5n = 105
5n/5 = 105/5
n = 21
This one tells us that Joe currently has 21 nickels. We need to check the two
numbers against what the problem originally stated.
0.05n + 0.10d = 2.15
0.05(21) + 0.10(11) = 2.15
1.05 + 1.10 = 2.15
2.15 = 2.15
Verified!
Answer: Joe has 21 nickels and 11 dimes.
