COIN WORD PROBLEM

Question

A newspaper carrier has $2.55 in change. He has 2 more quarters than dimes but 4 times as many nickels as quarters. How many of each coin does he have?

Answer 1

Renee, isn't this the 3rd or 4th question like this?

 

Have you learned nothing from the other solutions?

 

You need to try it yourself.

Answer 2

 

let Q= # of quarters

let Q-2= # of dimes

let 4Q= # of nickels

This should get you started. Look at the other solutions and try to set up an equation to solve.

 

Answer 3

Hi Renee;

A newspaper carrier has $2.55 in change. He has 2 more quarters than dimes but 4 times as many nickels as quarters. How many of each coin does he have?

x=quantity of dimes

quarters...(0.25)(x+2)=0.25x+0.50

dimes...(0.10)(x)=0.10x

nickles...(0.05)[(4)(x+2)]=0.05(4x+8)=0.20x+0.40

 

2.55=0.25x+0.50+0.10x+0.20x+0.40

2.55=0.55x+0.90

Let's subtract 0.90 from both sides...

2.55-0.90=0.55x+0.90-0.90

1.65=0.55x

Let's divide both sides by 0.55...

1.65/0.55=(0.55x)/0.55

3=x

 

quantity of quarters...3+2=5

quantity of dimes...3

quantity of nickles...(4)(5)=20

 

Let's check our work...

2.55=[(0.25)(5)]+[(0.10)(3)]+[(0.05)(20)]

2.55=1.25+0.30+1.00

2.55=2.55