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__SBI Clerk Prelims 2016- Practice Aptitude Questions (Time and Distance) Set-45__Dear Readers, Important Practice Aptitude Questions with solution for Upcoming SBI Clerk Exam, candidates those who are preparing for those exams can use this practice questions.

**1).**A man rides at the rate of 18 km / hr, but stops for 6 mins, to change horses at the end of every 7

^{th}km. The time that he will take to cover a distance of 90 km is

a) 6 hrs

b) 6 hrs. 12 min.

c) 6 hrs. 18 min.

d) 6 hrs. 24 min.

e) None of these

**2).**A man performs 2 / 15 of the total journey by train, 9 / 20 by bus and the remaining 10 km on foot. His total journey in km is

a) 15.6

b) 24

c) 16.4

d) 12.8

e) None of these

**3).**A train 200 m long running at 36 kmph takes 55 seconds to cross a bridge. The length of the bridge is

a) 375 m.

b) 300 m.

c) 350 m.

d) 325 m.

e) None of these

**4).**A train 100 metres long meets a man going in opposite direction at 5 km / hr and passes him in 7(1 / 5) seconds. What is the speed of the train in km / hr?

a) 45 km / hr

b) 60 km / hr

c) 55 km / hr

d) 50 km / hr

e) None of these

**5).**Two trains, each of length 125 metre, are running in parallel tracks in opposite directions. One train is running at a speed 65 km / hr and they cross each other in 6 seconds. The speed of the other train is

a) 75 km/hr

b) 85 km/hr

c) 95 km/hr

d) 105 km/hr

e) None of these

**6).**By walking at 3 / 4 of his usual speed, a man reaches his office 20 minutes later than usual. His usual time is

a) 30 min

b) 75 min

c) 90 min

d) 60 min

e) None of these

**7).**A train travelled at a speed of 35 km / hr for the first 10 minutes and at a speed of 20 km/hr for the next 5 minutes. The average speed of the train for the total 15 minutes is

a) 30 km/hr

b) 23 km/hr

c) 31 km/hr

d) 29 km/hr

e) None of these

**8).**A train goes from Ballygunge to Sealdah at an average speed of 20 km/hr and comes back at an average speed of 30 km/hr. The average speed of the train for the whole journey is

a) 27 km/hr

b) 26 km/hr

c) 25 km/hr

d) 24 km/hr

e) None of these

**9).**A certain distance is covered by a cyclist at a certain speed. If a jogger covers half the distance in double the time, the ratio of the speed of the jogger to that of the cyclist is

a) 1 : 4

b) 4 : 1

c) 1 : 2

d) 2 : 1

e) None of these

**10).**Two trains 180 metres and 120 metres in length are running towards each other on parallel tracks, one at the rate 65 km/hr and another at 55 km/hr. In how many seconds will they be clear of each other from the moment they meet?

a) 6

b) 9

c) 12

d) 15

e) None of these

__Answers__**:**

**1). b) 2). b) 3). c) 4). a) 5). b) 6). d) 7). a) 8). d) 9). a) 10). b)**

__Solution:__**1).**90 km = 12 × 7km + 6km

To cover 7 km total time taken = 7 / 18 hours + 6 min = 88 / 3 min.

So (12 × 7 km) would be covered in [12 × (88 / 3)] min and remaining 6 km is 6 / 18 hrs or 20 min

Total time = (1056 / 3) + 20

= 1116 / (3 × 60) hours

= 6(1 / 5) hours

= 6 hours 12 minutes

**Answer: b)**

**2).**If the total journey be of x km, then

(2x / 15) + (9x / 20) + 10 = x

x – (2x / 15) – (9x / 20) = 10

(60x – 8x – 27x) / 60 = 10

(25x / 60) = 10

x = (60 × 10) / 25 = 24 km

**Answer: b)**

**3).**Speed of train = 36 kmph

= 36 × (5 / 18) = 10 m / sec

If the length of bridge be x metre, then

10 = (200 + x) / 55

200 + x = 550

x = 550 – 200 = 350 metre.

**Answer: c)**

**4).**Speed of train = x kmph

Relative speed = (x + 5) kmph

Length of train = 100 / 1000 km = 1 / 10 km

(1 / 10) / (x + 5) = 36 / (5 × 60 × 60)

1 / 10(x + 5) = 1 / 500

x + 5 = 50

x = 45 kmph

**Answer: a)**

**5).**Length of both trains = 250 metre

Speed of second train = x kmph

Relative speed = (65 + x) kmph

= (65 + x) × (5 / 18) m/sec

Time = Sum of lengths of trains / relative speed

6 = 250 / [(65 + x) × (5 / 18)]

6 × (5 / 18) × (65 + x) = 250

65 + x = (250 × 3) / 5

65 + x = 150

x = 150 – 65 = 85 kmph

**Answer: b)**

**6).**Speed varies inversely as time

New speed = (3 / 4) × usual speed

New time = (4 / 3) × usual time

(4 / 3)t – t = 20 minutes

(1 / 3) × usual time = 20 minutes

Usual time = 3 × 20 = 60 minutes

**Answer: d)**

**7).**Distance covered = [ 35 × (10 / 60) + 20 × (5 / 60) ] km

= [(35 / 6) + (10 / 6)] = 45 / 6 km

Total time = 15 minutes = 1 / 4 hour

Required average speed = Distance covered / Time taken

= (45 / 6) × 4 = 30 kmph

**Answer: a)**

**8).**Required average speed = (2 × 30 × 20) / (30 + 20)

= (2 × 30 × 20) / 50 = 24 kmph

**Answer: d)**

**9).**Speed of cyclist = x kmph

Time = t hours

Distance = xt / 2 while time = 2t

Required ratio = xt / (2 × 2t) = 1 : 4

**Answer: a)**

**10).**Required time = Sum of the lengths of trains / Relative speed

Relative speed = 65 + 55 = 120 kmph = (120 × 5) / 18 m/sec

Required time = (180 + 120) / [(120 × 5) / 18]

= (300 × 18) / (120 × 5) = 9 seconds

**Answer: b)**