Speed Time And Distance - Study Mode

[#91] An old man driving bike at 80 km per hour. However being sugar patient, old man could not travel continuously. He takes small breaks each of 2 minutes for every 15 minute of his drive. How much distance the old man will cover in 90 minutes?
Correct Answer

(B) 104 Km

Explanation

Solution: For every 15 minutes he takes a rest of 2 minutes. Hence, for 90 minutes of drive he would require 12 minutes of rest. Thus, he will be traveling for 90 - 12 = 78 minutes. In 60 minutes he covers 80 Km. In 1 minute he would cover $$frac{{80}}{{60}}$$ Km. In 78 minutes he would cover $$frac{{80}}{{60}} imes 78$$ xa0 = 104 Km.

[#92] Two trains start simultaneously from two stations Howrah and Delhi, respectively towards each other on the same track. The distance between the two stations is 560 km and the speeds of trains are 30 kmph and 40 kmph.
Simultaneously with the trains, a sparrow sitting on the top of one of the train starts towards the other and reverses its direction on reaching the other train and so on. If the speed of sparrow is 80 kmph then the distance that the sparrow lies before being crushed between the train is :
Correct Answer

(C) 640 km

Explanation

Solution: Relative speed of the trains = 40 + 30 = 70 kmph. Time taken by trains to collide, = $$frac{{560}}{{70}}$$ = 8h. In 8h sparrow will cover, = 8 × 80 = 640 km.

[#93] 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?
Correct Answer

(A) 90 km

Explanation

Solution: Time taken to collide the trains, = one and half hour = $$frac{3}{2}$$ h. So, in $$frac{3}{2}$$ h bird travels, = $$frac{{3 imes 60}}{2}$$ = 90 km.

[#94] A man goes to the fair with his son 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?
Correct Answer

(D) 1000 m

Explanation

Solution: In 20 minutes the difference between man and son, = 20 × 20 = 400 m. Distance traveled by dog when he goes towards the child, = $$400 imes frac{{60}}{{40}}$$ = 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, = $$frac{{200}}{{100}}$$ = 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, = $$frac{{160}}{{40}}$$ = 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 = $$frac{{80}}{{100}}$$ = 0.8 min. Now remaining distance = 80 - (0.8 × 20) = 64 m.

[#95] A tiger is 50 of its own leaps behinds a deer. The tiger takes 5 leaps and per minutes to the deer's 4. If the tiger and the deer cover 8 m and 5 m per leap respectively, what distance will the tiger have to run before it caches the deer?
Correct Answer

(C) 800 m

Explanation

Solution: Speed of tiger = 40 m/min Speed of deer = 20 m/min. Relative speed = 40 - 20 = 20 m/min. Initial difference in distance = 50 × 8 = 400 m Time take to catch = $$frac{{400}}{{20}}$$ = 20 min. Distance traveled in 20 min, = 20 × 40 = 800 m