# Competitive Examination Question

### Quantitative Aptitude Question for Competitive Examinations

Try solving this questions which has appeared in CAT, GRE and GMAT papers and many other competitive examinations. So try your hands at this question from topic speed and distance. You can expect similar questions in competitive examinations.

Here is the Competitive Examination Question -

A thief escaped from police custody. Since he was a sprinter, he could run at a speed of 40 km/hr. The police realized it after 3 hr and started chasing him in the same direction at 50km/hr. The police had a dog, which could run at 60 km/hr. The dog would run to the thief and then return back to the police and then would turn back towards the thief. It kept on doing so till the police caught the thief. Find the total distance traveled by the dog in the direction of the thief?

Here are the options -

a) 720 km

b) 600 km

c) 660 km

d) 360 km

e) 230 km

I will provide the solution later. I urge everyone to try this. Please give explanation with your answers -

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time taken by police to catch Thief 50x = 40x+120… x=12 hrs.

Police will catch thief at 600 Km,

Dog would have traveled 600 Km, in direction of Thief..

(& total distance traveled by dog = 720 km are 600 Km in direction of Thief + 120 Km in direction of Police)

720 km.Police take 12 hrs to nab the thief.Hence dog ran for 12 hrs @ 60 km/hr.

People saying 720 missed the line “dog travelled in the direction of thief?” . total distance covered by the dog is 720 , but in the direction of the thief it’s 660.

Method of elimination is one important tool for competitive exams when one has to choose from a set of options. It can be used as follows-

1) Police takes 12 hrs to catch the thief (initial lead of 120 km covered by police with relative velocity of 10 km/hr)

The dog continues to chase for 12 hrs to and fro betwen police and thief. Hence, its total run towards the thief is less than 12 hrs

The dog can run 720 km in 12 hrs (12*60km/hr velocity of the dog)

Hence, actual distance covered by the dog towards the thief is less than 720 km

Option (a) is eliminated

2) The police covers 12 hr*50 km/hr=600 km during this chase. It may be noted that the dog has chased to and fro during this period, while police had chased only towards thief.

Hence, dog has chased more than the police

This implies the dog has run more than 600 km but less than 720 km. Out of the available options, the answer is (c) 660 km.

This method allows you to save time a guess a range and finally select the anaswer from given options through method of elimination.

Running of dog to and fro is some or less mechanical process which will be happening until police dont catch the thief.Even though Police and dog have chased approximately the same time period the distance covered by them would differ minutely.

Though dog runs for 12 hrs approximately he doesnt cover exactly 720 hours.

But 660 is a huge deviation from 720 kms and wont be considered as an authenticate answer.

So i think though your reasoning is correct, 720 seems to be a fair deal.

PLS Give your comments……

it is confirm 660.. i’ve checked it by plotting it on a graph.. i like your elimination method

The thief have already covered 40*3=120km before the police have started chasing him.

The distance covered by the police to catch the thief can be given by

50*x=40*x+120 (say x is the time taken)

hence x=12 hrs

Now, the dog with a spped of 60kmph travelled 60*12=720km.

its 720 kms the dog has travelled

the dog has traveled 720KM.

because.

in the initial 3 hrs the thief has escaped 120KM(40*3),

and then the police went to chase the thief who is 120KM away with 50kmph.

so relative speed will be 10KMPH(50-40)

so with this 10KMPH police will take 12H(120/10) to catch the thief.

during this 12H period the DOG IS CONSTANTLY RUNNING at 60KMPH, so the dog covered the distance of 720KM(60*12).

@Rajeev

In 6 hours, the thief has covered 40*6 = 240 km in addition to initial 120 km lead.

In the same 6 hours, the dog covers 60*6 = 360km.

At the distance of 360 km, the dog catches (meets) the thief for the first time. At that point of time, the police has covered 50*6 = 300 km.

The dog runs back towards the police.

The gap between police and the dog is 60km.

The net speed of the dog with respect to police is 60+50 = 110km/hr.

Time taken to cover this gap is (60/110) hr.

In this period the dog covers (60/110) *60 = 32.73km towards police.

By this time the thief has run further (60/110)*40 = 21.82km.

The gap between dog and thief is now 32.73+21.82 = 54.55km

The net speed of dog with respect to thief is 60-40 = 20km/hr

At this speed, it takes 54.55/20 = 2.728 hrs. (distance covered 2.728*60 = 163.64km towards thief)

After meeting the thief, the dog turns back and runs towards police.

This story goes on…

Thus the dog covers the following distances:

360km towards thief,

32.73km towards police,

163.64km towards thief,

14.88km towards police

and so on.

It is a long series of motions back and forth.

Ultimately at the end of 12 hours both police and dog meet the thief together.

I hope I am clear.

please explain me this logic

It is interesting to note that the forward and backward movements of the dog are a very long series. Considering an accuracy of 1m, there are about 15 backward movements of the dog.

The dog covers the following distances:

360, -32.73, 163.64, -14.88 and so on.

The forward distances are a geometric series with the first term as 360km and the ratio as 5/11. The sum of the series is 360/(1-5/11) = 660km.

The backward distances are again a geometric series with the first term as 32.73km and the ratio as 5/11. The sum of the series is 32.73/(1-5/11) = 60km.

Poor dog. If only it had slowed down to the speed of the police, it would not have oscillated like this!

How did you say that the series is in GP (is there any logic other than concluding it from the distance ratios which you got).

Thanks in advance.

Great, I never knew this, thanks.

It’s posts like this that keep me coming back and checking this site regularly, thanks for the info!

we can make an equation using the given information,the equation being 50t = 40t+120this equation is satisfied at t = 12 hrs. that means the time taken by the police to catch the thief is 12 hours. The dog runs at a speed of 6o kmph,Therefore the total distance covered by the dog = 60 kmph * 12 h = 720 km+1

it is 660 km in the direction of theif

Further to my earlier note:

It is interesting to note that the forward and backward movements of the dog are a very long series. Considering an accuracy of 1m, there are about 15 backward movements of the dog.

The dog covers the following distances:

360, -32.73, 163.64, -14.88 and so on.

The forward distances are a geometric series with the first term as 360km and the ratio as 5/11. The sum of the series is 360/(1-5/11) = 660km.

The backward distances are again a geometric series with the first term as 32.73km and the ratio as 5/11. The sum of the series is 32.73/(1-5/11) = 60km.

Poor dog. If only it had slowed down to the speed of the police, it would not have oscillated like this!

Initial distance of thief from police = 3hrs * 40km/hr = 120km.

In x hrs, police catch up with the thief.

50x = 40x + 120. So, x=12 hrs.

In 6 hrs, dog runs 60 *12 = 720.

Out of this, net 600km towards thief + 120 to & fro between thief and police.

Total distance towards thief = 600 + 120/2 = 660km.

got some different logic now…

continuing my from 1st post

time taken by police to catch Thief 50x = 40x+120… x=12 hrs.

Hence Police will catch thief at 600 Km,

So same has to be for Dog … i.e. Dog would have traveled 600 Km, in direction of Thief..

(& total distance traveled by dog = 720 km are 600 Km in direction of Thief + 120 Km in direction of Police)

plz explain how dog would have travelled 600…

we can make an equation using the given information,

the equation being

50t = 40t+120

this equation is satisfied at t = 12 hrs. that means the time taken by the police to catch the thief is 12 hours.

The dog runs at a speed of 6o kmph,

Therefore the total distance covered by the dog = 60 kmph * 12 h

= 720 km

time taken by police to catch Thief 50x = 40x+120… x=12 hrs.

therefor distance travel by dog km till the time police catches thief = 60*12 = 720