SOLUTION: I have $2.35 in nickles and dimes . If i have a total of 23 coins , how many of each coin do i have ?

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Question 251225: I have $2.35 in nickles and dimes . If i have a total of 23 coins , how many of each coin do i have ?
Answer by oberobic(2304) (Show Source):
You can put this solution on YOUR website!
In solving coin problems, you have to keep track of the number of coins and the value of the coins.
n = number of nickels
5n = value of the nickels in cents
d = number of dimes
10d = value of the dimes in cents
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We are told n+d = 23 coins.
We are told that 5n + 10d = 235 cents.
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We can solve these two equations in several ways. An obvious approach is to set n = 23-d and then substitute.
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5(23-d) + 10d = 235
115 - 5d + 10d = 235
5d = 120
d = 24
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n + d = 23
n = -1
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Checking our values:
5n = 5*-1 = -5 cents
24d = 10*24 = 240 cents
Adding: 235 cents.
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But this is a nonsensical answer. You cannot have -1 nickels.
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Using a 'brute force' approach, it is easy to see that the maximum value of 23 coins occurs with 22 dimes and 1 nickel...and it is 225 cents ($2.25), which is less than the stated amount.
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n d n+d 5n 10d 5n+10d 1 22 23 5 220 225 2 21 23 10 210 220 3 20 23 15 200 215 4 19 23 20 190 210 5 18 23 25 180 205 6 17 23 30 170 200 7 16 23 35 160 195 8 15 23 40 150 190 9 14 23 45 140 185 10 13 23 50 130 180 11 12 23 55 120 175 12 11 23 60 110 170 13 10 23 65 100 165 14 9 23 70 90 160 15 8 23 75 80 155 16 7 23 80 70 150 17 6 23 85 60 145 18 5 23 90 50 140 19 4 23 95 40 135 20 3 23 100 30 130 21 2 23 105 20 125 22 1 23 110 10 120

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So there is no solution to the problem.