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    Kamal Lohia

    @kamal_lohia

    Faculty and Content Developer at Tathagat | Delhi College of Engineering ( 1998 - 2002 )

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    Topics created by kamal_lohia

    • Quant Boosters - Kamal Lohia - Set 9
      Quant - Boosters • • kamal_lohia  

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      @zabeer sir.. Why have we considered numbers like 7 and such to proceed further? Could you explain the solution a bit more... ?
    • Quant Boosters - Kamal Lohia - Set 8
      Quant - Boosters • quant - mixed bag question bank • • kamal_lohia  

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      @kamal_lohia Option C - 3
    • Quant Boosters - Kamal Lohia - Set 7
      Quant - Boosters • quant - mixed bag question bank • • kamal_lohia  

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      Q30) Given f(x) = x - 1/x, solve the equation f(f(x)) = x.
    • Quant Boosters - Kamal Lohia - Set 6
      Quant - Boosters • quant - mixed bag question bank • • kamal_lohia  

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      @sumit-agarwal In such scenarios like finding distinct values of [x^2/n] where x can be from 1, 2, 3 ... n[1^2/n], [2^2/n] ... [(n/2)^2/n] will yield all numbers from 0 to [n/4] (means [n/4] + 1 distinct integers)Then the next set (from [(n/2 + 1)^2/n] till [n^2/n] will be all different integers (means [n/2] distinct integers)So the number of distinct integers would be [n/2] + [n/4] + 1 if n = 100,number of distinct integers would be [100/2] + [100/4] + 1 = 76 if n = 2014,number of distinct integers would be [2014/2] + [2014/4] + 1 = 1511 if n = 13number of distinct integers would be [13/2] + [13/4] + 1 = 10 Just trying to generalize a solution shared by Kamal sir (Quant Boosters - Set 1 - Q2). You can try out with various numbers (may be smaller numbers) so that this can be verified.
    • Quant Boosters - Kamal Lohia - Set 5
      Quant - Boosters • quant - mixed bag • • kamal_lohia  

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      If two numbers are a, b, then the new number you are putting in the place of these two numbers is = ab + a + b = (a + 1)(b + 1) - 1 = c (say) Now when you take c and d next, then the replacement number becomes = cd + c + d = (c + 1)(d + 1) - 1 = (a + 1)(b + 1)(d + 1) - 1. So final number remaining will be (1 + 1)(2 + 1)(3 + 1)....(39 + 1)(40 + 1) - 1 = 41! - 1.
    • Quant Boosters - Kamal Lohia - Set 4
      Quant - Boosters • quant - mixed bag • • kamal_lohia  

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      @kamal_lohia thank you @zabeer also , has anyone posted memory based LRDI questions from 2015 and 2016 ?
    • Quant Boosters - Kamal Lohia - Set 3
      Quant - Boosters • • kamal_lohia  

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      Note that 1855 = 7 mode 8 while all perfect squares are 0, 1 or 4 mode8. So it is impossible for 3 squares to sum up to 7 mod8. So no solutions are there.
    • Quant Boosters - Kamal Lohia - Set 2
      Quant - Boosters • quant - mixed bag • • kamal_lohia  

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      Sol: Remember that: If square of a number ends in same unit digit as that of number, then that unit digit can be 0, 1, 5 or 6 only. So possible values for G are 0, 1, 5 or 6. Now by checking you can easily negate that G can't be 0 as GOG will not remain a three digit number. It can't be 1 either as in this case the three digit number will be 1O1 and it'll be perfect square only when O is 2 i.e. T is 1 which is not possible as T and G are "DISTINCT" single digit positive integers. Similarly you can check for 5 also that no number of the form 5O5 is a perfect square. Only possible case is 26² = 676 i.e. O = 7.
    • Quant Boosters - Kamal Lohia - Set 1
      Quant - Boosters • quant - mixed bag • • kamal_lohia  

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      @kamal_lohia 270/N +3 = 270/0.8N 270+3N/ N = 2700/8N 270+3N = 337.53N= 67.5N=22.5
    • Modulus Function | x | - Kamal Lohia
      Quant Primer • functions • • kamal_lohia  

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    • Set Theory - Maxima & Minima concepts - Kamal Lohia
      Quant Primer • set theory • • kamal_lohia  

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    • Important Concepts in Geometry - Kamal Lohia
      Quant Primer • lines & angles • • kamal_lohia  

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    • Important Concepts in Number Theory - Kamal Lohia
      Quant Primer • • kamal_lohia  

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