# Coin Problem

• Mar 15th 2009, 06:48 PM
magentarita
Coin Problem
Joe has a collection of nickels and dimes that is worth \$6.05. If the number of dimes was doubled and the number of nickels was decreased by 10, the value of the coins would be \$9.85. How many dimes does he have?

My Set Up:

n + d = 6.05..............Equation A
(n - 10) + 2d = 9.85...Equation B

I realize that we have two equations in two unknowns.

What is wrong with my equations?

I keep getting a negative answer for coins and this does not make sense.

• Mar 15th 2009, 06:57 PM
e^(i*pi)
Quote:

Originally Posted by magentarita
Joe has a collection of nickels and dimes that is worth \$6.05. If the number of dimes was doubled and the number of nickels was decreased by 10, the value of the coins would be \$9.85. How many dimes does he have?

My Set Up:

n + d = 6.05..............Equation A
(n - 10) + 2d = 9.85...Equation B

I realize that we have two equations in two unknowns.

What is wrong with my equations?

I keep getting a negative answer for coins and this does not make sense.

Your equations seem fine to me.

rearrange equation A: \$\displaystyle n = 6.05 - d\$

6.05-d + 2d = 9.85

d = 3.80

and so n = 6.05-3.8 = 2.25.

Oddly when I tried d = 6.05-n I got a negative value of n
• Mar 15th 2009, 07:10 PM
magentarita
me too...
Quote:

Originally Posted by e^(i*pi)
Your equations seem fine to me.

rearrange equation A: \$\displaystyle n = 6.05 - d\$

6.05-d + 2d = 9.85

d = 3.80

and so n = 6.05-3.8 = 2.25.

Oddly when I tried d = 6.05-n I got a negative value of n

I also get a negative number using d = 6.05 - n.

I found that Joe has 38 dimes and 45 nickels.
• Mar 15th 2009, 07:27 PM
mollymcf2009
Quote:

Originally Posted by e^(i*pi)
Your equations seem fine to me.

rearrange equation A: \$\displaystyle n = 6.05 - d\$

6.05-d + 2d = 9.85

d = 3.80

and so n = 6.05-3.8 = 2.25.

Oddly when I tried d = 6.05-n I got a negative value of n

You left out the -10 from this part of the equation (above in red), probably why you are getting a negative number

n-10 +2d = 9.85

(6.05 - d)-10+2d = 9.85 = 13.80
• Mar 16th 2009, 12:53 PM
masters
Quote:

Originally Posted by magentarita
Joe has a collection of nickels and dimes that is worth \$6.05. If the number of dimes was doubled and the number of nickels was decreased by 10, the value of the coins would be \$9.85. How many dimes does he have?

My Set Up:

n + d = 6.05..............Equation A
(n - 10) + 2d = 9.85...Equation B

I realize that we have two equations in two unknowns.

What is wrong with my equations?

I keep getting a negative answer for coins and this does not make sense.

Hi magentarita,

If you're talking in "amounts", you need to multiply the number of dimes and nickels by their respective values.

Let n = the number of nickels

Let d = the number of dimes

Your first equation becomes:

[1] 5n + 10d = 605

Now, let's work on the changes.

Let n-10 = the new number of nickels

Let 2d = the new number of dimes

Your second equation becomes:

[2] 5(n - 10) + 10(2d) = 985

You can simplify both equations to the following:

[1] n + 2d = 121

[2] n + 4d = 207
-----------------

Subtract the two equations to get:

-2d = -86

d = 43

So the number of dimes = 43 to make \$4.30
The number of nickels can be found by subtracting \$4.30 from \$6.05 to get \$1.75 and dividing by \$.05, we get 35 nickels

Let's check it:

Original setup:

43 dimes = \$4.30
35 nickels = \$1.75
Total = \$6.05

Now, let's double the dimes, and reduce the number of nickels by 10.

86 dimes = \$8.60
25 nickels = \$1.25
Total = \$9.85
• Mar 18th 2009, 08:44 AM
magentarita
Thanks
Quote:

Originally Posted by masters
Hi magentarita,

If you're talking in "amounts", you need to multiply the number of dimes and nickels by their respective values.

Let n = the number of nickels

Let d = the number of dimes

Your first equation becomes:

[1] 5n + 10d = 605

Now, let's work on the changes.

Let n-10 = the new number of nickels

Let 2d = the new number of dimes

Your second equation becomes:

[2] 5(n - 10) + 10(2d) = 985

You can simplify both equations to the following:

[1] n + 2d = 121

[2] n + 4d = 207
-----------------

Subtract the two equations to get:

-2d = -86

d = 43

So the number of dimes = 43 to make \$4.30
The number of nickels can be found by subtracting \$4.30 from \$6.05 to get \$1.75 and dividing by \$.05, we get 35 nickels

Let's check it:

Original setup:

43 dimes = \$4.30
35 nickels = \$1.75
Total = \$6.05

Now, let's double the dimes, and reduce the number of nickels by 10.

86 dimes = \$8.60
25 nickels = \$1.25
Total = \$9.85

Thanks for the details and great reply.