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10 coins consistinig of nickels and dimes equal 80 cents how many coins does she have?
Wiki User
∙ 14y ago
You could (a) use trial and error; (b) write two equations; or (c) call the number of nickels "n", and the number of dimes "10-n" (since there are 10 coins in total). I will use the latter approach.
Number of nickels: n
Number of dimes: 10-n
For the main equation, multiply the number of coins by the value of each coin:
value of nickels + value of dimes = 80
5n + 10(10-n) = 80
Now solve the equation:
5n + 100 - 10n = 80
-5n = -20
5n = 20
n = 4
So, you have 4 nickels, and 6 dimes.
You could (a) use trial and error; (b) write two equations; or (c) call the number of nickels "n", and the number of dimes "10-n" (since there are 10 coins in total). I will use the latter approach.
Number of nickels: n
Number of dimes: 10-n
For the main equation, multiply the number of coins by the value of each coin:
value of nickels + value of dimes = 80
5n + 10(10-n) = 80
Now solve the equation:
5n + 100 - 10n = 80
-5n = -20
5n = 20
n = 4
So, you have 4 nickels, and 6 dimes.
You could (a) use trial and error; (b) write two equations; or (c) call the number of nickels "n", and the number of dimes "10-n" (since there are 10 coins in total). I will use the latter approach.
Number of nickels: n
Number of dimes: 10-n
For the main equation, multiply the number of coins by the value of each coin:
value of nickels + value of dimes = 80
5n + 10(10-n) = 80
Now solve the equation:
5n + 100 - 10n = 80
-5n = -20
5n = 20
n = 4
So, you have 4 nickels, and 6 dimes.
You could (a) use trial and error; (b) write two equations; or (c) call the number of nickels "n", and the number of dimes "10-n" (since there are 10 coins in total). I will use the latter approach.
Number of nickels: n
Number of dimes: 10-n
For the main equation, multiply the number of coins by the value of each coin:
value of nickels + value of dimes = 80
5n + 10(10-n) = 80
Now solve the equation:
5n + 100 - 10n = 80
-5n = -20
5n = 20
n = 4
So, you have 4 nickels, and 6 dimes.
Wiki User
∙ 14y ago
Wiki User
∙ 14y agoYou could (a) use trial and error; (b) write two equations; or (c) call the number of nickels "n", and the number of dimes "10-n" (since there are 10 coins in total). I will use the latter approach.
Number of nickels: n
Number of dimes: 10-n
For the main equation, multiply the number of coins by the value of each coin:
value of nickels + value of dimes = 80
5n + 10(10-n) = 80
Now solve the equation:
5n + 100 - 10n = 80
-5n = -20
5n = 20
n = 4
So, you have 4 nickels, and 6 dimes.
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