Question 1164758: Janice was given a piggy bank on her seventh birthday, and she put it to use immediately. Each time she puts one or more coins into the piggy bank, she keeps track of the number of coins she has collected to date and the accumulated value of her collection. Janice collects only nickels, dimes, and quarters. Six months after her seventh birthday, Janice looked at her record and ascertained that she had collected 240 coins, which were worth $36.
a.How many combinations of coins are possible in Janice's collection?
b.Janice counted 100 quarters in her savings. How many nickels and dimes are in her collection?
Answer by greenestamps(13198)   (Show Source):
You can put this solution on YOUR website!
Let's take care of the second question first, since it is much easier.
If there are 100 quarters, with a value of $25 = 2500 cents, then there are 240-100 = 140 nickels and dimes with a value of $36-$25 = $11 = 1100 cents:
n+d = 140
5n+10d = 1100
Multiply the first equation by 5 and compare to the second equation to eliminate n:
5n+5d = 700
5n+10d = 1100
5d = 400
d = 80
n = 140-80 = 60
ANSWER (question b): 100 quarters means 80 dimes and 60 nickels.
CHECK: 100(25)+80(10)+60(5) = 2500+800+300 = 3600
Now for the much harder first question....
(1)  [there are 240 total coins]
(2)  [the total value of the coins is $36 = 3600 cents]
We have 2 equations but 3 unknowns; nevertheless, we can find a finite number of discrete solutions knowing that all the unknowns are non-negative integers.
There are numerous ways to solve this kind of problem; and in many of those ways there are numerous different paths to solving the problem.
Here is one strategy:
>>Eliminate d between the two equations to get an expression for q in terms of n;
>>Eliminate n between the two equations to get an expression for q in terms of d;
>>Examine those two expressions with logical reasoning
Eliminate d:
(3) 
Eliminate n:
(4) 
Since q, d, and n are all non-negative, (3) tells us the minimum value for q is 80, and (4) tells us the maximum value for q is 120.
Now we need to determine if ALL the integer values from 80 to 120 inclusive are possible values for q.
To do that, we note that, if we have one solution to the problem, we can get another by adding 1 quarter and 3 nickels and taking away 4 dimes. That keeps both the number of coins and the total value of the coins unchanged.
So yes, every value of q from 80 to 120 inclusive is a valid number of quarters; and that means the number of possible combinations of coins is 120-80+1 = 41.
ANSWER: 41 possible combinations of coins.
We can arrive at the same answer informally, using the following strategy:
>> The maximum number of quarters will be when the other coins are all nickels; find that maximum number
>> The minimum number of quarters will be when the other coins are all dimes; find that minimum number
>> Use the strategy as in the above solution to show that every number of quarters between the minimum and maximum values is possible
Maximum possible number of quarters....
Minimum possible number of quarters....
As with the previous solution method, we have shown that the number of quarters is at least 80 and at most 120. Then, using the same reasoning as in the other method, we find the number of possible combinations is 41.
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