Question 670863: I am not sure how to figure out a coin problem without how much it's worth for example: A bank contains 44 coins (nickles, dimes, quarters). There are twice as many dimes as nickles and 8 fewer nickles than quarters. How much money is in the bank?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 44 coins in total.
n = number of nickels.
d = number of dimes
q = number of quarters.
there are twice as many dimes as nickels.
d = 2n
8 fewer nickels than quarters.
n = q - 8
how much money is in the bank.
the amount of money in the bank is determined as follows:
.05n = amount of money in nickels.
.1d = amount of money in dimes.
.25q = amount of money in quarters.
you have 2 equations.
n + d + q = 44
.05n + .1d + .25q = amount of money in the bank.
you need to solve for n, d, and q and then you can determine how much money is in the bank.
start with the equation of n + d + q = 44
you are given that d = 2n so you can substitute 2n for d which makes the equation becomes:
n + 2n + q = 44
you are given that n = q - 8 so you can substitute q - 8 for n which makes the equation become:
q - 8 + 2(q - 8) + q = 44
since you now have 1 unknown in 1 equation, you can solve for q.
simplify the equation to get:
q - 8 + 2q - 16 + q = 44
combine like terms to get:
4q - 24 = 44
add 24 to both sides of the equation to get:
4q = 44 + 24 = 68
divide both sides of the equation by 4 to get:
q = 68/4 = 17
since you know that n = q - 8, this means that n = 9
you have:
q = 17
n = 9
since you know that d = 2n, this means that d = 18
now you have:
q = 17
n = 9
d = 18
the first equation you had to solve is n + d + q = 44
replace q with 17 and n with 9 and d with 18 and you get:
9 + 18 + 17 = 44 which becomes 44 = 44 which is true so the numbers for n and d and q look good.
now to the money.
the equation is:
.05n + .1d + .25q = the money
replace n with 9 and d with 18 and q with 17 and you get:
.05(9) + .1(18) + .25(17) = .45 + 1.8 + 4.25 which is equal to 6.5 dollars that you have in the bank.
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