SOLUTION: A bag of nickels, dimes, quarters and loonies contain 55 coins with a total weight of 130, worth a total of $12.50. If dimes weigh 1, nickels weigh 2, quarters weigh 3, and loonies

Question 157859: A bag of nickels, dimes, quarters and loonies contain 55 coins with a total weight of 130, worth a total of $12.50. If dimes weigh 1, nickels weigh 2, quarters weigh 3, and loonies weigh 4, how many coins of each type are there?
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Let x=number of nickels
y=number of dimes
z=number of quarters
w=number of loonies
(Note: We know before we start that we have only three equations and four unknowns and this means that we may have to do some trial and error to come up with an answer)
We are told that:
x+y+z+w=55----------------------------------------------------eq1
and that: (let's deal in pennies)
5x+10y+25z+100w=1250 divide each term by 5
x+2y+5z+20w=250--------------------------------------------eq2
and that:
2x+y+3z+4w=130----------------------------------------------------eq3
subtract eq1 from eq2 and we get:
y+4z+19w=195--------------------------------------------------------eq1a
multiply eq2 by 2 and then subtract eq3 from it and we get:
3y+7z+36w=370--------------------------------------------------------eq2a
now multiply eq1a by 3 and then subtract eq2a from it and we get:
5z+21w=215 -------------------------------------------------------------eq3a
21w=215-5z and
w=(215-5z)/21-----a relationship between the number of loonies and the number of quarters.
Now we know several things:
(1) we cannot have fractions of coins (our answers must be whole numbers)
(2) we cannot have negative numbers (negative coins not allowed)
(3) we cannot have more than 12 loonies otherwise we break the bank
From (3) above, we have the following inequality:
(215-5z)/21<=12
If w is:
12-----no good. We get negative quarters
11-----no good. Ditto
10----BINGO!!!:----BUT NO GOOD
215-5Z=210; -5Z=-210; Z=42-------------NO GOOD--42 QUARTERS AND 10 LOONIES BREAKS THE BANK, SO LETS KEEP GOING
9 :-5Z=-26----------------------NO GOOD---FRACTIONS OF QUARTERS
8 :-5z=-47-----------------------DITTO---FRACTIONS
7 :-5Z=-68------------------------DITTO---FRACTIONS
6 :-5Z=-89--------------------------DITTO---FRACTIONS
5 :-5Z=-110 BINGO!!!! Z=22 THIS MAY WORK. BUT FIRST LETS SEE IF THERE ARE OTHER POSSIBILITIES
4 :-5Z=-131--------------FRACTIONS FOR QUARTERS
3 :-5Z=-152----------------DITTO---FRACTIONS
2 :-5Z=-173-----------------DITTO FRACTIONS
1 :-5Z=-194----------------DITTO FRACTIONS
LOOKS LIKE WE ONLY HAVE ONE SOLUTION:
w=5 ---LOONIES
z=22 ---QUARTERS
Lets plug these into eq1a and see what we get for y:
y+4z+19w=195
y+4*22+5*19=195
y=195-88-95=12 so:
y=12---DIMES
Now lets plug these into eq1 and see what we get for x:
x+y+z+w=55
x+12+22+5=55
x=55-39=16
x=16-----------NICKELS
CK weights
12*1+16*2+22*3+5*4=130
12+32+66+20=130
130=130
CK value
16*5+12*10+22*25+5*100=1250
80+120+550+500=1250
1250=1250

Hope this helps---ptaylor

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