SOLUTION: Colin has pennies, nickels and dimes in his piggybank. He has twice as many nickels as pennies, and he has ten more dimes than nickels. If he has 165 coins altogether, how many dim
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-> SOLUTION: Colin has pennies, nickels and dimes in his piggybank. He has twice as many nickels as pennies, and he has ten more dimes than nickels. If he has 165 coins altogether, how many dim
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Question 609354: Colin has pennies, nickels and dimes in his piggybank. He has twice as many nickels as pennies, and he has ten more dimes than nickels. If he has 165 coins altogether, how many dimes does he have? Answer by SwiftAlbatross(13) (Show Source):
You can put this solution on YOUR website! Let p = pennies
Let n = nickels
Let d = dimes
Twice as many nickels as pennies:
n = 2p
Ten more dimes than nickels:
d = n + 10
165 coins altogether:
p + n + d = 165
Now that you have 3 equations and 3 variables, you can solve:
n = 2p
d = n + 10
p + n + d = 165
Substitute the n value into the other two equations:
d = 2p + 10
p + 2p + d = 165 ... which simplifies to ---> 3p + d = 165
You have reduced the number of equations and variables to 2. Keep going:
Substitute d into the second equation:
3p + 2p + 10 = 165
5p + 10 = 165
5p = 155
p = 31
Then plug p back into the other equation
d = 2(31) + 10
d = 62 + 10
d = 72
Then plug p and d back into one of the original equations:
p + n + d = 165
31 + n + 72 = 165
103 + n = 165
n = 62
Check:
n = 2p
62 = 2(31)
d = n + 10
72 = 62 + 10
p + n + d = 165
31 + 62 + 72 = 165
You're good to go. ^_^
