IAS 2019 — CSAT Maths

Correct Option (D)

The problem requires determining the total length of an original sheet of paper, which is divided into three pieces. The lengths of these pieces are defined as follows:

• The single-digit odd prime numbers are 3, 5, and 7.
• Length of the first piece = Average of (3, 5, 7) = $\frac{3 + 5 + 7}{3} = \frac{15}{3} = \textbf{5 units}$33+5+7​=315​=5 units.

Let the length of the second piece be $L_2$L2​ and the length of the third piece be $L_3$L3​.

• According to the problem statement, $L_2 = 5 + \frac{1}{3}L_3$L2​=5+31​L3​.
• Also, $L_3 = 5 + L_2$L3​=5+L2​.

Substitute the expression for $L_3$L3​ into the equation for $L_2$L2​:

$$L_2 = 5 + \frac{1}{3}(5 + L_2)$$L2​=5+31​(5+L2​)

$$L_2 = 5 + \frac{5}{3} + \frac{1}{3}L_2$$L2​=5+35​+31​L2​

$$L_2 - \frac{1}{3}L_2 = 5 + \frac{5}{3}$$L2​−31​L2​=5+35​

$$\frac{2L_2}{3} = \frac{15 + 5}{3}$$32L2​​=315+5​

$$\frac{2L_2}{3} = \frac{20}{3}$$32L2​​=320​

$$2L_2 = 20$$2L2​=20

$$L_2 = \textbf{10 units}$$L2​=10 units.

Now, calculate the length of the third piece:

• $L_3 = 5 + L_2 = 5 + 10 = \textbf{15 units}$L3​=5+L2​=5+10=15 units.

The total length of the original sheet of paper is the sum of the lengths of the three pieces:

• Total length = Length of Piece 1 + Length of Piece 2 + Length of Piece 3
• Total length = 5 + 10 + 15 = 30 units.

Therefore, the correct option is D.

Incorrect Options:

Options A, B, and C are incorrect because they do not represent the sum of the calculated lengths of the three pieces of paper. The individual lengths are 5 units, 10 units, and 15 units, respectively. Any sum other than 30 units would be erroneous based on the problem's conditions and mathematical derivation.

• Option A (13 units): This value is incorrect as the sum of the calculated lengths is 30 units.
• Option B (15 units): This value is incorrect. It corresponds to the length of the third piece only, not the total length of the original sheet.
• Option C (16 units): This value is incorrect as the sum of the calculated lengths is 30 units.