A wallet 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

Put the value of y into x = 2y + 4 and solve for x

x = 2 (14) + 4

x = 32

Put the value of y into z = y - 6 and solve for z

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

There are 14 $ 10 bills, 8 $ 20 bills and 32 $ 5 bills.

Step-by-step explanation:

Let x be the number of $10 bills,

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

So, the number of $ 20 bills = x - 6,

Also, the number of $5 bills exceeds twice the number of $10 bills by 4,

So, the number of $ 5 bills = 2x + 4

Thus, the total amount = $ 10 * x + $ 20 (x - 6) + $ 5 (2x + 4)

= 10x + 20x - 120 + 10x + 20

= (40x - 100) dollars

According to the question,

40x - 100 = 460

40x = 460 + 100

40x = 560

⇒ x = 14

Hence, the number of $ 10 bills is 14, $ 20 bills is 8 and $ 5 bills is 32.