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1. Walking at the rate of 4 kmph a man cover certain distance in 2hr 45 min. Running at a speed of 16.5 kmph the man will cover the same distance in.

A. 12 min

B. 25 min

C. 40 min

D. 60 min

**Answer : ****40 min**

From the problem we understand that the distance is same in both scenarios. Only speed and time vary.

While walking, distance1 = speed1 x time1

While running, distance2 = speed2 x time2

distance1 = distance2

speed1 x time1 = speed2 x time2

4 x 2 hr 45 min = 16.5 x time2

4 x 165 = 16.5 x time2 (Changing time in minutes)

time2 = 40 min.

2. A travels from B to C a distance of 250 miles in 5.5 hours. He returns to B in 4 hours 40 minutes. His average speed is

A. 44

B. 46

C. 48

D. 50

**Answer : ****50**

Total distance = 250 + 250 = 500 miles

Total time = 5 hr 30 min + 4 hr 40 min = 10 hr 10 min = 10 (1/6)

Speed = distance / time

Speed = 500 / 10(1/6) = 3000 / 61

Average speed = 49.18 ≈ 50

(**Note :** Average speed stands for actual speed in this context. We calculated the speed finally considering two cases since the distance is same)

3. In a 4000 meter race around a circular stadium having a circumference of 1000 meters, the fastest runner and the slowest runner reach the same point at the end of the 5th minute, for the first time after the start of the race. All the runners have the same staring point and each runner maintains a uniform speed throughout the race. If the fastest runner runs at twice the speed of the slowest runner, what is the time taken by the fastest runner to finish the race ?

A. 20 min

B. 15 min

C. 10 min

D. 4. 5 min

**Answer : ****10 min**

The ratio of the speeds of the fastest and the slowest runners is 2 : 1.

Hence they should meet at only one point on the circumference i.e. the starting point (As the difference in the ratio in reduced form is 1).

For the two of them to meet for the first time, the faster should have completed one complete round over the slower one.

Since the two of them meet for the first time after 5 min, the faster one should have completed 2 rounds (i.e. 2000 m) and the slower one should have completed 1 round. (i.e. 1000 m) in this time.

Thus, the faster one would complete the race (i.e.
4000 m) in **10 min**.

4. A snail is at the bottom of a 20 meters deep pit. Every day the snail climbs 5 meters upwards, but at night it slides 4 meters back downwards. How many days does it take before the snail reaches the top of the pit?

A. 4 days

B. 8 days

C. 16 days

D. 20 days

**Answer : ****16 days**

Since the snail moves up 1 meter a day (5 meters upwards, but at night it slides 4 meters back downwards) so it will reach 15 meters in 15 days.

Next day it will again climb 5 meters upward and reaches the top. (During daytime itself)

**Note:**

This is a very common problem in all competitive examinations. On 16^{th} day, it will climb 5 meters and reach the top during the daytime itself and there is no point of sliding down since it has already reached the top.

5. Walking at the rate of 4 kmph a man cover certain distance in 2 hr 45 min. Running at a speed of 16.5 kmph the man will cover the same distance in.

A. 12 min

B. 25 min

C. 40 min

D. 60 min

**Answer : ****40 min**

Distance = Speed * time

2 hr 45 min --> 2 (3/4) hr --> 11/4 hr

4* (11/4) = 11km

New Speed = 16.5 kmph

Therefore time in hrs = D/S = 11/16.5 = 0.6666.. hr

Time in mins = 0.666.. x 60 = 40 min

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