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51. 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 ?

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Solution:
The distance covered in two hour,
= 2 × 25 = 50 km
Time taken by first individual = (3h + 2h) = 5h
Then, the speed of predecessor,
= 505 = 10 kmph.
52. A 6 cm long cigarette burns up in 15 minutes if no puff is taken.For every puff, it burns three times as fast during the duration of the puff.If the cigarette burns itself in 13 minutes, then how many puffs has the smoker taken if the average puff lasted 3 seconds?

Discuss
Solution:
Let the number of puffs abe,
3x×(3×1150)+(13×60−3x)×1150=6
On solving, We get, x = 20 puffs.
53.  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?

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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 8060 Km.
In 78 minutes he would cover 8060×78   = 104 Km.
54. 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 :

Discuss
Solution:
Relative speed of the trains = 40 + 30 = 70 kmph.
Time taken by trains to collide,
= 56070
= 8h.
In 8h sparrow will cover,
= 8 × 80
= 640 km.
55. 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?

Discuss
Solution:
Time taken to collide the trains,
= one and half hour = 32 h.
So, in 32 h bird travels,
= 3×602
= 90 km.
56. 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?

Discuss
Solution:
In 20 minutes the difference between man and son,
= 20 × 20
= 400 m.
Distance traveled by dog when he goes towards the child,
= 400×6040
= 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,
= 200100
= 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,
= 16040
= 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 = 80100 = 0.8 min.
Now remaining distance = 80 - (0.8 × 20) = 64 m.
57. 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?

Discuss
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 = 40020 = 20 min.
Distance traveled in 20 min,
= 20 × 40
= 800 m
58. A candle of 6 cm long burns at the rate of 5 cm in 5 hour and an another candle 8 cm long burns at the rate of 6 cm in 4h. What is the time required to each candle to remain of equal lengths after burning for some hours, when they starts to burn simultaneously with uniform rate of burning?

Discuss
Solution:
(6 - x) = (8 - 1.5x)
Or, x = 4 cm.
So, It will take 4 hours to burn it in such a way that they will remain equal in lengths.
59. A man walking at the speed of 4 km/hr,cross a square field diagonally in 3 minutes.The area of the field is?

Discuss
Solution:
Speed of man,
= 4 kmph = 4×518 m/sec
In 3 min (180 sec) man will go = 20×1808 = 200 m.
That means the diagonal of the square field = 200 m.
Diagonal of square,
= Side of Square × 2  = 200 m.
→ Side of Square = 2002
Area of Square = Side2
Area = 200×2002   = 20000 sqm.
60. A train approaches a tunnel AB. Inside the tunnel a cat located at a point i.e. 512 of the distance AB measured from the entrance A. When the train whistles the Cat runs. If the cat moves to the exit B, the train catches the cat exactly the exit. The speed of the train is greater than the speed of the cat by what order ?

Discuss
Solution:
Train(T)__________ A_____5k____CAT__________B

T<-------x--------------><----12k-------------------------->

Let the speed of train be u and the speed of Cat be v and train whistles at a point T, X km away from A. Let AB = 12k and Cat was 5k distance away from A. Time was constant for both, then
⇒vu=x5k=x+12k7k⇒7x=5(x+12k)⇒xk=301Thus,⇒uv=305=61
  or,6:1

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