A and B run a 5$$\frac{1}{2}$$​ km race on a 500 m long circular track starting from the same point, simultaneously, in the same direction. The speed of A is 50% faster than B.

Which of the following statements is/are correct?

Statement I: A and B meet for the first time after starting the race when A covers 1200 m.

Statement II: A meets B three times during the race.

Correct option is D

Given:
A and B run a 5.5 km (i.e., 5500 m) race on a 500 m circular track
They start simultaneously from the same point, in the same direction
Speed of A is 50% faster than B, i.e.,
If speed of B = x, then speed of A = 1.5x

We are to evaluate the correctness of:
Statement I: A and B meet for the first time when A has covered 1200 m
Statement II: A meets B three times during the race

Formula Used:

$$\text{Relative Speed} = \text{Speed of A} - \text{Speed of B}$$

Time for First Meeting: $$\text{Time} = \frac{\text{Track Length}}{\text{Relative Speed}}$$​​

​$$\text{Distance} = \text{Speed} \times \text{Time}$$

Solution:

Step 1: When do A and B meet for the first time?

When moving in the same direction, the relative speed = (Speed of A − Speed of B) = 1.5x − x = 0.5x

They will meet when A has gained one full lap (500 m) on B

Time to meet = $$\frac{\text{Distance}}{\text{Relative Speed}} = \frac{500}{0.5x} = \frac{1000}{x}\ \text{seconds}$$​​

Distance covered by A in that time = speed × time $$= 1.5x \times \left( \frac{1000}{x} \right) = 1500\ \text{m}$$​​

So, Statement I is incorrect (It says A covers only 1200 m)

Step 2: How many times does A meet B during 5500 m?

Each meeting occurs when A gains 1 lap = 500 m on B

Total number of such gains = A's lead distance / 500
But we calculate this by:

Relative distance A gains over B in total = (1.5x − x) × t = 0.5x × t

We find how many times this equals 500, within the race:

Total distance A runs = 5500 m
Each time A gains 500 m, one meeting happens.

From above, A meets B every 1000 m (we already calculated when A covered 1500 m, they met; so per 1000 m relative, 1500 m of A).

So, how many such 1500 m fits in 5500 m?

​$$\frac{5500}{1500} = 3.\overline{6}$$​​

So they meet 3 full times

Statement II is correct

Final Answer: (D) II only