Daniel C.

Period 7

Math

Computer Project
 
Christy has a coin collection consisting of nickels and dimes. She has 28 coins worth $2.25. How many of each coin does Christy have?

 

 

Let n = number of nickels                     Let d = number of dimes

 

Equations

n+d = 28          .05n+.1d = 2.25

 

Elimination

            n+d = 28                 Multiply by -.05

            .05n+.1d = 2.25

 

            -.05(n+d = 28)

           

           

-.05n-.05d = -1.4

.05n+.1d = 2.25                  Add or subtract

 

.05d = .85                  Divide by .05

.05       .05

      d = 17                    Number of dimes

 

n+d = 28                      Substitute 17 in as d to find n, the number of nickels

n+17 = 28

   -17   -17

n = 11                          Number of nickels

 

Check

            .05(11) + .1(17) = 2.25              Substitute numbers as n and d.

            .55+1.7 = 2.25                         Add

2.25 = 2.25

 

Chart

Nickels

Dimes

Value

1

27

2.75

3

25

2.65

5

23

2.55

7

21

2.45

10

18

2.30

11

17

2.25

12

16

2.20

 

Solution: Christy has 11 nickels and 17 dimes.

 

 

Explanation: To solve this problem, I first formed the equations. n would represent nickels and d would represent dimes. The equations are, n+d = 28 because the amount of nickels and dimes added together would equal 28 coins. The other equation is .05n+.1d = 2.25. .05 is how much a nickel is worth and .10 is how much a dime is worth. When you add up the values of the nickels and dimes, they should equal $2.25. Next I used the elimination method to solve the problem and for this process, you have to eliminate a variable. So I multiplied the whole first equation by -.05 in order for the variable n to be eliminated from both equations. The equation was now -.05n-.05d = -1.4 and then I got rid of the -.05n in this equation and the .05n in the other because the two cancel each other out. After that I added the leftover numbers together and turned out to be this; .05d= .85.  At that point I divided each side by .05 to let d stand alone. Remaining was d = 17. Afterwards, I found what n was by substituting 17 as d in the equation n+d = 28. n+17 = 28. Subtract 17 from each side and you have n = 11. I checked my answer by substituting the numbers into the other equation and answer was correct. Last of all, I drew a chart  showing the dimes and nickels as another method and that proved the answer was correct too. If I was to count change that I had, this could be a way I’d count it because it seems to me to be a lot quicker and much more interesting way to do it.