Neat work!

SDT#7.

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When running towards the thief, the relative speed of dog is 60-40=20 kmph and when running back towards the police, the relative speed is 60+50=110 kmph!

Now, the total distance travelled by dog i.e. 720 km is travelled with two speeds of 20 kmph and 110 kmph in 12 hrs! Let x be the time for which he runs in the thieves direction!

Hence, 110(12-x)+20x=720

x=20/3

Therefore, distance travelled in the direction of thief is 60*20/3=400 km !! ]]>

In 3 hrs the thief has run 120 km. Relative speed of thief and police is 10 kmph. So, police will take 12 hrs to catch the thief and thus the dog will run for the same amount of time (no matter in what direction!)!

Hence, the total distance travelled by the dog is 12*60=720 km!

720 is the total distance covered by dog, the question asks the distance covered by dog in direction of the thief... you have to remove the distance which dog covered coming back to police each time after seeing the thief...

]]>Hence, the total distance travelled by the dog is 12*60=720 km!]]>

SDT # 6

A man riding a cycle at 12 km/hour can reach his village in 4½ hours. If he is delayed by 1½hour at the start, in order to reach his destination in time, at what speed should he ride?

[Solution]

Total distance = 12 X 4.5 = 54 Kms

If he starts 1.5 hrs late, then he has only (4.5 - 1.5) = 3 Hrs to commute 54 Kms.

So he has to cycle at speed of :

**V = 54/3 = 18 Km/hour**

SDT # 4

X is picked up by his father by car from school everyday and they reach home at 5.00 p.m. One day, since the school got over an hour earlier than usual, he started walking towards home at 3 km/hr. He met his father on the way and they reached home 15 minutes earlier than their usual time. What is the speed of the car?

[Solution]

H=Home, S=School, M=Father and Son meet

H========================================M=============S

(Time = ?) (Car arrives at 5.00 pm)

The car arrives at 5.00 pm at point 'S'. But since Mr X started walking at

4.00 pm, they will meet at a point 'M'.

The car saved travel distance M -> S and S -> M. This resulted in 15 minutes

savings. Assuming uniform speed, it takes 7 1/2 minutes from 'M' to 'S' and

vice-versa. So both meet at 4:52 1/2 pm. The Son had been walking from 4.00 pm

till this time, which is 52 1/2 min.

The same distance is covered by the car in just 7 1/2 min. Therefore the car

travels at 52.5/7.5 times the walking speed of Mr X (3 km/hr).

********************************************************

Speed of the car = 3 X 52.5/7.5 = 21 km/hr

Distance from S = 21 X 7.5/60 = 2 5/8 km (2.625 km)

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A man riding a cycle at 12 km/hour can reach his village in 4½ hours. If he is delayed by 1½hour at the start, in order to reach his destination in time, at what speed should he ride?

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