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    Sibanand Pattnaik

    @sibanand_pattnaik

    QA/DILR Mentor | Be Legend

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    Topics created by sibanand_pattnaik

    • Quant Boosters - Nitin Gupta - Set 2
      Quant - Boosters • quant - mixed bag • • sibanand_pattnaik  

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      Each angle is 180(p-2)/p. 180-{360}/{p} = k So 360/p has to be an integer. 360 = 2^3 * 3^2 * 5^1 So there are 4 * 3 * 2 = 24 possibilities, but we exclude 1 and 2, because p > = 3 So , 24 -2 = 22 Hence, choice (c) is the right answer
    • Quant Boosters - Nitin Gupta - Set 1
      Quant - Boosters • • sibanand_pattnaik  

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      Warning - There might be an easier method! f(1) + f(6) = 0f(1) + f(4) + 24 = 0f(1) + f(2) + 8 + 24 = 0f(1) + f(0) = -32 f(1) + f(2) + f(3) + f(4) + f(5) + f(6)= f(1) + f(6) + f(0) + f(1) + 3 + f(2) + 8 + f(3) + 15= 0 + f(0) + f(1) + 3 + f(0) + 8 + f(1) + 3 + 15= 2 (f(0) + f(1)) + 29= 2 (-32) + 29= -64 + 29 = -35Option D
    • Question Bank - Nitin Gupta, Alphanumeric
      Quant BBQ - Best of Best Questions • • sibanand_pattnaik  

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      Let the two digit number be represented as ab. a^2 – b^2 = (a – b)(a + b) Now, |a^2 – b^2| will be a prime number only when |a – b| = 1 and (a + b) comes out to be a prime number Two digit numbers in which |a – b| = 1 are 10, 12, 21, 23, 32, 34, 43, 45, 54, 56, 65, 67, 76, 78, 87,89, 98 From these numbers, the numbers in the form of 'ab' in which (a + b) comes out to be a prime number are 12, 21, 23, 32, 34, 43, 56, 65, 67, 76, 89 and 98. There are 12 two digit numbers such that absolute difference of the square of the two digits results in a prime number.
    • Sequence & Series Concepts & Solved Examples for CAT - Nitin Gupta, AlphaNumeric
      Quant Primer • sequence & series • • sibanand_pattnaik  

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    • Maxima & Minima Concepts & Solved Examples for CAT - Nitin Gupta, AlphaNumeric
      Quant Primer • maxima & minima • • sibanand_pattnaik  

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    • Time, Speed & Distance Concepts & Solved Examples for CAT - Nitin Gupta, AlphaNumeric (Part 2/2)
      Quant Primer • time speed & distance • • sibanand_pattnaik  

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      Practice Problem - 5 : Seven children A, B, C, D, E, F and G started walking from the same point at the same time, with speeds in the ratio of 1 : 2 : 3 : 4 : 5 : 6 : 7 respectively and they are running around a circular park. Each of them carry flags of different colours and whenever two or more children meet, they place their respective flag at that point. However nobody places more than 1 flag at a same point. They are running in anti-clockwise direction. How many flags will be there in total, when there will be no scope of putting more flags? Solution : When running in the same direction : If the ratio of speeds of two athletes (in the most reducible form) is a : b, the number of distinct meeting points on the track would be would be |a – b| A and B will meet at |1 - 2| = 1 point.A and C will meet at |1 - 3| = 2 pointsA and D will meet at |1 - 4| = 3 pointsA and E will meet at |1 - 5| = 4 pointsA and F will meet at |1 - 6| = 5 pointsA and G will meet at |1 - 7| = 6 pointsSo A will put 1 + 2 + 3 + 4 + 5 + 6 = 21 flags. similarly B and C will meet at |2 - 3| = 1 pointB and D will meet at |2 - 4| = 2 pointsB and E will meet at |2 - 5| = 3 pointsB and F will meet at |2 - 6| = 4 pointsB and G will meet at |2 - 7| = 5 pointsSo B will put 1 + 2 + 3 + 4 + 5 = 15 flags Similarly find for C, D, E and F. We will get 21 + 15 + 10 + 6 + 3 + 1 = 56 flags
    • Time, Speed & Distance Concepts & Solved Examples for CAT - Nitin Gupta, AlphaNumeric (Part 1/2)
      Quant Primer • time speed & distance • • sibanand_pattnaik  

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    • Conditional Probability & Baye's Theorem - Nitin Gupta, AlphaNumeric (Part 2/2)
      Quant Primer • probability • • sibanand_pattnaik  

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    • Probability Concepts & Solved Examples For CAT - Nitin Gupta, AlphaNumeric (Part 1/2)
      Quant Primer • probability • • sibanand_pattnaik  

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    • Quant Boosters - Sibanand Pattnaik - Set 2
      Quant - Boosters • quant - mixed bag • • sibanand_pattnaik  

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      1000 = 2^3 * 5^ When we say HCF of factors (x,y,z) = 1 ; then we mostly talk of 3 cases .. Case -1 ; ( 1, 2^3, 5^3) ..now here ordered arrangements = 3^2*3! = 54 Case - 2 ( 1, 1, 2^3) ; (1, 1, 5^3) ; (1, 1, 2^3 * 5^3)now total arrangement of these cases is 3!/2! ( 15) = 45 case -3 when (1 , 1, 1) ; as we are taking of factors of 1000 with HCF 1 , so this has to be a case ... And it can be arranged in only n only 1 way .. so 1 So finally 54 + 45 + 1 = 100
    • Quant Boosters - Sibanand Pattnaik - Set 1
      Quant - Boosters • quant - mixed bag • • sibanand_pattnaik  

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      @sibanand_pattnaik number of scores possible = 251 - 15 + 1 ...= 237 Could anyone explain this last step ?
    • Question Bank - Number theory - Sibanand Pattnaik
      Quant BBQ - Best of Best Questions • question bank • • sibanand_pattnaik  

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      @sibanand_pattnaik b18081
    • Time & Work Concepts - Sibanand Pattnaik
      Quant Primer • time & work • • sibanand_pattnaik  

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    • Quant Boosters - Sibanand Pattnaik - Set 3
      Quant - Boosters • question bank • • sibanand_pattnaik  

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      Q30)