SOLUTION: A bag contains 15 coins. Each coin is a penny, nickel, dime or quarter. The total value of these coins is $2.16. There is at least one coin of each denomination, and the number

Algebra ->  Customizable Word Problem Solvers  -> Coins -> SOLUTION: A bag contains 15 coins. Each coin is a penny, nickel, dime or quarter. The total value of these coins is $2.16. There is at least one coin of each denomination, and the number       Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 947343: A bag contains 15 coins. Each coin is a penny,
nickel, dime or quarter. The total value of these
coins is $2.16. There is at least one coin of
each denomination, and the number of pennies,
nickels, dimes and quarters are all distinct. If
there are 5 dimes, how many nickels are in the bag?

Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
SEE BELOW, IMPROVEMENT/ALTERNATIVE

p, n, d, q;
p+n+d+q=15.
0.01p+0.05n+0.1d+0.25q=2.16.
d=5.

Simplify the money count.
p%2B5n%2B10d%2B25q=216
and substitute for d;
p%2B5n%2B10%2A5%2B25q=216
p%2B5n%2B25q=166

You have a system,
system%28p%2B5n%2B25q=166%2Cp%2Bn%2B5%2Bq=15%29
which is
system%28p%2B5n%2B25q=166%2Cp%2Bn%2Bq=10%29

You can assume any one of p, or n, or q is constant, and then solve for the other two variables in terms of the chosen constant. This is because you are assuming a two-variable, linear system.


As a guess, the number of quarters is probably not too large compared to the 15 count of coins. Assign k=q.

-
system%28p%2B5n=166-25k%2Cp%2Bn=10-k%29
-
E1-E2,
4n=%28166-25k%29-%2810-k%29
4n=166-10-25k%2Bk
4n=156-24k
n=39-6k
-
Try to find p based on this formula for n.
p%2Bn=10-k
p=10-k-n
p=10-k-%2839-6k%29
p=10-39-k%2B6k
p=-29%2B7k
-
-
What will make sense for k? In summary, we have a formula for p and a formula for n:
system%28n=39-6k%2Cp=7k-29%29.
We must have natural numbers for each of n, p, k. None of them be equal. None of them over 13.

Choose and find.

k__________n________p
1__________33_______-22
2__________27_______-15
3__________21_______-8
4__________15_______-1
5__________9________6, still not good because need some nickels, 9+6=15.
6__________3________13
7__________-3
FAILED TO RESOLVE THE PROBLEM...........................



IMPROVEMENT/ALTERNATIVE:
Re-examining the description, the two key equations are
system%28p%2B5n%2B25q=166%2Cp%2Bn%2Bq=10%29
which are based on the given assignment that d=5.
-
Subtracting the coin count equation from the money count equation (both being in their just stated simplified form), we have
4n%2B24q=156
highlight_green%28n%2B6q=39%29
and p has been eliminated.

Using this simple equation relating n and q, try choosing different natural number values for q and find resulting n, and look for results which makes sense.

pick q____________find n
1_________________33
2_________________27
3_________________21
4_________________15
5_________________9
6_________________3--------THERE!!! Their sum is only 9, allowing the....five dimes.


ANSWER RESULT
-----------------------------------------------
Combination appears to be:
p=?
n=3
d=5
q=6
-
p=15-3-5-6=1, because total coins must be 15
-
Notice all counts are unequal.
One penny, three nickels, five dimes, six quarters.
----------------------------------------------