Hi all

Here is the answer and explanation to the DS question sent to you a week back. For the benefit of those who joined late here is the question.

**Question**

The following data sufficiency question is a question that tests your knowledge about circles in geometry.

Directions to answer the DS question

1. Statement (A) ALONE is sufficient, but statement (B) alone is not sufficient to answer the question asked.

2. Statement (B) ALONE is sufficient, but statement (A) alone is not sufficient to answer the question asked.

3. BOTH statements (A) and (B) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.

4. EACH statement ALONE is sufficient to answer the question asked.

5. Statements (A) and (B) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

Question

A right circular cone has height H and radius R. A small cone is cut off at the top by a plane parallel to the base. At what height above the base the section has been made?

A. H = 20 cm

B. Volume of small cone: volume of large cone :: 1:15The following data sufficiency question is a question that tests your knowledge about circles in geometry.

**Correct Answer is choice (3).**

**Explanation**

**From Statement A**: H = 20; we do not know the height at which the section has been cut. Hence, we can eliminate choices (1) and (4). We are down to choices (2), (3) or (5).

**Statement B**: Let the volume of the small cone be (1/3) pi*r*r*h; r is radius and h is the height.

Then (1/3) pi*r*r*h = (1/15)* (1/3) pi*R*R*H

=> h = (1/15) (R*R*H/r*r) . From this information, we cannot find the answer for h. Hence, statement B alone is not sufficient. We can eliminate choice (2).

**Combining both the statements**: When a section is made the two cones are similar triangles. so h/H = r/R

Therefore, R = rH/h

We know H = 20

h = (1/15) ( rH/h) ( rH/h)*H /r*r)

=> h^3 = (1/15)(H^3)

h can be found.

Hence, choice 3 is the answer.

An IIM alumni venture