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Thread: Quant Thread For CAT 2011

  1. #21
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    Quote Originally Posted by grimreaper View Post
    DUde.... sum of "squares " of 3 consecutive odd numbers

    hehehe...try again
    oops sorry,

    Op3)5555

    the 3 nos are 41,43,45
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  2. #22
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    12n(n+1)+11 = Sum

    Only 5555-11=5544 is div by 12

    Also,then n=21
    Hence the nos are 41,43,45
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    Quote Originally Posted by shaunak_87 View Post
    oops sorry,

    Op3)5555

    the 3 nos are 41,43,45
    Let the numbers be n-4 , n-2 ,n
    so i get,
    (n-4)^2+(n-2)^2+n^2 = 5555
    so
    Solving this i found n value to be 45.

    Some1 can post efficient method than this if exists

  4. #24
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    Quote Originally Posted by madhu89 View Post
    Let the numbers be n-4 , n-2 ,n
    so i get,
    (n-4)^2+(n-2)^2+n^2 = 5555
    so
    Solving this i found n value to be 45.

    Some1 can post efficient method than this if exists
    well how did u take it to be 5555 in the first place... ?
    n is least of my concern...... post the method as to how you got 5555 as the answer .... cause if you get 5555 u get the numbers too...

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    let the no be n-2 , n ad n+2

    now (n-2)^2 + (n)^2 + (n + 2)^2 = 3n^2 + 8
    Now only one options satisfies the condition. Hence 5555

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    Quote Originally Posted by aloknitw View Post
    let the no be n-2 , n ad n+2

    now (n-2)^2 + (n)^2 + (n + 2)^2 = 3n^2 + 8
    Now only one options satisfies the condition. Hence 5555
    correct
    i hope u applied logic to select 5555 and not checking each number one by one......

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    @GrimSaar

    Some tips to improve Speed
    CAT2011 : 99.97%ile
    XAT2012 : 99.07%ile
    FMS 2012-14
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  10. #28
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    Quote Originally Posted by shaunak_87 View Post
    @GrimSaar

    Some tips to improve Speed
    lol naam ki rape kardi be :P :P....

    where are the mods by the way?? i mean... long time..havent seen them online at all......

    and well hope more ppl start visiting the forum... its really suna suna for now

  11. #29
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    Default official solution for Q3 (my method)

    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......


    well, first of all let the numbers be n-2,n and n+2
    so sum of squares would be (n-2)^2 + n^2 + (n+2)^2 = 3n^2+8

    now see the options...
    only two numbers that is 2222 and 5555 when 8 is sub are divisible by 3
    that is 2214 and 5547,
    when divided by 3 we get... 738 and 1849

    clearly we knw that a square can never end in 8 hence 1849 is the answer...
    which is nothing but 43^2

    hence your answer... 41,43,45

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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......
    Let the three consecutive odd numbers be X-2, X and X+2.
    Now given that N= (X-2)^2 + X^2+(X+2)^2
    N= X^2 + 4-4X +X^2 + X^2 + 4 + 4X
    => N= 3X^2 + 8
    If we put N=1111 then
    N=3X^2+8
    => 3X^2+8 =1111
    => 3X^2=1111-8
    X^2= (n-8)/ 3 --------------(1)

    The values of N which are not divisible by 3 will be eliminated since we want the values of X as positive odd integers.
    We will substitute values of N one by one now and see which one satisfes the equation (1)
    Options (a),(b) don't satisfy it.

    If N= 2222 then
    X^2 =(2222-8)/3
    = > 2214/3
    X^2 = 738

    But since we know that a perfect square can't end in 8
    so N can't be 2222

    if N=5555 then
    X^2=(5555-8)/3
    = > X^2 = 1849
    so X= 43
    Now X-2 = 43-2= 41

    and X+2 = 43+2 = 45

    thus the three consecutive odd integers are 41, 43 and 45

    so N= (41)^2 + (43)^2 + (45)^2
    = > 1681 +1849 + 2025
    hence N = 5555

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    Quote Originally Posted by grimreaper View Post


    well, first of all let the numbers be n-2,n and n+2
    so sum of squares would be (n-2)^2 + n^2 + (n+2)^2 = 3n^2+8

    now see the options...
    only two numbers that is 2222 and 5555 when 8 is sub are divisible by 3
    that is 2214 and 5547,
    when divided by 3 we get... 738 and 1849

    clearly we knw that a square can never end in 8 hence 1849 is the answer...
    which is nothing but 43^2

    hence your answer... 41,43,45

    hi

    only two numbers that is 2222 and 5555 when 8 is sub are divisible by 3
    what do you mean by " when 8 is sub "
    please explain

  16. #32
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    Quote Originally Posted by anjalig View Post
    hi

    only two numbers that is 2222 and 5555 when 8 is sub are divisible by 3
    what do you mean by " when 8 is sub "
    please explain
    oh... when 8 is subtracted.... sorry about the typo

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    Quote Originally Posted by grimreaper View Post
    oh... when 8 is subtracted.... sorry about the typo
    oh thanks !
    btw what about my solution
    is there any other approach to solve this ques.

  18. #34
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    Quote Originally Posted by anjalig View Post
    oh thanks !
    btw what about my solution
    is there any other approach to solve this ques.
    your solution is correct.... u did almost the same thing as me... only last line different......
    there are many ways...but this is by far the shortest method to do it
    post some Qs here too.... that way you cld get diff solutions to it......

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    Default Q 4

    Q4) 3*2^p + 5^q + 7^r + 11^s =2008 what is the value of p+q+r+s, where all are greater than 0
    a)9
    b)10
    c)24
    d)15
    e)unique solution does not exist

  21. #36
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    Quote Originally Posted by grimreaper View Post
    Q4) 3*2^p + 5^q + 7^r + 11^s =2008 what is the value of p+q+r+s, where all are greater than 0
    a)9
    b)10
    c)24
    d)15
    e)unique solution does not exist
    op e)unique solution does not exist

    p q r s
    0 4 2 3
    9 0 3 2
    Last edited by shaunak_87; 29-Mar-11 at 2:37 PM.
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    Quote Originally Posted by shaunak_87 View Post
    op e)unique solution does not exist

    p q r s
    0 4 2 3
    8 0 3 2
    the second set.... 8 0 3 2 where does that satisfy the condition ???
    how did you get the second set O.O

    am i missing something ??? O.o

  23. #38
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    Quote Originally Posted by grimreaper View Post
    the second set.... 8 0 3 2 where does that satisfy the condition ???
    how did you get the second set O.O

    am i missing something ??? O.o
    edited the post
    9 0 3 2

    Made a mistake had calculated 512 = 2^8
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  24. #39
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    Quote Originally Posted by shaunak_87 View Post
    edited the post
    9 0 3 2

    Made a mistake had calculated 512 = 2^8
    then also its coming upto only 2001

    i mean how can there be another pair ?? i mean the very basis of the solution is that only one pair can exist.... think think u making a silly mistake

  25. #40
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    Quote Originally Posted by grimreaper View Post
    then also its coming upto only 2001

    i mean how can there be another pair ?? i mean the very basis of the solution is that only one pair can exist.... think think u making a silly mistake
    True
    Just 1 possible, peeking a boo in office and trying to increase speed...and accuracy goes for a toss...

    That's y, how to be accurate and increase speed??
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