Andrew has 50 coins (all nickles, dimes, and quarters) worth $4.85. He has 12 more nickles than dimes, How many quarters does Andrew have?

Question

I NEED HELP ASAP!!! STEP BY STEP PLEASE!!!

Answer 1

I would use a table...

 

nickles      dimes      quarters     total

     n     +    d      +     q            = 50 (number of coins)

 

 .05n     + .1d      +  .25q          =  4.85  (money value)

     n     =  d+12   (Because if he has 12 more nickles, he would have to add 12 to the dimes to make them equal.)

 

Now, solve the system of equations.

 

Since n=d+12, we can substitute d+12 into the other equations. 

 

Number of Coins

n + d + q = 50

d+12 + d + q = 50

    12 + 2d + q = 50

    12 + 2d + q -12 = 50 -12

            2d + q = 38

 

Money Value

.05n + .1d + .25q = 4.85

.05(d+12) + .1d + .25q = 4.85

.05d+0.6 + .1d + .25q = 4.85

       0.6 + .15d + .25q = 4.85

0.6 + .15d + .25q -0.6= 4.85-0.6

         .15d + .25q = 4.25

 

Now, combine the Number of Coins equation with the Money Value equation.

   2d + 1q = 38

.15d + .25q = 4.25

 

We don't care about d, so solve each equation for d.

  2d  = 38 -1q
.15d  = 4.25 - .25q

 

d = (38 -1q)/2
d = (4.25 - .25q)/.15

As long as d=d, we can do this:

(38 -1q)/2 = (4.25 - .25q)/.15

19   -.5q =  4.25/.15  - .25q/.15

19 -(4.25/.15) = - .25q/.15  +.5q

-28/3  =  -7/6 q

-28/3 *(-6/7) = -7/6 q *(-6/7)

-28/3 *(-6/7) = q 

8 = q

 

Now, to check to see if it makes sense.

8 quarters

12 coins that are nickles (20 coins, so far.), and then nickles and dimes can be counted in nickle-dime pairs. Or we could take the remaining 30 coins and say that 15 need to be nickles and the other 15 need to be dimes.  All the coins are accounted for.

8q  =$2.00

12n =$0.60

15n =$0.75

15d =$1.50

total=$4.85