how many of each coin does she have?

Question

jennifer has nickels, dimes and quarters in her piggy bank. In total, she has 49 coins, with a value of $5.20. If she has five more dimes than all the nickels and quarters combined, how many of each type of coins does she have

Answer 1

This requires solving a system of simultaneous equations several times.

 

Let N equal #nickels, D equal #dimes, Q equal #quarters.....

 

So, based on given information.....

Eq1.  N+D+Q=49 coins

Eq2.  (N+Q)+5=D.....(5 more dimes than nickels and quarters combined.

Eq3.  5N+10D+25Q=520....(value of coins in cents)

 

Solve Eq1 and Eq2 simultaneously by adding....

Eq1.  N+D+Q=49

Eq2.  N -D+Q=-5....(variables left of = sign, constants right of = sign).  

        --------------

        2N +2Q=44....(divide both sides by 2) get Eq4....

Eq4.    N + Q=22

 

Substitute N+Q=22 into Eq2, get....

Eq2.  22 + 5 = D

                 D=27

        

Now substitute D=27 into Eq3, simplify...

Eq3.  5N+10(27)+25Q=520

         5N+  270 + 25Q=520...(subtr 250 both sides) get...

                     5N+25Q=250...(solve simultaneously with Eq4)

 

Eq3.  5N+25Q=250

Eq4. -5N - 5Q=-110....(multiply Eq4 both sides by -5 and add), get...

        ----------------

               20Q=140....(divide by 20), get...

                   Q=  7

 

Substitute Q=7 into Eq4, get....

Eq4.  N+Q=22

         N+7=22

             N=15

 

Therefore 15nickels (N), 27dimes (D), 7quarters (Q) is the answer.

Make sure you substitute these values into Eqns1, 2, 3 to check original assumptions!!!  Always check!!!

 

 

Answer 2

This problem requires that you form a system of equations to solve it. You have 3 unknowns and information for 3 equations, so it can be solved. I will help you establish equations and then you can try to solve. If you need further assistance, please ask.

Equations:

n+d+q=49

0.05n+0.1d+0.25q=5.20

d=n+q+5

 

Good luck!