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The speed of a bus increases by 22 kmph after every one hour. If the distance travelled in the first one hour was 3535 km, what was the total distance travelled in 1212 hours?

A.

422 km

B.

552 km

C.

502 km

D.

492 km

Answer with explanation

Answer: Option BExplanation

Distance travelled in 1st1st hour =35=35 km

Speed of the bus increases by 22 kmph after every one hour. Hence,

distance travelled in 2nd2nd hour =37=37 km

distance travelled in 3rd3rd hour =39=39 km

and so on

Total distance travelled

=[35+37+39+⋯(12 terms)]=122[2×35+(12−1)2]=6(70+22)=6×92=552

Workspace

A train covers a distance in 50 min, if it runs at a speed of 48 kmph on an average. The speed at which the train must run to reduce the time of journey to 40 min will be.

A.

45 kmph

B.

60 kmph

C.

75 kmph

D.

None of these

Answer with explanation

Answer: Option BExplanation

Workspace

The distance between two cities A and B is 330330 km. A train starts from A at 88 a.m. and travels towards B at 6060 km/hr. Another train starts from B at 99 a.m. and travels towards A at 7575 km/hr. At what time will they meet?

A.

10.30 a.m.

B.

10 a.m.

C.

12 noon

D.

11 a.m.

Answer with explanation

Answer: Option DExplanation

Workspace

A and B walk around a circular track. A and B walk at a speed of 2 rounds per hour and 3 rounds per hour respectively. If they start at 8 a.m. from the same point in opposite directions, how many times shall they cross each other before 9.30 a.m.?

A.

5

B.

6

C.

7

D.

8

Answer with explanation

Answer: Option CExplanation

Relative speed = Speed of A + Speed of B (∴ they walk in opposite directions)

=2+3=5=2+3=5 rounds per hour

Therefore, they cross each other 55 times in 11 hour and 22 times in 1212 hour

Time duration from 88 a.m. to 9.309.30 a.m. =1.5=1.5 hour

Hence they cross each other 77 times before 9.309.30 a.m.

Workspace

A.

2 Km/h

B.

3 Km/h

C.

4 Km/h

D.

5 Km/h

Answer with explanation

Answer: Option BExplanation

Let he walked for t hour on first day, so he walked for (t + 2) hour on second day.

Let s km/h be the speed of the man on first day. Then (s + 1) km/h be the speed of the man on second day.

So, according to the question,

t + t + 2 = 18

=> 2t = 16

=> t = 8 h

And st + (s + 1)(t + 2) = 64

=> 8s + 10s + 10 = 64

=> 18s = 54

=> s = 3 km/h

Workspace

A.

2.14 km

B.

1.34 km

C.

2.54 km

D.

1.39 km

Answer with explanation

Answer: Option DExplanation

Let speed of the man be x km/h then speed of the train is (145-x) km/h.

According to the question,

2400/1000 = (76-x)*10/60

24*6 = 760 – 10x

10x = 760 – 144

10x = 616

x = 61.6 km/h

speed of train =83.4km/h

Length of the first train = 70*1/60 = 1.39 km

Workspace

Ravi has a roadmap with a scale of 1.5 cm for 18 km. He drives on that road for 72 km. What would be his distance covered in that map.

A.

4 cm

B.

6 cm

C.

7 cm

D.

8 cm

Answer with explanation

Answer: Option BExplanation

18 km on road = 1.5 cm on map

1 km on road = 1.5/18 cm on map

72 km on road = 1.5 * 72/18 = 6 cm on map

Workspace

A dog is 50 of its own leaps behind a cat. The dog takes 5 leaps per minute to the cat’s 4. If the dog and the cat cover 8m and 5m per leap respectively, what distance will the dog have to run before it catches the cat?

A.

600 m

B.

700 m

C.

800 m

D.

1000 m

Answer with explanation

Answer: Option CExplanation

Distance covered by dog in one minute = 5 X 8 = 40 m

Distance covered by dog in one leap = 8 m

Distance between dog and cat = 50 leaps of dog = 50 X 8 = 400 m

Speed of dog = 40 metre/minute

Speed of cat = 20 metre/minute

Time to cover 400 m distance by dog = 400/ (40-20) = 20 minute (distance / relative speed)

Distance travelled by dog to catch the cat = 40 X 20 = 800 m (speed X time)

Workspace

Ankit goes to his school for extra classes by cycle at 10 km/h and reached at 1pm. If he cycles at 15 km/h, he will reach at 11:00 am. At what speed must he cycle to get there at noon?

A.

11 km/h

B.

12 km/h

C.

13 km/h

D.

14 km/h

Answer with explanation

Answer: Option BExplanation

Since the distance is fix

So if the speed be increased by half then time is also reduced by 1/3 hours

If initial time is t hours

t/3 = 2 => t = 6 hours

Or we may calculate time as

10 X t = 15 X (t-2) => t = 6 hours

Now the distance travelled = 10 X 6 or 15 X 4 = 60 km

But the new time is 5 hours

So, the required speed = 60/5 = 12 km/h

Workspace

A train started from point A at a speed of 60 km/hr and after 2 hours another train of same length started from A at a speed of 80 km/hr in the same direction as the first one. After how much time the second train will meet the first train?

A.

5 hours

B.

3 hours

C.

6 hours

D.

8 hours

Answer with explanation

Answer: Option CExplanation

Let after x hours the second train will meet the first train.

Because distance is same,

S_{1} t_{1} = S_{2} t_{2}

60 (x + 2) = 80 × x

60x + 120 = 80x

80x – 60x = 120

20x = 120

x = 6 hours

Workspace

Two persons start running simultaneously around a circular track of length 300 m from the same point at speeds of 15 km/hr and 25 km/hr. When will they meet for the first time anywhere on the track if they are moving in opposite directions?

A.

25 sec

B.

27 sec

C.

30 sec

D.

23 sec

Answer with explanation

Answer: Option BExplanation

Time taken to meet for the first time anywhere on the track

=> length of the track / relative speed

=> 300 / (15 + 25)5/18

=> 300x 18 / 40 x 5 = 27 seconds.

Workspace

A walks around a circular field at the rate of one round per hour while B runs around it at the rate of six rounds per hour. They start at the same point at 7:30 am. They shall first cross each other at?

A.

7:30 am

B.

7:15 am

C.

7:50 am

D.

7: 42 am

Answer with explanation

Answer: Option DExplanation

Relative speed between two = 6-1 = 5 round per hour

They will cross when one round will complete with relative speed,

which is 1/5 hour = 12 mins.

So 7:30 + 12 mins = 7:42

Workspace

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