SOLUTION: There are 237 coins in a bag that total $52.90. There are nickels and dimes in the bad. How many of each are in the bag?
This is how I set up the problem, but I keep getting eit
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This is how I set up the problem, but I keep getting eit
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Question 276683: There are 237 coins in a bag that total $52.90. There are nickels and dimes in the bad. How many of each are in the bag?
This is how I set up the problem, but I keep getting either negative numbers or really large numbers for my answer :-/
n + d = 237
0.05n + 0.10d = 52.90
n = 237 - d
0.05(237 - d) + 0.10d = 52.90
11.85 - 0.05d + 0.10d = 52.90
11.85 + 0.05d = 52.90
0.05d = 41.05
d = 821????
n = -584????
I tried using the substitution method, but it doesn't seem to work out when I do it. Answer by solver91311(24713) (Show Source):
That's because your problem has no solution. If you have 237 coins that are some mixture of either nickels or dimes, the least amount of money you could have would be $11.85 (237 nickels and 0 dimes) and the most you could have would be $23.70 (0 nickels and 237 dimes). Hence, there is no way that the total amount of money could be as much as $52.90.
