Sunita cuts a sheet of paper into three pieces. Length of first piece is equal to the average of the three single digit odd prime numbers. Length of the second piece is equal to that of the first plus one-third the length of the third. The third piece is as long as the other two pieces together. The length of the original sheet of paper is

The correct option is 30 units.

Let us break down the problem step-by-step to find the length of each piece of paper and then the total length of the original sheet.

Step 1: Find the length of the first piece
The first piece's length is equal to the average of the three single-digit odd prime numbers.
Let's identify the single-digit odd prime numbers. The prime numbers are 2, 3, 5, 7, 11, etc. Among these, the single-digit prime numbers are 2, 3, 5, and 7. The odd single-digit prime numbers are:
3, 5, and 7
Now, we calculate their average:

$$\text{Average}=\frac{3+5+7}{3}=\frac{15}{3}=5$$

Thus, the length of the first piece (let's denote it as L₁) is:
L₁ = 5 units

Step 2: Set up equations for the second and third pieces
Let L₂ be the length of the second piece, and L₃ be the length of the third piece.
According to the problem:
1. The length of the second piece is equal to that of the first plus one-third the length of the third:

$$L_2=L_1+\frac{1}{3}{L}_3$$

Substituting L₁ = 5:

$$L_2=5+\frac{1}{3}$$

2. The third piece is as long as the other two pieces together:

$$L_3=L_1+L_2$$

Substituting L₁ = 5:

$$L_3=5+L_2$$

Step 3: Solve for the lengths of the second and third pieces
We can substitute the expression for L₂ into the equation for L₃:

$$L_3=5+\left(5+\frac{1}{3}{L}_3\right)$$

Simplify the equation:

$$L_3=10+\frac{1}{3}$$

Subtract $\frac{1}{3}{L}_3$ from both sides:

$$L_3-\frac{1}{3}$$

$$\frac{2}{3}{L}_3=10$$

Multiply both sides by 3 and divide by 2:

$$L_3=10\times \frac{3}{2}=15\text{units}$$

Now, find the length of the second piece (L₂) using L₃ = 15:

$$L_2=5+\frac{1}{3}\times 15=5+5=10\text{units}$$

Step 4: Find the total length of the original sheet
The original sheet length is the sum of the lengths of all three pieces:

$$\text{Total Length}=L_1+L_2+L_3$$

$$\text{Total Length}=5+10+15=30\text{units}$$

Therefore, the length of the original sheet of paper is indeed 30 units.