Money problem

Question

Jamie has a collection of 78 nickels, dimes and quarters worth 12.40. If the number of quarters is doubled, the value becomes 22.15. How many of each coin does Jamie have?

Answer 1

n + d + q = 78      {equation 1}

 

0.05n + 0.10d + 0.25q = 12.40

multiply through by 100 to clear decimals

5n + 10d + 25q = 1240    {equation 2}

 

0.05n + 0.10d + 0.25(2q) = 2215

5n + 10d + 50q = 2215   {equation 3}

 

This is your system of equations as given by the problem information:

n + d + q = 78                  {equation 1}

5n + 10d + 25q = 1240       {equation 2}

5n + 10d + 50q = 2215     {equation 3}

 

We can pare this down to two equations with 2 unknowns.  Let's eliminate

the n variable.  Multiply equation 1 through by -5 and add it to equation 2.

-5n - 5d - 5q = -390

5n + 10d + 25q = 1240

----------------------------

       5d + 20q = 850     {equation 1a}

 

Do the same with equation 1 and equation 3

-5n - 5d - 5q = -390

5n + 10d + 50q = 2215 

--------------------------------

        5d + 45q = 1825    {equation 2a}

 

Subtract equation 1a from equation 2a to eliminate d

 

  5d + 45q = 1825

-(5d + 20q = 850)

-----------------------

          25q = 975

           q = 39

 

5d + 20q = 850

5d + 20(39) = 850

5d + 780 = 850

5d = 70

d = 14

 

we started with 39 quarters and 14 dimes

Plug that into n + d + q = 78 and solve for n