Awallet contains $460 in $5, $10, and $20 bills. the number of $5 bills exceeds twice the number of $10 bills by 4, and the number of $20 bills is 6 fewer than the number of $10 bills. how many of each type are there?

Number of $5 bills = 32

Number of $10 bills = 14

Number of $20 bills = 8

Step-by-step explanation:

Let x number of $5, y number of $10 and z number of $20

The number of $5 bills exceeds twice the number of $10 bills by 4.

Therefore, x = 2y + 4

The number of $20 bills is 6 fewer than the number of $10 bills.

Therefore, z = y - 6

A wallet contains $460 in $5, $10, and $20 bills.

Therefore,

5x + 10y + 20z = 460

Substitute x and y into equation

    5(2y+4) + 10y + 20(y-6) = 460

10y + 20 + 10y + 20y - 120 = 460

                           40y - 100 = 460

                                    40y = 460 + 100

                                    40y = 560

                                      y = 14

                                      x = 2(14) + 4

                                      x = 32

                                      z = 14 - 6

                                      z = 8

Hence, the each type of bills,

Number of $5 bills = 32

Number of $10 bills = 14

Number of $20 bills = 8

14.3

step-by-step explanation:

add all the numbers together.

9 + 12 + 15 + 16 + 17 + 17 = 86

now, divide 86 by how many numbers we have: 6.

86 / 6 = 14.3