Question

A cashier has 54 bills, all of which are $10 or $20 bills. The total value of the money is $910. How many of each type of bill

does the cashier have?

Answer 1

20x + 10(54-x) = 910
x = 37 = number of $20 bills
54-x=54-37=17=number of $10 bills

Answer 2

the number of $10 bills = 17

the number of $20 bills = 37

Explanation:

Let x be the number of $10 bills and y be the number of $20 bills

Total bills = 54

So
`x+y= 54`

Total value of money is $910

`10x+20y= 910`

now we solve for x and y

Solve the first equation for y

`x+y= 54`

`y=54-x`

Now replace y in second equation

`10x+20y= 910`

`10x+20(54-x)= 910`

`10x+ 1080-20x= 910`

`1080-10x= 910`

Subtract 1080 from both sides

`-10x= -170`

Divide both sides by -10

x= 17

`y=54-x`

Replace x with 17

`y=54-17=37`

the number of $10 bills = 17

the number of $20 bills = 37