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Thread: Quant Thread For CAT 2011

  1. #1
    Available on PM sumit2goody's Avatar
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    Default Quant Thread For CAT 2011

    Hi People,

    Please use this thread for Quant discussions (for CAT 2011 and other MBA exams of the coming season)

    Some very simple rules

    1. Please do not spam.
    2. Try to post a solution and not just the key.
    3. Post any and every doubt you have, it helps everyone learn.
    4. Be polite and civil

    Good Luck!
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    Can any1 please tell me hw to go about this problem? what will be the remainder when 2^133 is divided by 133?

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    Quote Originally Posted by shon.n View Post
    Can any1 please tell me hw to go about this problem? what will be the remainder when 2^133 is divided by 133?
    You must be using Euler's theorm for solving these kind of questions.

    I guess you can find related info in one of TF's thread.

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    Default Time Speed distance

    Here`s an interesting question -

    A man leaves office daily at 7pm, when a driver comes to pick him up from his home. One day he gets free at 5.30pm and instead of waiting for the driver, he starts walking towards his home. On the way, he meets the car and returns home. He reaches home 20min earlier than usual. Had he become free at 6.00pm, how much early would he have reached ?

    a. 11 mins
    b. 13 mins
    c. 40/3 min
    d. 40/7 mins
    e. None of these.

    Kindly post your solution alongwith the method. Assume constant speed for both the man and the car

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    Default TSD question

    C`mon guys.... looking forward to some innovative solutions ! It would be great to compare approaches !!...

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    Quote Originally Posted by mohasan02 View Post
    C`mon guys.... looking forward to some innovative solutions ! It would be great to compare approaches !!...
    Answer : 40/3 minutes

    I just took few variables here!

    a is the speed of the man , b is the speed of the car
    t = time taken by the car

    T = in next case to total time taken by man and car ( when he starts at 6:00 pm in evening )

    a ' is the next time speed of the man and b' is the next time speed of the car


    for first case we got this equation :

    a/v + b/x = t -20

    t = l/x l = total length of the road.

    if you solve above equation we get this the ratio of v and x

    i.e v/x = 11/9 ( given a=90v)

    Now , in next case :

    T = a'/v+b'/x ( a'=60v and b'=l-60v)
    if we solve this eqaution we get this :

    T = l/x - 40/3

    which is why man will take 40/3 minutes lesser time to reach in next case.

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    i did it like this..
    133=2^7 + 2^2 + 2^0
    now... 2^133 / (2^7+2^2+2^0)...
    now 2^133= (2^7)^19
    4m denominator we take 2^7 as common..and we get deno in form
    2^7(1+2^-5+2^-7)..
    now..remainder when 2^133 is divided by 2^7 is zero..so
    the overall remainder will be zero..

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    Quote Originally Posted by Mandreak2k3 View Post
    Answer : 40/3 minutes

    I just took few variables here!

    a is the speed of the man , b is the speed of the car
    t = time taken by the car

    T = in next case to total time taken by man and car ( when he starts at 6:00 pm in evening )

    a ' is the next time speed of the man and b' is the next time speed of the car


    for first case we got this equation :

    a/v + b/x = t -20

    t = l/x l = total length of the road.

    if you solve above equation we get this the ratio of v and x

    i.e v/x = 11/9 ( given a=90v)

    Now , in next case :

    T = a'/v+b'/x ( a'=60v and b'=l-60v)
    if we solve this eqaution we get this :

    T = l/x - 40/3

    which is why man will take 40/3 minutes lesser time to reach in next case.
    i appreciate your response. The answer is correct. However the concept is a bit wrong !!.... To help you understand the concept , try another question on tha same lines.

    Q. Every Day Rupa's Husband meets he at the bus stop at 6:30 p.m sharp and drives her back home. Today Rupa left Office early, caught an earlier bus and reach the bus stop at 6:00 pm. Not seeing her husband, she began walking back home. Her husband met her on the way, and they reach the home 10 minutes earlier than usual.
    1> How Long had Rupa been Walking?
    2.>What is the Speed Of Rupa And her husband.(can we determine this or not?)

    I will post the solution for both the question and the explanation tomorrow. Looking forward to some good responses !

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    Default Explanation to Rupa`s question

    Q. Every Day Rupa's Husband meets he at the bus stop at 6:30 p.m sharp and drives her back home. Today Rupa left Office early, caught an earlier bus and reach the bus stop at 6:00 pm. Not seeing her husband, she began walking back home. Her husband met her on the way, and they reach the home 10 minutes earlier than usual.
    1> How Long had Rupa been Walking?
    2.>What is the Speed Of Rupa And her husband.(can we determine this or not?)


    Explanation :

    Let bus stand be at point A and home be at point B.

    A .................................................................................................B

    The sequence of events are - the car reaches A at 6.30pm and takes, say t mins, to reach B.
    If Rupa leaves early i.e. at 6.00pm, which means the car meets Rupa while its on its way to the bus stand i.e. to say, car meets Rupa before 6.30pm. Now the question is - how much minutes before 6.30pm, does the car meet Rupa ???? We have two ways to calculate this --

    Method 1

    Let,

    speed of car = x
    speed of Rupa = v
    & let car meet Rupa t` mins before 6.30pm.

    Therefore, the time saved as compared to previous days = 10 mins.
    Corresponsding distance saved = 10x (since the time taken for the journey is evaluated from 6.30pm onwards. Any distance covered from 6.00pm to 6.30pm is a bonus !)

    Distance saved = distance travelled by Rupa in (30-t`)min + distance travelled by car (carrying Rupa) in t` mins.

    10x = (30-t`)v + xt` .................... (i)

    Refering to the diagram below -

    A........................X........................................................................B



    a = AX= distance travelled by Rupa
    b = BX= distance travelled by car after picking up Rupa at point X.

    Thus 'a' was travelled by Rupa in 30-t` mins, whereas 'a' would have been covered by car in t` mins. Thus for constant distance case -

    (x/v) = (30-t`)/t` ....................... (ii)

    Solving (i) & (ii) ---
    we get t` = 5 mins

    Thus Rupa travelled for 25 mins.

    Method 2

    Now lets look at the shorter method !

    Total time saved = 10mins
    As seen from above -- the car first reaches the bus stand and then starts home bound with Rupa. Having intercepted Rupa on the way itself, it saved some portion of to & fro travelling. As speed is constant, thus, refering to the above diagram... car was saved from covering distance 'a' twice due to the reduced to & fro travelling. This translates into a saving of 5 min in each direction !As the question deals with Rupa`s travel to home (one way travel), hence the car catches Rupa at 6.25pm !!... i.e. time travelled by Rupa = 25mins !!!

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    Quote Originally Posted by mohasan02 View Post
    Q. Every Day Rupa's Husband meets he at the bus stop at 6:30 p.m sharp and drives her back home. Today Rupa left Office early, caught an earlier bus and reach the bus stop at 6:00 pm. Not seeing her husband, she began walking back home. Her husband met her on the way, and they reach the home 10 minutes earlier than usual.
    1> How Long had Rupa been Walking?
    2.>What is the Speed Of Rupa And her husband.(can we determine this or not?)

    I will post the solution for both the question and the explanation tomorrow. Looking forward to some good responses !

    well.... this is a simple prob... simply see she starts at 6... while husband wld reach at 6 30.. but he saves ten min in all... so 5 in each direction.... so in effect they meet at 6 25.. so she has travelled for 25 min

    as for speeds... u wld only know the ratio of their speeds and not the value in itself... here... speed of car to rupa is 5:1

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    Quote Originally Posted by sameerkishan View Post
    i did it like this..
    133=2^7 + 2^2 + 2^0
    now... 2^133 / (2^7+2^2+2^0)...
    now 2^133= (2^7)^19
    4m denominator we take 2^7 as common..and we get deno in form
    2^7(1+2^-5+2^-7)..
    now..remainder when 2^133 is divided by 2^7 is zero..so
    the overall remainder will be zero..
    O.O how can overall remainder be zero here..... ???? since when did an odd number divide an even number.... well i think u made some mistake in between ...check again....... this is more about euler theory... n^p-1 form were p is a prime number....

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    Smile Sequence & series

    Heres another question from my side -

    If S1 = (1/1-x); S2 = (1/1-S1); S3 = (1/1-S2); & Sn = (1/1-Sn-1) [for n more than equal to 2; x not equal to 0]

    Then S204 is ?

    a. 1/(1 - x)
    b. 1/x
    c. x
    d. (x - 1)/x

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    Quote Originally Posted by mohasan02 View Post
    Heres another question from my side -

    If S1 = (1/1-x); S2 = (1/1-S1); S3 = (1/1-S2); & Sn = (1/1-Sn-1) [for n more than equal to 2; x not equal to 0]

    Then S204 is ?

    a. 1/(1 - x)
    b. 1/x
    c. x
    d. (x - 1)/x
    Op c.x
    Period of 3, and as 204 is a mult of 3, S204= S3=x
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    Default Sequence & series

    Quote Originally Posted by shaunak_87 View Post
    Op c.x
    Period of 3, and as 204 is a mult of 3, S204= S3=x
    Bravo !!... Correct

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    Quote Originally Posted by mohasan02 View Post
    Heres another question from my side -

    If S1 = (1/1-x); S2 = (1/1-S1); S3 = (1/1-S2); & Sn = (1/1-Sn-1) [for n more than equal to 2; x not equal to 0]

    Then S204 is ?

    a. 1/(1 - x)
    b. 1/x
    c. x
    d. (x - 1)/x
    well well.....
    periodicity 3, so S204 = S3
    so option c

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    Default Q3

    Lets number the Question ok.... so that way its easier to keep track.....

    Q3) N is the sum of squares of 3 consecutive odd numbers such that all digits of N are same. If N is a four digit no. then N is -
    a)9999
    b)7777
    c)2222
    d)5555
    also what are the three consecutive numbers ?

    p.s- when someone posts a good solution feel free to show yo appreciation by clicking on the thanks button......

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    Quote Originally Posted by grimreaper View Post
    Lets number the Question ok.... so that way its easier to keep track.....

    Q3) N is the sum of squares of 3 consecutive odd numbers such that all digits of N are same. If N is a four digit no. then N is -
    a)9999
    b)7777
    c)2222
    d)5555
    also what are the three consecutive numbers ?

    p.s- when someone posts a good solution feel free to show yo appreciation by clicking on the thanks button......
    Op a)9999

    The nos are 3331,3333,3335
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    Quote Originally Posted by shaunak_87 View Post
    Op a)9999

    The nos are 3331,3333,3335
    DUde.... sum of "squares " of 3 consecutive odd numbers

    hehehe...try again

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    Quote Originally Posted by grimreaper View Post
    Lets number the Question ok.... so that way its easier to keep track.....

    Q3) N is the sum of squares of 3 consecutive odd numbers such that all digits of N are same. If N is a four digit no. then N is -
    a)9999
    b)7777
    c)2222
    d)5555
    also what are the three consecutive numbers ?

    p.s- when someone posts a good solution feel free to show yo appreciation by clicking on the thanks button......

    Option d : 5555
    The numbers are 41,43,45.

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    Quote Originally Posted by madhu89 View Post
    Option d : 5555
    The numbers are 41,43,45.
    correct... post your method....

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