Question 1117809: A piggy bank contains 30 coins worth $1.90.
(A) If the bank contains only nickels and dimes, how many coins of
each type does it contain?
(B) If the bank contains nickels, dimes, and quarters, how many
coins of each type does it contain?
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39620)   (Show Source):
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
The total value of the 30 coins, all dimes or nickels, is $1.90, or 190 cents.
If all were nickels, the value would be 30*5 = 150 cents; that is 40 cents short of the actual total.
Exchanging a nickel for a dimes keeps a total of 30 coins but increases the total value by 5 cents.
To make up the 40 cents you were short using only nickels, the number of times you would need to exchange a nickel for a dime would be 40/5 = 8.
So there are 8 dimes and 30-8 = 22 nickels.
One way to solve the problem when the coins include quarters is to look at different cases with different numbers of quarters.
Using only n coins that are all nickels and dimes, you can make any total number of cents between 5n and 10n. So, since the total value of the coins is 190 cents, it will always be possible to make a total of 190 cents with any number of quarters, making up the difference with dimes and nickels. The task is to find such a solution in which the total number of coins is 30.
If there is 1 quarter (25 cents), then the problem becomes making a total of 165 cents using 29 coins that are dimes and nickels. Using the same kind of reasoning as before, we find out that with 1 quarter we need 4 dimes and 25 nickels to make the total of 190 cents.
If there are 2 quarters (50 cents), then the problem becomes making a total of 140 cents using 28 coins that are dimes and nickels. We can't do that unless we use 28 nickels and 0 dimes; and the problem implies that we have to have coins of each type.
So by examining cases, we find that there is only one way to make $1.90 using 30 coins that are quarters, dimes, and nickels: 1 quarter, 4 dimes, and 25 nickels.
It is useful to know how to solve this second problem using algebraic processes.
We have two equations but three unknowns:
 a total of 30 coins
 a total value of 190 cents
To obtain a unique solution in a problem with three variables, we usually need three equations. But when, as in this problem, the variable values are positive integers, we can solve the problem with only two equations.
The general process is to use elimination to reduce the system of two equations in three unknowns to a system of one equation in two unknowns; then find positive integer solutions to that equation.
For this problem, we have
 first equation, multiplied by 5
Subtracting the two equations gives us
Starting with q=1, we find a solution: q=1; d=8-4(1) = 4; n = 30-(1+4) = 25.
And when we try the next possible solution with q=2, we get d = 8-4(2) = 0 -- telling us that there are no other solutions.
Note that this kind of problem often has more than one solution....
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