Question 1199919: A piggy bank contains only nickels, dimes, and quarters. There is a total of 105 coins and the total value of those coins is $10.75.
Find a general solution of the number of each coin.
Use the general solutions from part 1 to list 3 specific solutions (for n, d, & q).
n=nickel
d=dime
q=a quarter
Found 4 solutions by josgarithmetic, math_tutor2020, MathTherapy, greenestamps:
Answer by josgarithmetic(39623)   (Show Source):
Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!
n = number of nickels
d = number of dimes
q = number of quarters
These are nonnegative integers {0,1,2,3,etc}
Given info: There is a total of 105 coins and the total value of those coins is $10.75
Break that into 2 pieces
Given info: There is a total of 105 coins and the total value of those coins is $10.75
The first part lets us say
n+d+q = 105
and the second fact lets us say
0.05n+0.10d+0.25q = 10.75
which is equivalent to
5n+10d+25q = 1075
after moving the decimal point two spots to the right for each item.
We multiplied both sides by 100.
We have this system of equations
n+d+q = 105
5n+10d+25q = 1075
We have 2 equations, but 3 unknowns.
Unfortunately there isn't enough information to determine the values of n,d,q.
The next section goes into greater detail as to why this is the case.
--------------------------------------------
n+d+q = 105 solves to n = 105-d-q
Substitute that into the other equation
5n+10d+25q = 1075
5(105-d-q)+10d+25q = 1075
525-5d-5q+10d+25q = 1075
525+5d+20q = 1075
5(105+d+4q) = 1075
105+d+4q = 1075/5
105+d+4q = 215
d = 215-105-4q
d = 110-4q
If q = 10, then,
d = 110-4q
d = 110-4*10
d = 70
and,
n = 105-d-q
n = 105-70-10
n = 25
Showing that (n, d, q) = (25, 70, 10) is one ordered triple solution.
Check:
n+d+q = 25+70+10 = 105 coins total
5n+10d+25q = 5*25+10*70+25*10 = 1075 cents total value
1075 cents = 1075/100 = $10.75
If q = 11, then,
d = 110-4q
d = 110-4*11
d = 66
and,
n = 105-d-q
n = 105-66-11
n = 28
Showing that (n, d, q) = (28, 66, 11) is another ordered triple solution.
Check:
n+d+q = 28+66+11 = 105 coins total
5n+10d+25q = 5*28+10*66+25*11 = 1075 cents total value
Try out other values of q to see what happens.
Here are all the ordered triple solutions (n,d,q) such that we focus on positive integers only.
(n, d, q) = (1, 102, 2)
(n, d, q) = (4, 98, 3)
(n, d, q) = (7, 94, 4)
(n, d, q) = (10, 90, 5)
(n, d, q) = (13, 86, 6)
(n, d, q) = (16, 82, 7)
(n, d, q) = (19, 78, 8)
(n, d, q) = (22, 74, 9)
(n, d, q) = (25, 70, 10)
(n, d, q) = (28, 66, 11)
(n, d, q) = (31, 62, 12)
(n, d, q) = (34, 58, 13)
(n, d, q) = (37, 54, 14)
(n, d, q) = (40, 50, 15)
(n, d, q) = (43, 46, 16)
(n, d, q) = (46, 42, 17)
(n, d, q) = (49, 38, 18)
(n, d, q) = (52, 34, 19)
(n, d, q) = (55, 30, 20)
(n, d, q) = (58, 26, 21)
(n, d, q) = (61, 22, 22)
(n, d, q) = (64, 18, 23)
(n, d, q) = (67, 14, 24)
(n, d, q) = (70, 10, 25)
(n, d, q) = (73, 6, 26)
(n, d, q) = (76, 2, 27)
I used computer software to quickly generate this list.
This is to do so efficiently and to avoid error.
It's also to ensure that the list is fully exhaustive.
Once again, there isn't enough information to nail down the single exact value of each n, d, and q value.
Answer by MathTherapy(10555)  (Show Source):
You can put this solution on YOUR website!
A piggy bank contains only nickels, dimes, and quarters. There is a total of 105 coins and the total value of those coins is $10.75.
Find a general solution of the number of each coin.
Use the general solutions from part 1 to list 3 specific solutions (for n, d, & q).
n=nickel
d=dime
q=a quarter
Since number of nickels, dimes, and quarters are n, d, and q, we get the following COIN-COUNT equation:
n + d + q = 105 --- eq (i), and total value-equation, .05n + .1d + .25q = 10.75, or 20(.05n + .1d + .25q) = 20(10.75), or
n + 2d + 5q = 215 --- eq (ii)
d + 4q = 110 ----- Subtracting eq (i) from eq (ii)
4q = 110 - d
 ----- Solving for q ----- eq (iii)
The SMALLEST number of dimes (d) that can be substituted The next "d" integer-value that'll create the next
into the above equation, and which also yields an smaller integer in numerator that's a MULTIPLE of 4 is 6,
INTEGER-number of quarters is 2, since 2 leads to a since 6 leads to a numerator of 104, a multiple of 4, and
numerator of 108, a MULTIPLE of 4 (the denominator). which also yields an INTEGER-VALUE for q.
We then get:  We then get: 
Therefore, for the above, we get: Number of dimes: 2 Therefore, for the above, we get: Number of dimes: 6
Number of quarters: 27 Number of quarters: 26
Number of nickels, 105 - (2 + 27) = 105 - 29 = 76 Number of nickels, 105 - (6 + 26) = 105 - 32 = 73
This gives us, (n, d, q) = (76, 2, 27) This gives us, (n, d, q) = (73, 6, 26)
So far, we have: (n, d, q) = (76, 2, 27)
(n, d, q) = (73, 6, 26)
Looking at the above SEQUENCE, it's quite obvious that the nickels decrease by 3, the
dimes INCREASE by 4, and the
quarters DECREASE by 1.
Thus, we get the following: (n, d, q) = (76, 2, 27)
(n, d, q) = (73, 6, 26)
(n, d, q) = (70, 10, 25)
(n, d, q) = (67, 14, 24)
(n, d, q) = (64, 18, 23)
You don't need that many because you were asked for just 3, but you should get the "picture." CAPISCHE?
Answer by greenestamps(13203)  (Show Source):
You can put this solution on YOUR website!
I note that none of the responses you have received so far show the general solution that the problem asks for....
The number of coins is 105:
 [1]
The total value of the coins in cents is 1075:
 [2]
Simplify [2]:
 [3]
Find the difference between [1] and [3]:
 [4]
Solve [4] for d in terms of q:
 [5]
Use [1] and [5] to find an expression for n in terms of q:
 [6]
The general solution using parameter t is...
Confirm this general solution by verifying that the original equations [1] and [2] are satisfied.
The general solution is correct; choose any value for the parameter t and find the resulting numbers of quarters, nickels and dimes.
The solutions are of course restricted to values of t that yield non-negative integer values of n, d, and q. We can determine the range for the valid numbers of quarters by looking at the expressions for the numbers of dimes and nickels.
dimes: 110-4t>=0; 110>=4t; t<=27.5; the maximum number of quarters is 27
nickels: 3t-5>=0; 3t>=5; t>=5/3; the minimum number of quarters is 2
Now find 3 solutions to the problem using parameter values between 2 and 27.
t=4: q=4; d=110-4(4) =94; n=3(4)-5 = 7 (n,d,q) = (7,94,4)
t=15: q=15; d=110-4(15)=50; n=3(15)-5=40 (n,d,q) = (40,50,15)
t=22: q=22; d=110-4(22)=22; n=3(22)-5=61 (n,d,q) = (61,22,22)
etc....
Note that the parametric equations for the numbers of quarters, dimes, an nickels shows you how to get from one particular solution to the "next" one:
These equations show that, when the number of quarters is increased by 1, the number of dimes decreases by 4 and the number of nickels increases by 3. That of course makes sense, because adding 1 quarter and 3 nickels and taking away 4 dimes keeps the total value of the coins unchanged.
So if you have, for example, the particular solution (q,d,n) = (15,50,40), you can find the "next" solution is (16,46,43) -- by adding one quarter and 3 nickels and taking away 4 dimes.
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