Question

A customer goes to a bank and gets change for a $100 bill. The change is to be in $1, $5, and $10 bills. There were four times as many $5 bills as $10 bills. If there are 25 bills in all, how many are $5 bills?

Answer 1

Answer:

12

Step-by-step explanation:

x = number of $1 bills

y = number of $5 bills

z = number of $10 bills

We know that:

x + y + z = 25 (total amount of bills)

x + 5y + 10z = 100 (total amount of money)

y = 4z (Given)

Substitute the third equation into the first one.

x + 4z + z = 25

x = 25 - 5z

Put this into the second equation.

(25 - 5z) + 5 (4z) + 10z = 100

Simplify and solve.

25 - 5z + 20z + 10z = 100

25z + 25 = 100

25z = 75

z = 3

Substitute this into y = 4z

y = 4 * 3

y = 12

So, there are 12 $5 bills.

Answer 2

Answer:

12 $5 bills

Step-by-step explanation:

Represent the number of each kind of bill by x, y and z:  there are x $1 bills, y $5 bills and z $10 bills.

According to the problem statement,

x + y + z = 25 and y = 4z (or z = y/4).  Also, the values of the bills add up to $100:

($1)x + ($5)y + ($10)z = $100.  Eliminate z by typing y/4 in its place:

x + y + (y/4) = 25 and

x  + 5y + 10(y/4) = 100

Doing this has reduced the number of variables to two:  x and y.  Combining like terms in both equations, we get:

1x + (5/4)y = 25 and

1x +  (30/4)y       =100

Subtracting the 1st equation from the 2nd, we get:

(25/4)y = 75, or

(4/25)(25/4)y = (4/25)(75) , or y = 12.

If y = 12, z = 12/4, or 3.

Then x + 12 + 3 = 25 coins in all; therefore, x = 15 = 25, and x = 10

There are 10-$1 bills, 12-$5 bills and 3-$10 bills.