a jar contains $8.75 in nickels, dime, quarters.

Question

If there are two times as many dimes as there are nickels, and 10 more dimes than quarters, how many of each coins are in the jar?

Answer 1

You can easily get 3 equations from this:

 

1. .25Q+.10D+.05N=8.75 (that's the straight money equation, Q=quarters, etc)
2. D=2N (twice as many dimes as nickels)
3. D=Q+10 (10 more dimes than quarters

 

Note that equations 2 and 3 both contain just one variable in addition to D, so we can easily rewrite those in terms of D only.  Then we can substitute back in the 1st equation to get the number of dimes.

 

2. N=D/2 (divide by 2)

3. Q=D-10 (subtract 10)

 

1. .25(D-10)+.10D+.05(D/2)=8.75 (substituting the rearranged equations into the 1st equation)

 

Many students like to get rid of those decimals.  You can solve that as is, or multiply everything by 100:

 

• 25(D-10)+10D+5D/2=875 (substitution)
• 25D-250+10D+2.5D=875 (distribute)
• 37.5D=1125 (combine like terms)
• D=30 (divide by 37.5)

 

Once you have one number, the others easily follow:

 

• N=D/2=30/2=15
• Q=D-10=30-10=20

 

Therefore you have 30 dimes, 15 nickels, and 20 quarters.