Systems of equations

Question

A piggy bank contains 30 coins worth $.1.90. If the bank contains nickels, dimes and quarters, how many coins of each types does it contain?

Answer 1

At first this looks as if there are more unknowns than equations. You have 3 unknowns, call them N, D, and Q, for the number of nickels, dimes and quarters. There are 30 total coins, so N+D+Q=30 is one equation. .05N+.10D+.25Q=1.90 is the 2nd equation. The units are in dollars. The 2nd equation can be simplified by multiplying both sides by 100, giving: 5N+10D+25Q=190 Now the units are in cents. Take the first equation and multiply it by 5 giving 5N+5D+5Q=150. Subtract this from 5N+10D+25Q=190. That gives 5D+20Q=40. Q cannot be larger than 2 without making D negative. The number of each of the coins cannot negative. So Q is 2, 1 or 0. IF Q is 2, then D=0. Total coins are 30, so N = 30-2=28. This is one solution, 28 nickels, zero dimes and 2 quarters. If Q is 1 then 5D+20=40 or 5D=20 and D=4. Substitute D=4 and Q=1 into N+D+Q=30. Then N+4+1=30 and N=25. This is a 2ndsolution to the problem. 25 nickels, 4 dimes and 1 quarter. However, there is a 3rd solution, with Q=0. IF there are no quarters, then the equations reduce to 5D+0=40 or D=8. 5N+10D+25Q=190 becomes with substitution for D=8 and Q=0 as 5N+10(8)+25(0)=190. Simplifying leaves 5N=190-80 or 5N=110 or N=110/5=22. A 3rd solution is 22 nickels and 8 dimes, with zero quarters. Because there were more unknowns than equations, there was likely no one unique solution, and possibly infinite solutions. Here, there were 2 equations with 3 unknowns but also 3 possible third equations Q=2, 1 or 0. Each value of Q gives a different solution. Using Q=2 combined with the other two equations gives N=28, D=0, Q=2. Using Q=1 combined with the other two equations gave another solution. Using Q=0 combined with the other two equations gave a 3rd solution. (N,D,Q)=(28,0,2), (25,4,1) or (22,8,0) More solutions would have been possible were the number of any of the coins allowed to be negative, but as a practical matter that makes no sense.

Answer 2

The answer has been given, but I want to offer a general strategy for this type of problems. I'ts a system of equations with more variables than equations, which mean infinite solutions, but there is one important restriction the variables are integers, this is called a Diophantine equation (in this particular case the variables are greater than 0).

x - nickles, y = dimes, z = quarters

0.05x + 0.1y + 0.25z =1.90 (multiply by 20)

x + y + z = 30 (multiply by 2)

then:

x +2y +5z = 38

2x +2y +2z = 60 (Subtract term by term)

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-x +3z = -22

3z= x-22

And now comes the interesting part:

z= (x-22)/3 (z is a whole number, thus x-22 needs to be a multiple of 3, also 0<= x,y,z <=30)

and we have 3 possibilities: x=22, x=25, and x=28

If x=22, then z=0, y=8

If x=25, then z=1, y=4

If x=28, then z=2, y=0

Answer 3

x - nickles

y = dimes

z = quarters

minimum value is 28X1.05 = 1.40 + (1)10 + (1) 25 = 1.75. At z=2, y=1, x=27 = 1.95 there for z = 1

for each dime substituted, we gain $.05 so y =4; z=1, and therefore x = 25.