## 18 May 2016

### SBI Clerk Prelims 2016- Practice Aptitude Questions (Time and Distance)

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 7th 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

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
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
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.
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
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
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
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
8). Required average speed = (2 × 30 × 20) / (30 + 20)
= (2 × 30 × 20) / 50 = 24 kmph
9). Speed of cyclist = x kmph
Time = t hours
Distance = xt / 2 while time = 2t
Required ratio = xt / (2 × 2t) = 1 : 4
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