COIN WORD PROBLEM

Question

A newspaper carrier has $4.10 in change. He has 4 more quarters than dimes but 3 times as many nickels as quarters. How many of each coin does he have.

He has this many Quarters=

He has this many Dimes=

He has this many nickels=

Answer 1

quarters = 9

 

dimes = 5

 

nickels = 27

Answer 2

Hi Renee;

total=$4.10

quantity of dimes=x

dimes=0.10x

quarters=0.25(x+4)=0.25x+1.00

nickles=0.05(3)(x+4)=0.15x+0.60

 

4.10=0.10x+0.25x+1.00+0.15x+0.60

Combine like terms...

4.10=0.50x+1.60

Let's subtract 1.60 from both sides...

4.10-1.60=0.50x+1.60-1.60

2.50=0.50x

Divide both sides by 0.50...

1.50/0.50=(0.50x)/0.50

5=x

 

quantity of dimes=5

quantity of quarters=9

quantity of nickles=27

 

Let's check our work...

4.10=[(0.10)(5)]+[(0.25)(9)]+[(0.05)(27)]

4.10=0.50+2.25+1.35

4.10=4.10

Answer 3

See my corrected answer here:

http://www.wyzant.com/resources/answers/26973/coin_word_problem