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# Time and Distance Questions for SSC-CGL Set-2 PDF

Download SSC CGL Time and Distance Questions with answers set-2 PDF based on previous papers very useful for SSC CGL exams. Very important Time and Distance Questions for SSC exams.

Question 1: Two cars travel from city A to city B at a speed of 60 and 108 km/hr respectively. If one car takes 2 hours lesser time than the other car for the journey, then the distance between City A and City B?

a) 270 km

b) 324 km

c) 405 km

d) 216 km

Question 2: Aman and Kapil start from Delhi and Gwalior respectively towards each other at the same time. They meet at Mathura and then take 196 minutes and 225 minutes respectively to reach Gwalior and Delhi. If speed of Aman is 30 km/hr, then what is the speed (in km/hr) of Kapil?

a) 28

b) 30

c) 225/7

d) 392/15

Question 3: A train leaves Delhi at 10 a.m. and reaches Jaipur at 4 p.m. on the same day. Another train leaves Jaipur at 12 p.m. and reaches Delhi at 5 p.m. on the same day. What is the time of day (approximately) when the two trains will meet?

a) 1:42 p.m.

b) 1:27 p.m.

c) 2:04 p.m.

d) 1:49 p.m.

Question 4: Two people A and B are at a distance of 260 km from each other at 9:00 a.m. A immediately starts moving towards B at a speed of 25 km/h and at 11:00 a.m. B starts moving towards A at a speed of 10 km/hr. At what time (in p.m.) will they meet each other?

a) 5:00

b) 6:00

c) 6:30

d) 7:00

Question 5: The ratio of speed of three racers is 3 : 4 : 6. What is the ratio of time taken by the three racers to cover the same distance?

a) 3:4:6

b) 6:4:3

c) 4:3:2

d) 2:3:5

Question 6: If I walk at 7/6 of my usual speed, then I reach my office 15 minutes early. What is the usual time taken (in minutes) by me to reach the office?

a) 60

b) 75

c) 90

d) 105

Question 7: 37 trees are planted in a straight line such that distance between any two consecutive trees is same. A car takes 20 seconds to reach the 13th tree. How much more time (in seconds) will it take to reach the last tree?

a) 36

b) 40

c) 57

d) 60

Question 8: After repairing a scooter runs at a speed of 54 km/h and before repairing runs at speed of 48 km/h. It covers a certain distance in 6 hours after repairing. How much time will it take to cover the same distance before repairing?

a) 6 hours 15 minutes

b) 6 hours 45 minutes

c) 7 hours

d) 7 hours 30 minutes

Question 9: A runner starts running from a point at 6:00 am with a speed of 8 km/hr. Another racer starts from the same point at 8:30 am in the same direction with a speed of 10 km/hr. At what time of the day (in p.m.) will the second racer will overtake the other runner?

a) 8:00

b) 4:00

c) 6:30

d) 5:30

Question 10: If a person walks at 15 km/hr instead of 9 km/hr, he would have walked 3 km more in the same time. What is the actual distance (in kms) travelled by him?

a) 5.5

b) 6.5

c) 4.5

d) 7.5

Let the distance between City A and City B = $d$ km

Speed of first car = 60 km/hr and speed of second car = 108 km/hr

Let time taken by first car = $t$ hrs and time taken by second car = $(t – 2)$ hrs

Using, speed = distance/time for first car :

=> $\frac{d}{t} = 60$

=> $d = 60t$ ————–(i)

For second car, => $\frac{d}{t – 2} = 108$

Substituting value of $d$ from equation (i), we get :

=> $60t = 108t – 216$

=> $108t – 60t = 48t = 216$

=> $t = \frac{216}{48} = 4.5$ hrs

From equation (i), => $d = 60 \times 4.5 = 270$ km

=> Ans – (A)

Time taken by Aman from Mathura to reach Gwalior = 196 minutes and 225 minutes by Kapil to reach Delhi from Mathura.

=> Time taken by them to reach Mathura from their respective starting points = $t=\sqrt{196\times225}$

=> $t=14\times15=210$ minutes

Now, total time taken by Aman to reach Gwalior from Delhi = $210+196=406$ minutes

and total time taken by Kapil = $210+225=435$ minutes

Also, speed of Aman = 30 km/hr

Thus, total distance between Delhi and Gwalior = $d=(30\times406)$ km

$\therefore$ Speed of Kapil = $\frac{30\times406}{435}=28$ km/hr

=> Ans – (A)

Total time taken by first train = 6 hours and second train = 5 hours

$\because speed \propto\frac{1}{time}$

=> Let speed of first train = $5x$ km/hr and speed of second train = $6x$ km/hr

=> Distance between Delhi and Jaipur = $5x\times6=30x$ km

At 10:00 am, first train starts moving, => distance covered by it in 2 (12-10) hours = $5x\times2=10x$ km

Distance left between the two trains at 12:00 pm = $30x-10x=20x$ km

Now, at 12:00 pm both move towards each other, => relative speed = $5x+6x=11x$ km/hr

=> Time taken to meet each other = $(\frac{20x}{11x}\times60)$ minutes $\approx 109$ minutes

= 1 hour and 49 minutes

$\therefore$ Time at which they will meet = 12:00 pm + 1 hour and 49 minutes = 1:49 pm

=> Ans – (D)

Speed of A = 25 km/hr and speed of B = 10 km/hr

Distance between A and B = 260 km

At 9:00 am, A starts moving, => distance covered by A in 2 (11-9) hours = $25\times2=50$ km

Distance left between A and B at 11:00 am = $260-50=210$ km

Now, at 11:00 am both move towards each other, => relative speed = $25+10=35$ km/hr

=> Time taken to meet each other = $\frac{210}{35}=6$ hours

$\therefore$ Time at which they will meet = 11:00 am + 6 hours = 5:00 pm

=> Ans – (A)

Ratio of speed = 3 : 4 : 6

Also, $speed\propto\frac{1}{time}$

=> Ratio of time taken = $\frac{1}{3}:\frac{1}{4}:\frac{1}{6}$

Multiplying by L.C.M.(3,4,6) = 12

= $12(\frac{1}{3}:\frac{1}{4}:\frac{1}{6})$

= $4:3:2$

=> Ans – (C)

Let my usual speed = $6x$ km/min and my usual time to reach office = $t$ minutes

New speed = $\frac{7}{6}\times6x=7x$ km/min and new time = $(t-15)$ minutes

$\because speed\propto\frac{1}{time}$

=> $\frac{6x}{7x}=\frac{(t-15)}{t}$

=> $\frac{6}{7}=\frac{t-15}{t}$

=> $6t=7t-105$

=> $7t-6t=t=105$ minutes

=> Ans – (D)

Let the distance between any two consecutive trees = $d$ metres

=> Total distance between the 37 trees = $36d$ metres

Time taken to reach the 13th tree = 20 seconds

=> Speed of car = distance/time

= $\frac{12d}{20}=\frac{3d}{5}$ m/s

Thus, time taken to reach the last tree from the first tree = $\frac{36d}{\frac{3d}{5}}$

= $36\times\frac{5}{3}=60$ seconds

$\therefore$ The car will take (60-20) = 40 seconds more to reach the last tree

=> Ans – (B)

Speed before repairing = $s’=48$ km/hr and after repairing = $s=54$ km/hr

Let time taken to cover the distance before repairing = $t’$ hours and after repairing = $t=6$ hours

Using, distance = speed x time

=> $s’\times t’=s\times t$

=> $48\times t’=54\times6$

=> $t’=\frac{54}{8}=6.75$ hours

=> $t’=6.75\times60=405$ minutes

$\therefore$ Time taken to cover the same distance before repairing = 405 minutes = 6 hours 45 minutes

=> Ans – (B)

Speed of first runner = 8 km/hr starting at 6:00 am

Speed of second runner = 10 km/hr starting at 8:30 am

=> Distance covered by the first runner till 8:30 am (2.5 hours) = $8\times2.5=20$ km

Now, at 8:30 am, the first runner is 20 km ahead of second runner who just started running and since they are running in same direction, => relative speed = 10 – 8 = 2 km/hr

Time taken to overtake the first runner = distance/speed

= $\frac{20}{2}=10$ hours

$\therefore$ Time of the day when the second racer will overtake the other runner = 8:30 am + 10 hours = 6:30 pm

=> Ans – (C)