Time and Distance questions Set 1(Hard)

1.A man goes to the fair with his and dog. Unfortunately man misses his son which he realizes 20 minutes later. The son comes back towards his home at the speed of 20 m/min and man follows him at 40 m/min. The dog runs to son and comes back to the man to show him the direction of his son. He keeps moving to and fro at 60 m/min between son and father, till the man meets the son. What is the distance traveled by the dog in the direction of the son?

A. 800 m
B. 1675 m
C. 848 m
D. 1000 m
E. None of these
Answer: Option D
Solution

In 20 minutes the difference between man and son,
= 20*20
= 400m.
Distance traveled by dog when he goes towards the child,
= 400*60/40
= 600 m and time required is 10 minutes.
In 10 min remaining distance between man and child,
= 400 - (20*10)
= 200 m.
Time taken by dog to meet the man,
= 200/100
= 2 min

(Relative speed of dog with child is 40 m/min and same with man is 100 m/min.)

Remaining distance in 2 min,
= 200-(2*20),
= 160 m.
Now, the time taken by dog to meet the child again,
= 160/40
= 4 min.
In 4 min he covers = 4*60 = 240 m.
Now, remaining distance in 4 min = 160-(4*20) = 80 m.
Time required by dog to meet the man once again = 80/100 = 0.8 min.
Now remaining distance = 80-(0.8*20) = 64 m.<

2.Due to the technical snag in the signal system two trains start approaching each other on the same track from two different stations, 240 km away each other. When the train starts a bird also starts moving to and fro between the two trains at 60 kmph touching each train each time. The bird initially sitting on the top of the engine of one of the trains and it moves so till these trains collide. If these trains collide one and half hour after start, then how many kilometers bird travels till the time of collision of trains?

A. 90 km
B. 130 km
C. 120km
D. 95 km
E. None of these
Answer: Option A
Solution

Time taken to collide the trains,
= one and half hour = 3/2 h.
So, in 3/2 h bird travels,
= (3*60)/2
= 90km.


3.A train met with an accident 60 km away from Anantpur station. It completed the remaining journey at 5/6 th of the previous speed and reached Barmula station 1 hour 12 in late. Had acciden taken place 60 km further , it would have been only 1 hour late .A.) What is the normal speed of train?B.) What is the distance between Anantpur and Barmula?

A. 60 km/hr, 420km
B. 80km/hr, 800km
C. 8km/hr, 80km
D. None of these
Solution

First Method:

Distance = Speed * Time.
When distance is constant, speed is inversely proportional to Time that is,
Speed is proportional to 1/Time.
Let normal time is T from the point of accident and normal speed is S.
If speed become (5/6)*S then time (6/5)T.
Extra Time = (6/5)T – T = T/5.
Here, given that train was late = 1 hour, 12 minutes = 6/5 hours. (Minutes are converted into hours.)
Now,
T/5 = 6/5
T = 6 hour.

Again,
If accident took place 60 km further, train would be just 1 hour late.
So,
T/5 = 1
T = 5 hours.
It means train travels 60 km in hour.
Now,
Total distance,
 = 60 *6 + distance from Annantpur to accident point.
= 360 + 60 = 420 km.
Normal speed = 60 kmph.

Second Method:
Let normal speed of the train be x km/hr and distance between Anantpur and Barmula be y km.
normally train takes y/x hrs to reach Trent Bridge.
Case 1:
Time taken till accident =60/x hours.
Time taken 2 cover remaining distance= 6 * (y-60)/5x
According to question,
60/x + 6 * (y-60)/5x= y/x + 1.2 (converting minute into hour)
on solving this equation we get,
y-6x = 60 .................(1)
Case 2:
Time taken till accident =(60+60)/x hr=120/x hr
Time taken 2 cover remaining distance= 6 * (y-120)/5x
Now,
120/x + 6 * (y-120)/5x= y/x + 1
on solving this equation we get,
y-5x=120 .................(2)
On solving (1) and (2) we get
x=60
y=420

4.An individual is cycling at a speed of 25 km per hour. He catches his predecessor who had started earlier in two hours. What is the speed of his predecessor who had started 3 hours earlier ?

A. 15 kmph
B. 12 kmph
C. 10 kmph
D. 8 kmph
Answer: Option C
Solution

The distance covered in two hour,
= 2*25 = 50 km
Time taken by first individual = (3h +2h) = 5 h
Then, the speed of predecessor,
= 50/5 = 10 kmph.

5.A racetrack is in the form of a right triangle. The longer of the legs of track is 2 km more than the shorter of the legs (both these legs being on a highway). The start and end points are also connected to each other through a side road. The escort vehicle for the race took the side road and rode with a speed of 30 km/h and then covered the two intervals along the highway during the same time with a speed of 42 km/h. find the length of the race track.

A. 14 km
B. 10 km
C. 24 km
D. 36 km
E. 30 km
Answer: Option A
Solution

The given conditions are met on a Pythagoras triplet 6, 8, 10.
Since, the racetrack only consists of the legs of the right angle triangle the length must be,
= 6+8 = 14 km.