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Quantitative Aptitude Question for Competitive Examinations

July 6th, 2010 Posted in Uncategorized

Quantitative Aptitude Question: Topic – Time, Speed and Distance

Try solving this question 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 –

35 Responses to “Quantitative Aptitude Question for Competitive Examinations”

  1. vkm108 Says:

    ans is 360


  2. debs Says:

    yep, the answer is 360


  3. ANKIT Says:

    CAN SOMEBODY POST THE DETAILED SOLUTION PLEASE


  4. avinash Says:

    ans: 360
    relative velocity b/w thief and police = 10 km/hr
    distance = distance traveled by thief in 3hr = 120 km
    therefore time = 12hr
    in 12 hr dog would cover 720km
    in the direction of thief = 360km


  5. hari Says:

    correct me if im wrong

    logical approach

    Hour Km by Thief Km by Cop Km by Dog
    3 120 0 0
    4 160 50 60
    5 200 100 120
    6 240 150 180
    7 280 200 240
    8 320 250 300
    9 360 300 360

    so after 9 hours (ie 6 hrs travel) the dog reached the theif, right? at this point itself it ran 160 km. after this it started running towards the cop for app 30-33 mins to 33 km(ie at 9.5 hr).At this time the theif was at 380 km(ie 53 km far).So it have 2 run again for 120km to reach the theif and again it will start towards the cop.

    Den how come u guys say it ran only 360km?????

    By my approach it ran (360+30+150+15+45)= around 600 km….

    Sorry if im wrong.Waiting 4 ur reply.


  6. hari Says:

    correct me if im wrong
    logical approach

    H-hour;
    T-Km by Thief
    C-Km by Cop
    D-Km by Dog

    H T C D
    3 120 0 0
    4 160 50 60
    5 200 100 120
    6 240 150 180
    7 280 200 240
    8 320 250 300
    9 360 300 360
    so after 9 hours (ie 6 hrs travel) the dog reached the theif, right? at this point itself it ran 160 km. after this it started running towards the cop for app 30-33 mins to 33 km(ie at 9.5 hr).At this time the theif was at 380 km(ie 53 km far).So it have 2 run again for 120km to reach the theif and again it will start towards the cop.
    Den how come u guys say it ran only 360km?????
    By my approach it ran (360+30+150+15+45)= around 600 km….
    Sorry if im wrong.Waiting 4 ur reply.


  7. Alka Says:

    3 120 0 0
    4 160 50 60
    5 200 100 120
    6 240 150 180
    7 280 200 240
    8 320 250 300
    9 360 300 360
    9.5 380 325 330 dog: 360+30
    10 400 350 360 dog: 360+30+30
    11 440 400 420 dog: 360+30+30+60
    12 480 450 480 dog: 360+30+30+60+60
    12.3 492 465 465 dog: 360+30+30+60+60+15
    13 520 500 525 dog: 360+30+30+60+60+15+45 = 600 KM

    dog: 360+30+30+60+60+15+45 = 600 KM


  8. hari Says:

    hi alka,
    I think u missed a point in the question.Did you see the question properly? It asked to find the distance traveled by the dog in the direction of the thief.
    I think you added the distance it traveled in the direction of cop with this distance it travelled in the direction of thief. Am i right?

    So it ll come as 360+30+60+60+45=555km which is weird and approx to 600 km.
    I think your answer is right even u misunderstood the question….
    Sorry if im wrong…..
    Thank You


  9. namrata Says:

    the thief’s speed is 40 and police realises this in 3 hours so in that time he has travelled 120 kms. now the distance between police and thief is 120 kms and relative speed is (50-40)= 10 kmph. so police will reach thief in 12 hours.

    now the dog’s speed is 60 kmph. so it will travell 12 *60 =720 kms in 12 hours out of which 720 /2 = 360 kms will be in the direction of thief :)ans : 360


  10. sridevi Says:

    In 3hrs the thief can travel 120km.
    Time taken for the police to catch the thief is D/u-v
    =120/50-40=12hrs.
    Time taken by the dag to catch the thief=
    120/60-40=6hrs.
    therefore in the first 6 hrs the dag can travel 360 km n catches the thief.
    And in the next 6 hrs it travels 360 km from the thief to the police.
    Ans=360km


  11. Bhimraj Meena Says:

    ans=660km
    here D=dog, P=police, T=thief
    initially (D,P)T and D(dog) will take 6 hr time to reach to T(thief) n will travel d1=6*60km in thief’s direction. distance travel by P in this time is=10*6=60km. now distance between P n T is 120-60=60km & situation is
    P(D,T)
    D will return toward P with relative speed of 110kmph wrt P n will take 60/110=6/11 hr to reach P, & in this time P will move by distance=6*10/11=60/11km towards T. thus disance between P & T will remain 60-60/11=10*60/11km
    (P,D)T
    now D will take time to reach T(thief) =60*10/(11*20)hr
    & distance travel by D toward T this time is d2=60*10*60/(11*20)=60*6*(5/11)km .
    in this time P also travel by 60*10*10/(11*20)km so nw dist betn P & D wud remain 60*10/11-60*10/(11*2)=60*10/(11*2)km.
    nw situation is
    P(D,T) again D wil go toward P & calculating in the same manner
    (P,D)T n from this
    d3=6*60*(5/11)^2km n similarly d4,d5,…can b calculated which form geometrical series.
    so total distance travel by D toward T is= 6*60*{1+5/11+(5/11)^2+….}
    =6*60*1/(1-5/11)
    =660km


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  13. Satya Says:

    the answer is 660. The dog takes 6 hrs to reach the thief the first time. Distance traveled is 360kms. In this time, the police has traveled 300kms.
    Then Dog travels back (360-300) = 60kms to meet the police at a relative speed of 60+50 = 110kmph. time taken for this is 60/110 = 6/11 hrs.
    Dog travels 60×6/11 = 360/11km in this time.
    Thief travels 40×6/11 = 240/11km in this time.
    Next point at which dog catches thief = (360+240)/(11×20) = 30/11 hrs.
    Distance traveled by Dog in this time = 60×30/11 = 1800/11.

    Calculating further, we get a geometric progression as:
    360, 1800/11, 9000/121,…..
    this sums to 660kms.


  14. Abhi Says:

    EASY SOLUTION:
    Answer is 720KM
    HOW?
    See the thief is running at constant speed of 40kmh.Now three hours are passed when the Police realizes this(Usual Indian Police!!!).During this three hours the thief had traveled
    3*40=120KM
    Now the Police start chasing him at constant speed of 50kmph.But since thief is also running at speed of 40kmph hence the relative speed of Police will boil down to ONLY 50-40=10kmph.Now at this rate the distance to be covered by Police is 120kmph, hence he will take time of 120/10=12 hours.
    Now here comes the dog….Take the dog simply as a runner whose speed is 60kmph.No matter it has to perform to and fro motion( so sad for the poor dog!!) it will keep on running till the time Police catches the thief which we have calculated as 12 hours…
    Hence the distance the dog will run during this time is 12*60=720KM…..
    Any clarifications???


  15. Ritanshu Says:

    ans 660.
    i calculated same as Bhimraj meena calculated in his post….
    police caught the thief after 600 kms…..on 15th hour, in the mean time dog also travelled 60kms in the direction of thief…


  16. Jaydip Raval Says:

    say, after x hours
    50x=40x+120(thief ran three hours before police)
    i:e x=12
    after 12 hours police will caught to thief.
    since, dog is running in both directions
    i:e total distance traveled by dog =12/2*60(dog speed)=360


  17. Swapnil Bahekar Says:

    As all of us have got that police will take 12 hours to catch a thief, now in these 12 hours the dog just keeps running from thief to police and from police to thief again and again till the police catches the thief. so time for which the dog runs = 12 hrs and its speed is 6o km/hr.
    So the distance travelled by dog = 720 km


  18. Swapnil Bahekar Says:

    Sorry guys we have to divide the total distance travelled by dog by 2 to get the required answer i.e 720/2 = 360, as we are asked to calculate the distance of the dog in the direction of thief..


  19. Swapnil Bahekar Says:

    Guys can somebody explain me where i m wrong if i do like this :
    since we need to calculate the dog’s distance in the direction of the thief and we know its speed. so lets try to find out the time it travels in the direction of the thief.We wil assume the thief to be stationary and use relative velocities of the dog and the police.
    so the first time the dog meets the thief is when it travels 120 km i.e 120 / 20 = 6 hrs
    now in these 6 hrs the police covered 10*6 = 60 km and hence distance between the dog and the police is 60 kms
    now dog will run towards police and will meet him after 60 /(10+20) = 2 hrs.. now again he will run towards the thief for 2 hours…in this way we can come up with a series of dog’s time in hrs in the direction of the thief as
    6 + 2 + 2/3 + 2/9 + …so on which is equal to 9.
    Hence the dog travels 9 hrs in the direction of the thief, consequently giving the distance to be 60*9 = 540 km, which is not there in the options!!!!!
    I m really sorry if i m confusing you people but not able to understand why this is not working…Help needed!


  20. bhimraj meena Says:

    ans is 660km …….
    720km is the total distance travelled by dog in both direction bt we hv to calculate only in direction of thief..ok any clarification is required….


  21. bhimraj meena Says:

    hey Swapnil u r making mistake in the 2nd step…that is from ur para…”now dog will run towards police and will meet him after 60 /(10+20) = 2 hrs”.. here speed of the dog wrt police wil be 50+60kmph….so the time wil be 60/110hrs…same wil follow…


  22. Swapnil Bahekar Says:

    hi bhimraj, can u tell me how u got 660… i know what was wrong in my method before..but that is not what u pointed out above…actually we cant assume the thief to be stationary there..thinking differently we can say that the police takes 12 hours to catch the thief i.e he travels 600 km to catch the thief, now to cover this 600 km when the dog is also running in the same direction, it takes 10 hrs and the total distance of the dog covered in both the directions in 12 hrs is 720, so our required ans must be between 600 and 720.


  23. FIREUP Says:

    Thanks for the wonderful discussion.

    Some of you got the answer right.

    Answer and explanation given Bhimraj Meena and Satya are correct and can be referred to.

    For any further clarification mail us at info@fireup.co.in


  24. Shivani Says:

    I agree with all those people who say that the answer is 360 km.


  25. Vigneshwar Says:

    i feel its 660

    because net distance traveled by the dog is 12*60=720
    while chasing rel. speed bt. thief and dog is just 60-40=20
    and while returning bt. police and dog is 50+60=110

    dog takes more time to reach the thief and lesser time to return back to police so dividing the net distance into two equal halves wont give the right ans..


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