You are given $893 in one, five, and ten dollar bills. There are 165 bills. There are twice as many five dollar bills as there are ones and tens combined. How many bills of each type are there?

There are 23 bills of $1 type.

There are 110 bills of $5 type.

There are 32 bills of $10 type.

According to the given situation

You are given $893 in one, five, and ten dollar bills.

Let the number of $1 bills be "x"

Let the number of $5 bills be "y"

Let the number of $10 bills be "z"

Total number of bills = x + y + z  

Also it is given that there are twice as many five dollar bills as there are ones and tens combined

hence we can write the equations in the form of 3 variables

These are three linear equations  with 3 variables that can be solved for x, y and z .

On solving equations (1) , (2),  and (3) we get

So we can conclude that there are 23 bills of $1 type.

There are 110 bills of $5 type.

There are 32 bills of $10 type.

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Given:

Total number of bills of one, five and ten = 165

Total amount = $893

There are twice as many five dollar bills as there are ones and tens combined.

To find:

Number of bills of each type.

Solution:

Let number of bills of one, five and ten are x, y and z respective.

Total number of bills of is 165.

        ...(i)

Total amount is $893.

        ...(ii)

There are twice as many five dollar bills as there are ones and tens combined.

    ...(iii)

From equation (iii), we get

      ...(iv)

Putting in (i), we get

Putting y=110 in (iv), we get

      ...(v)

Putting y=110 and x=55-z in (ii), we get

Divide both sides by 9.

Putting z=32 in (v), we get

Therefore, the number of one, five and ten bills are 23, 110 and 32 respectively.

8121 because

130,000 divided by 16 is 8121