# Time and Distance Practice Test for SBI PO Prelims

TIME AND DISTANCE PRACTICE TEST
(PREVIOUS YEAR QUESTION)
1.The speed of a boat in still water is 6 kmph and that of current is 3 kmph. The boat starts from point A and rows to point B and comes back to point A. It takes 12 hours during this journey. How far is point A from point B?
a)27 km
b)25 km
c)20 km
d)30 km
e)None of these

2.A 360 metre long train runs at the speed of 80 kmph. In what time will the train cross a man running at 8 kmph in the same direction of train?
a)16 seconds
b)18 seconds
c)20 seconds
d)15 seconds
e)None of these

3.The speed of a boat in still water is 9.5 kmph while that of current is 2.5 kmph. If the boat takes 114 minutes in rowing from point A to B and coming back to point A, what is the distance between A and B?
a)8.4 km
b)4.8 km
c)8.8 km
d)7.4 km
e)None of these

4.A car starts at 11 am from point A towards point B at 36 kmph, while another car starts at 1 pm from point B towards A at 44 kmph. They cover a distance of 592 km till meeting. At what time will they meet each other?
a)8 pm
b)6:30 pm
c)7:30 pm
d)5:30 pm
e)None of these

5.Train A crosses a pole and platform in 18 seconds and 39 seconds respectively. The length of platform is 157.5 metre. What will be the length of the train B if it is equal to the sum of half of the length of train A and twice the length of the platform?
a)382.5 metre
b)328.5 metre
c)238.5 metre
d)315 metre
e)None of these

6.An employee may claim Rs.7.00 for each km when he travels by taxi and Rs.6.00 for each km if he drives his own car. If in one week he claimed Rs.675 for travelling 90 km, how many kms did he travel by taxi?
a)135
b)155
c)162
d)170
e)None of these

7.Ravi can walk a certain distance in 40 days when he rests 9 hours a day. How long will he take to walk twice the distance, twice as fast and rest twice as long each day?
a)40 days
b)80 days
c)50 days
d)100 days
e)None of these

8.A plane left 30 minutes later than the scheduled time and in order to reach the destination 1500 km away in time, it had to increase the speed by 250 km/hr from the usual speed. Its usual speed is:
a)720 km/hr
b)730 km/hr
c)740 km/hr
d)750 km/hr
e)None of these

9.The speed of a boat in still water is 17.5 kmph and that of current is 2.5 kmph. The boat goes from X to Y in downstream and returns to point Z. The whole journey takes 429 minutes. The distance between Z and Y is 2/5th of the distance between X and Y. Find the total distance covered by the boat. (Approximated to the nearest integer)
a)130 km
b)140 km
c)160 km
d)120 km
e)None of these

10.The speed of the boat in still water is 16 kmph and speed of the current is 2 kmph. It takes a total of 6.5 hours to row up-stream from point A to point B and downstream from point B to point C. If the distance from point A to point B is two-third the distance between point B and C, what is the total distance travelled by the boat (both up-stream and downstream) ?
a)112 km
b)98 km
c)124 km
d)90 km
e)105 km

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

Solution
1.Rate of downstream = (6 + 3) = 9 kmph
Rate of upstream = (6 - 3) = 3 kmph
If the required distance be x km, then
(x/9) + (x/3) = 12
or, 4x/9 = 12
or, x = 27 km

2.Relative speed of train = (80 - 8) kmph = (72 × 5/18) m/sec = 20 m/sec
Required time = (360/20) = 18 seconds

3.Rate downstream of boat = (9.5 + 2.5) = 12 kmph
Rate upstream of boat = (9.5 - 2.5) = 7 kmph
Distance between A and B = x km (let)
According to the question,
(x/7) + (x/12) = 114/60
or, 19x/7 = 114/5
or, x = 8.4 km

4.Let both cars meet each other after t hours from 11 am.
now, 36 × t + 44(t - 2) = 592
or, 80t = 680
or, t = 8.5 hours i.e, 7:30 pm

5.Let, length of train A = x metre
so, (x/18) = (x + 157.5)/39
or, x = 135 metre
Therefore, length of train B = (135/2 + 2 × 157.5) metre = 382.5 metre

6.Let the employee travelled x kms by taxi.
Therefore, distance covered by him by his own car = (90 - x) km
According to the question,
7x + 6(90 - x) = 675
or, x = 135

7.Walking hours per day = (24 - 9) = 15 hours
Total time = (15 × 40) = 600 hours
Decrease in walking hours per day after increase in hours of rest = 24 - (2 × 9) = 6 hours
Required time = (600/6) = 100 days

8.Usual speed of plane = x kmph
Therefore, new speed = (x + 250) kmph
so, (1500/x) - 1500/(x + 250) = 30/60
or, x(x + 250) = 750000
or, x = 750 kmph

9.Rate downstream of boat = (17.5 + 2.5) = 20 kmph
Rate upstream of boat = (17.5 - 2.5) = 15 kmph
Distance XY = x km.
Therefore, distance YZ = (2x/3) km.
Total time = 429 minutes = 143/20 hours
So, (x/20) + 2x/(5 × 15) = 143/20
or, 23x = 143 × 15
or, x = 93 km (approx)
Therefore, total distance = x + 2x/5 = 7x/5 = (7 × 93)/5 = 130 km (approx)

10.Length of BC = x km (let)
length of AB = 2x/3 km
Rate downstream = (16 + 2) = 18 kmph
Rate upstream = (16 - 2) = 14 kmph
According to the question,
2x/(3×14) + (x/18) = 6.5
or, 13x = 6.5 × 126
or, x = 63
Total distance = 2x/3 + x = 5x/3 = (5 × 63)/3 = 105 km. 