Question 360319: A piggy bank contains $2 made from 24 coins which are nickels, dimes and quarters. What combinations of coins are possible?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 2 dollars = 200 cents.
1 dime = 10 cents
1 nickel = 5 cents
1 quarter = 25 cents.
let:
d = number of dimes
n = number of nickels
q = number of quarters
you will wind up with 2 equations that need to be solved simultaneously.
n + d + q = 24 (number of coins)
5n + 10d + 25q = 200 (number of coins times the value of each coin)
multiply the first equation by 5 to get:
5n + 5d + 5q = 120 (first equation multiplied by 5)
5n + 10d + 25q = 200 (second equation)
subtract first equation from second equation to get:
5d + 20q = 80
subtract 20q from both sides of the equation to get:
5d = 80 - 20q
divide both sides of the equation by 5 to get:
d = 16 - 4q
go back to your original equations and substitute for d.
original equations are:
n + d + q = 24 (number of coins)
5n + 10d + 25q = 200 (number of coins times the value of each coin)
substitute for d to get:
n + (16-4q) + q = 24
5n + 10*(16-4q) + 25q = 200
simplify to get:
n + 16 - 4q + q = 24
5n + 160 - 40q + 25q = 200
combine like terms to get:
n + 16 - 3q = 24
5n + 160 - 15q = 200
subtract 16 from both sides of the first equation, and subtract 160 from both sides of the second equation, to get:
n - 3q = 8
5n - 15q = 40
multiply both sides of first equation by 5 to get:
5n - 15q = 40
5n - 15q = 40
subtract second equation from first equation to get:
0 = 0
since this is true, then there appears to be any number of solutions to this equation as long as d = 16 - 4q provides us with a valid answer.
if we let q = 0, then we get d = 16
if we let q = 1, then we get d = 12
if we let q = 2, then we get d = 8
if we let q = 3, then we get d = 4
if we let q = 4, then we get d = 0
those are the only valid solutions because (d or q) can't be negative.
d is the number of dimes and q is the number of quarters.
since n + d + q = 24 from our first equation, then we get:
1 = 0, d = 16, n = 8, and 25q + 10d + 5n = 0 + 160 + 40 = 200
q = 1, d = 12, n = 11, and 25q + 10d + 5n = 25 + 120 + 55 = 200
q = 2, d = 8, n = 14, and we get 25q + 10d + 5n = 50 + 80 + 70 = 200
q = 3, d = 4, n = 17, and we get 25q + 10d + 5n = 75 + 40 + 85 = 200
q = 4, d = 0, n = 20, and we get 25q + 10d + 5n = 100 + 100 = 200
there are other combinations possible to get to 200 cents, but those combinations do not get us 24 coins in total.
for example:
1 quarter and 1 dime and 33 nickels add up to 5 + 10 + 165 = 200 cents but the number of coins add up to 35.
while this satisfies the value equation (has to add up to 200), it doesn't satisfy the number of coins equation (has to add up to 24).
value = 200 (200 is satisfied), number of coins = 35 (24 is not satisfied).
both equations have to be satisfied, not just one.
the fact that your final 2 equations resolved to 0 = 0 meant that you have any number of possible solutions as long as d = 16 - q provides you with a valid answer, meaning that the number of dimes has to be positive and the number of quarters has to be positive.
that's because 0 = 0 is true.
if those last 2 equations resolved to something like 0 = 15, then you would have had no no possible number of solutions because 0 = 15 is false (not a true statement).
the fact that you had 2 equations in 3 unknowns means that you had to get the value of one of the variables in terms of the other variables and substitute so you could wind up with 2 equations in 2 unknowns.
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