A newspaper carrier has $5.80 in change. He has seven more quarters than dimes but three times as many nickels as quarters. How many coins of each type does he

Question

A newspaper carrier has $5.80 in change. He has seven more quarters than dimes but three times as many nickels as quarters. How many coins of each type does he have? 

Answer 1

Hi Racheal,

 

This is a system of equations problem.  Let's use Q for quarters, D for dimes, and N for nickels.

 

The amount of money he has is $5.80, so let's represent this in terms of the number of quarters, dimes and nickels.

 

Now first you must decide if you are going to work this problem in cents  ($5.80 = 580 cents, 1Q = 25 cents, 1D = 10 cents, 1N = 5 cents)  or if you are going to work it in dollars ($5.80 is already in dollars, 1Q = $0.25, 1D = $0.10, 1N = $0.05).  I think it is way easier to work in whole numbers, so I would choose to work it in cents.

 

So the number of quarters*25 + the number of dimes*10 + the number of nickels*5 is the amount of money you will have, so that has to equal 580.  Or in math-speak, 

 

25Q + 10D + 5N = 580

 

 

"seven more quarters than dimes" translates to

 

Q = D + 7

 

 

"three times as many  nickels as quarters" translates to

 

N = 3Q

 

Substitute "D+7" for "Q" in the previous equation, so now

 

N=3(D+7) or N = 3D + 21

 

Now into the first equation, replace N with "3D + 21" and replace Q with "D + 7".

 

 

This looks like

 

25(D+7) + 10D + 5(3D+21) = 580

 

Distribute on the left

 

25D + 175 + 10D + 15D + 105 = 580

 

Combine like terms

 

50D + 280 = 580

 

Subtract 280 from both sides

 

50D = 300

 

Divide both sides by 50

 

D = 6

 

So he/she has 6 dimes, Q=D+7, so he/she has 6+7 or 13 quarters.  

 

N = 3Q, so he/she has 3(13) or 39 nickels.

 

 

Now, let's check our work:

 

6 dimes = 60 cents

13 quarters = 325 cents

39 nickels = 195 cents

 

Add these up, do  you get 580 cents?   YES!  so we got the right answer!  :-)

Answer 2

Let x = the number of dimes

Let x + 7 = the number of quarters

Let 3(x + 7) = 3x + 21 = the number of nickels

.10x + .25(x+7) + .05(3x+21) = 5.80       Multiply both sides of the equation by 100 to get rid of the decimal points.

10x + 25(x+7) + 5(3x+21) = 580

10x + 25x + 175 + 15x + 105 = 580

50x + 280 = 580

50x = 300

x = 6 dimes

x+7 = 13 quarters

13(3) = 39 nickels