A question is given, followed by three statements labelled I, II and III. Identify which of the statements is/are sufficient to answer the question.

Question:

What is the time (in second) taken by A to run the race?

Statements:

I. A beats B by 30 seconds and B beats C by 60 seconds.

II. A beats C by 90 seconds.

III. B's speed is the average of A's speed and C's speed.

Correct option is D

Given:

Question:
What is the time (in second) taken by A to run the race?

Statements:
I. A beats B by 30 seconds and B beats C by 60 seconds.
II. A beats C by 90 seconds.
III. B's speed is the average of A's speed and C's speed.

Formula Used:

$$\text{Time} = \frac{\text{Distance}}{\text{Speed}} \\\\\text{If A beats B by } t \text{ seconds:} \\\frac{D}{S_B} = \frac{D}{S_A} + t\\[10pt]$$

Solution:

Let:
T_A = time taken by A
T_B = time taken by B = T_A + 30
T_C = time taken by C = T_B + 60 = T_A + 90

From Statement I:
We know the time differences:
T_B = T_A + 30 and T_C = T_A + 90
But we do not know the actual value of T_A
=> Statement I alone is not sufficient

​From Statement III:

B's speed is the average of A's and C's speeds:

​$$\\S_B = \frac{S_A + S_C}{2} \\\text{Since speed is inversely proportional to time (for same distance):} \\S_A = \frac{1}{T_A}, \quad S_B = \frac{1}{T_B}, \quad S_C = \frac{1}{T_C}\\[10pt]\text{Substitute into the average formula:} \\\frac{1}{T_B} = \frac{1}{2} \left( \frac{1}{T_A} + \frac{1}{T_C} \right)\\[10pt]\text{Now substitute } T_B = T_A + 30 \text{ and } T_C = T_A + 90: \\\frac{1}{T_A + 30} = \frac{1}{2} \left( \frac{1}{T_A} + \frac{1}{T_A + 90} \right)\\[10pt]$$​​

This is a solvable equation in one variable T_A.

=> Statements I and III together are sufficient to find T_A.

Conclusion:

• Statement I gives time differences
• Statement III gives a relationship between the speeds
• Combining both, we get a solvable equation in T_A​

Final Answer: (D) Statements I and III are sufficient

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