1) A man completed a certain journey by a car. If he covered 30% of the distance at the speed of 20km/hr, 60% of the distance at 40km/hr and the remaining distance at 10km/hr, his average speed for the whole journey was

B) 28 Km/hr

C) 30 Km/hr

D) 33 Km/hr

**A) 25 Km/hr**B) 28 Km/hr

C) 30 Km/hr

D) 33 Km/hr

*Solution :*Let total distance =100 km.

Then total time = 30/20 + 60/40+10/10 = 1.5 + 1.5 + 1= 4 hrs.

Average speed = 100/4 = 25 km/hr.

——————————————————————————————————————–

2. The ratio of lengths of two trains is 5 : 3 and the ratio of their speeds is 6 : 5. The ratio of times taken by them to cross a pole is…

A) 5:6

B) 11:8

D) 27:16

A) 5:6

B) 11:8

**C) 25:18**D) 27:16

*Solution :*To cross a pole, train has to travel distance equal to its length

Let the lengths of the trains be 5x and 3x and their speeds be 6y and 5y.

Ratio of the time taken 5x/6y : 3x/5y= 25:18

——————————————————————————————————————–

3) If a train takes 6 seconds to cross a pole and if the length of train is 120 metres, then the speed of the train in km/hr will be

A) 60 Km/hr

C) 70 Km/hr

D) 80 km/hr

A) 60 Km/hr

**B) 72 Km/hr**C) 70 Km/hr

D) 80 km/hr

**Time = 6 sec**

*Solution :*Distance = 120m (as length of pole can be taken negligible)

Speed = Distance/Time = 120/6 = 20 m/sec = 20 x 18/5 = 72km/hr

Speed = Distance/Time = 120/6 = 20 m/sec = 20 x 18/5 = 72km/hr

——————————————————————————————————————–

4) Two trains of equal length, running in opposite directions, pass a pole in 18 and 12 seconds. The trains will cross each other in…

B) 15.5 sec

C) 18.8 sec

D) 20.2 sec

**A) 14.4 sec**B) 15.5 sec

C) 18.8 sec

D) 20.2 sec

*Solution :* Let length of each train be x meter.

Then speed of first train = x /18 m/sec

Speed of 2nd train = x/12 m/sec

When both trains cross each other,

Time taken = (length of train 1 + length of train 2)/(speed of train 1+ speed of train 2)

= [(2x/(x/(18) + (x/12)]

= 2x/[(2x +3x)/36] = 2x * 36/5x = 72/5 = 14.4 seconds

——————————————————————————————————————–

5) A man rides at the rate of 18 km/hr, but stops for 6 minutes to change horses at the end of every 7th km. The time that he will take to cover a distance of 90 km

A) 6 hours

C) 6 hours 18 minutes

D) 6 hours 24 minutes

A) 6 hours

**B) 6 hours 12 minutes**C) 6 hours 18 minutes

D) 6 hours 24 minutes

**Total riding time to cover 90 km= 90/18= 5hrs**

*Solution :*Number of times horse changed = 90/7=12 times (since he stops after every 7 kms)

Time taken in changing hours = 12 X 6 = 72min = 1 Hr 12 Min

Total time taken = 5 Hr+1 Hr 12 Min = 6 hours 12minutes

——————————————————————————————————————–

6) From two places, 60 km apart, A and B start towards each other at the same time and meet each other after 6 hours. Had A travelled with 2/3 of his speed and B travelled with double of his speed, they would have met after 5 hours. The speed of A is…

A) 4 Km/hr

C) 10 Km/hr

D) 12 Km/hr

A) 4 Km/hr

**B) 6 Km/hr**C) 10 Km/hr

D) 12 Km/hr

*Solution :* Let speeds of A and B be x and y km/hr respectively

Then x + y = 60/6 =10………..(1)

And 2x/3+2y=60/5=12?12x+6y=36………(2)

And 2x/3+2y=60/5=12?12x+6y=36………(2)

From (1) and (2)

4x =24 ? x = 6 km/hr.

——————————————————————————————————————–

7) Two men start together from the some place in the same direction to go round a circular path. If one takes 10 minutes and the other takes 15 minutes to make one complete round they will meet after…

B) 33 min

C) 40 min

D) 45 min

**A) 30 min**B) 33 min

C) 40 min

D) 45 min

*Solution :*LCM of 10 and 15 = 30

They will meet after 30 minutes

——————————————————————————————————————–

8. Two trains started at the same time, one from A to B and the other from B to A. if they arrived at B and A respectively 4 hours and 9 hours after they passed each other, the ratio of the speeds of the two trains was.

A) 2:1

C) 4:3

D) 5:4

A) 2:1

**B) 3:2**C) 4:3

D) 5:4

**If two trains (or bodies) start at the same time from points A and B towards each other and after crossing they take a and b units of time in reaching B and A respectively,**

*Solution :*then: (A’s speed) (B’s speed)- vb : va

Therefore, Ratio of the speeds of two trains = V90 : V4 = 3:2

——————————————————————————————————————–

9) By walking at 3/4 of his usual speed, a man reaches his office 20 minutes later than his usual time. The usual time taken by him to reach his office is…

A) 75 min

C) 40 min

D) 30 min.

A) 75 min

**B) 60 min**C) 40 min

D) 30 min.

*Solution :* Let the original speed be S and time be T

If new speed Sx then new time would be Tx4/3 (D = ST = Constant)

Given 4T/3 -T = 20

T = 60 Minutes = 1 hour

——————————————————————————————————————–

10) A train starts from A at 7 a.m. towards B with speed 50 km/h. Another train from B starts at 8 a.m. with speed 60 km/h towards A. Both of them meet at 10 a.m. at C. The ratio of the distance AC to BC is…

A) 5:6

C) 6:5

D) 4:5

A) 5:6

**B) 5:4**C) 6:5

D) 4:5

*Solution :* >A———- C———–B<

Distance travelled by first train starting at 7 AM = AC = 50×3 =150Km

Distance travelled by second train starting at 8 AM = AC = 60×2 =120Km

Ratio of distance covered = AC : BC = 150:120 = 5:4

——————————————————————————————————————–

11) A thief is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km and 11 km per hour respectively. The distance (in metres) between them after 6 minutes is ?

A) 190 m

B) 200 m

D) 150 m

A) 190 m

B) 200 m

**C) 100 m**D) 150 m

*Solution :* Relative speed of the thief and policeman = (11-10) km/hr = 1 km/hr

Distance covered in 6 minutes

= 1/60×6 = 1/10 kms = 100 meters

So distance between them after 6 minutes = 200-100 =100 meters

——————————————————————————————————————–

12) Shri X goes to his office by scooter at a speed of 30 km/h and reaches 6 minutes earlier. If he goes at a speed of 24 km/h, he reaches 5 minutes late. The distance of his office is..

A) 20 Km

B) 21 Km

D) 24 km

A) 20 Km

B) 21 Km

**C) 22 Km**D) 24 km

*Solution :* Let the distance of Office = X Km

Right time to reach office = T min

@ 24 km/h time taken to reach office = (T+5)/60=X/24 ……..(1)

@ 30 km/h Time taken to reach office = (T-6)/60=X/30 ……..(2)

Subtracting Equitation 2 from Equitation 1

11/60 = X/24 -X/30

X = 22 km

Please note that I have divided time by 60 to convert it into hour and I have taken Speed in Km/hr

——————————————————————————————————————–

13) Two trains travel in the same direction at the speeds of 56 km/h and 29 km/h respectively. The faster train passes a man in the slower train in 10 seconds. The length of the faster train (in metres) is.

A) 100 m

B) 80 m

D) 120 m

A) 100 m

B) 80 m

**C) 75 m**D) 120 m

*Solution :*Relative speed of faster train & man in slower train = 56 – 29 = 27 km per hour = 27 x 5/18 = 15/2 m/ second

To pass a man, train has to cover distance equal to it’s own length.

Distance travelled = (time * speed) = 10 * 15/2 = 75 meters

Thus the length of faster train is 75 meters