1. An athlete runs 200 meters race in 24 seconds. His speed in km/h is

a) 20

b) 24

c) 28.5

d) 30

Explanation:

$$\eqalign{ & {\text{Speed of athlete}} \cr & = \frac{{200\,{\text{m}}}}{{24\,{\text{secs}}}} \cr & = \frac{{200 \times 18}}{{5 \times 24}} \cr & = 30\,{\text{km/hour}} \cr} $$

2. If A travels to his school from his house at the speed of 3 km/h, then he reaches the school 5 minutes late. If he travels at the speed of 4 km/h, he reaches the school 5 minutes earlier than school time. The distance of his school from his house is:

a) 1 km

b) 2 km

c) 3 km

d) 4 km

Explanation: Let the distance between school and home be x km.

The difference of time when A goes school to school with these two different speed is 10 min

$$\eqalign{ & = \frac{{10}}{{60}}\,{\text{hour}} \cr & {\frac{x}{3}} - {\frac{x}{4}} = \frac{{10}}{{60}} \cr & \frac{x}{{12}} = \frac{1}{6} \cr & x = \frac{{12}}{6} \cr & = 2\,{\text{km}} \cr} $$

3. A man starts climbing a 11 m high wall at 5 pm. In each minute he climbs up 1 m but slips down 50 cm. At what time will he climb the wall?

a) 5:30 pm

b) 5:21 pm

c) 5:25 pm

d) 5:27 pm

Explanation: Man climbs 1m and slips down 50 cm (0.5m) in one minute

i.e. he climbs (1 - 0.5 = 0.5 m) in one minute.

But in the last minute he will be climbing 1m as he gets on the top so no slip.

Time taken to climb 11 meter = $$ {\frac{{10}}{{0.5}} + 1} $$ = 21 minutes.

He climbs the wall at 5:21 pm

4. A monkey climbs a 60 m high pole. In first minute he climbs 6 m and slips down 3 m in the next minute. How much time is required by it to reach the top?

a) 35 minutes

b) 33 minutes

c) 37 minutes

d) 40 minutes

Explanation: Monkey climbs 6m in 1

^{st}minute and slips down 3 m in next minute

i.e. Monkey climbs 3 meter in 2 minute then he climbs in one minute,

= $$\frac{3}{2}$$ m.

But in the last minute he climbs 6 m as he gets on the top so there is no slip.

Time required = $$2 \times \frac{{54}}{3} + \frac{6}{6}$$ = 37 minutes

5. An ant climbing up a vertical pole ascends 12 meters and slips down 5 meters in every alternate hour. If the pole is 63 meters high how long will it take it to reach the top?

a) 18 hours

b) 17 hours

c) 16 hours 35 min

d) 16 hours 40 min

Explanation: Since it climbs 7 meter in every 2 hr. At the end of 14 hours it would have climbed up to 49 meter. In 15th hour it will reach 61 meter. Then drops back to 56 meter in the 16th hour. It will take another 35 min to travel remaining 7 meter since it will be ascending in the 17th hour. Thus, 16 hr and 35 min.

6. Two buses start from a bus terminal with a speed of 20 km/h at interval of 10 minutes. What is the speed of a man coming from the opposite direction towards the bus terminal if he meets the buses at interval of 8 minutes?

a) 3 km/h

b) 4 km/h

c) 5 km/h

d) 7 km/h

Explanation: Let Speed of the man is x kmph.

Distance covered in 10 minutes at 20 kmph = distance covered in 8 minutes at (20 + x) kmph.

$$20 \times \frac{{10}}{{60}} = \frac{8}{60}\times\left(20+x\right) $$

200 = 160 + 8x

8x = 40

x = 5kmph.

7. Walking $$\frac{3}{4}$$ of his normal speed, Rabi is 16 minutes late in reaching his office. The usual time taken by him to cover the distance between his home and office:

a) 48 min.

b) 60 min.

c) 42 min.

d) 62 min.

Explanation: $$\frac{4}{3}$$ of usual time = Usual time + 16 minutes;

Hence, $$\frac{1}{3}{\text{rd}}$$ of usual time = 16 minutes;

Usual time = 16 × 3 = 48 minutes.

8. Two trains for Mumbai leave Delhi at 6 am and 6.45 am and travel at 100 kmph and 136 kmph respectively. How many kilometers from Delhi will the two trains be together:

a) 262.4 km

b) 260 km

c) 283.33 km

d) 275 km

Explanation: Difference in time of departure between two trains = 45 min. = $$\frac{{45}}{{60}}$$ hour = $$\frac{{3}}{{4}}$$ hour.

Let the distance be x km from Delhi where the two trains will be together.

Time taken to cover x km with speed 136 kmph be t hour

and time taken to cover x km with speed 100 kmph (As the train take 45 mins. more) be

$$ {{\text{t}} + \frac{3}{4}} $$

= $$ {\frac{{4{\text{t}} + 3}}{4}} $$

Now,

100 × $$ {\frac{{4{\text{t}} + 3}}{4}} $$ = 136t

25(4t + 3) = 136t

100t + 75 = 136t

36t = 75

t = $$\frac{{75}}{{36}}$$ = 2.083 hours

Distance x km = 136 × 2.083 ≈ 283.33 km.

9. A man takes 6 hours 15 minutes in walking a distance and riding back to starting place. He could walk both ways in 7 hours 45 minutes. The time taken by him to ride back both ways is:

a) 4 hours

b) 4 hours 30 min.

c) 4 hours 45 min.

d) 5 hours

Explanation: Time taken in walking both the ways = 7 hours 45 minutes ----- (i)

Time taken in walking one way and riding back = 6 hours 15 minutes ----- (ii)

By the equation (ii) × 2 - (i),

Time taken by the man in riding both ways,

= 12 hours 30 minutes - 7 hours 45 minutes

= 4 hours 45 minutes

10. A man completes a certain journey by a car. If he covered 30% of the distance at the speed of 20kmph. 60% of the distance at 40km/h and the remaining of the distance at 10 kmph, his average speed is:

a) 25 km/h

b) 28 km/h

c) 30 km/h

d) 33 km/h

Explanation:

$$\eqalign{ & {\text{Let the total distance be 100 km}}. \cr & {\text{Average speed}} \cr & = \frac{{{\text{total}}\,{\text{distance}}\,{\text{covered}}}}{{{\text{time}}\,{\text{taken}}}} \cr & = \frac{{100}}{{ { {\frac{{30}}{{20}}} + {\frac{{60}}{{40}}} + {\frac{{10}}{{10}}} } }} \cr & = \frac{{100}}{{ { {\frac{3}{2}} + {\frac{3}{2}} + 1 } }} \cr & = \frac{{100}}{{ {\frac{{ {3 + 3 + 2} }}{2}} }} \cr & = \frac{{ {100 \times 2} }}{8} \cr & = 25\,\text{kmph} \cr} $$