Quant Primer

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  Mission IIMpossible 2018 - Quant Study Plan For Your MBA Admission Tests
  • mbatious

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  Number Theory : Topic Tests Question Bank - Number Theory CAT Level Question Bank - Co Primes
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  Gyan Room - Modern Math - Concepts & Shortcuts
  mission iimpossible • • Mission IIMpossible

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  Credits @gaurav_sharma There is a frog who could climb either 1 stair or 3 stairs in one shot. In how many ways he could reach at 10th stair ? Fibonacci with a gap of 1 : 1, 1, 2, 3, 4, 6, 9, 13, 19, 28 Answer : 28 Method 2 :1 step in 1 way2 steps in 1 way3 steps in 2 ways4 steps in 3 ways5 steps in 4 ways6 steps in 6 ways7 steps in 9 ways8 steps in 13 ways9 steps in 19 ways10 steps in 28 waysSo 28 should be the answer Method 3 : x + 3y = 10(1 , 3 ) -> 4(4 , 2 ) -> 6!/4!2! = 15(7 , 1) -> 8!/7! = 8(10 , 0) -> 1TOTAL = 15 + 4 + 8 + 1 = 28
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  Installments – Various Cases and Questions including Simple and Compound Interest
  simple & compound interest • • handakafunda

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  Simple Interest and Compound Interest – Basic Concepts and Tricks for solving CAT Questions
  simple & compound interest • • handakafunda

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  Mixtures & Alligations Concepts for CAT
  mixture & alligation • • handakafunda

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  Profit and Loss – Basic Concepts for CAT Preparation
  profit & loss • • handakafunda

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  Dealing with Percentage – Ravi Handa
  percentage • • handakafunda

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  Gyan Room - Number Theory - Concepts & Shortcuts
  mission iimpossible • • Mission IIMpossible

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  Kaprekar's constant 6174 is known as Kaprekar's constant. Take any four-digit number, using at least two different digits. (Leading zeros are allowed.)Arrange the digits in descending and then in ascending order to get two four-digit numbers, adding leading zeros if necessary.Subtract the smaller number from the bigger number.Go back to step 2 and repeat.The above process will always yield 6174, in at most 7 iterations. The only four-digit numbers for which Kaprekar's routine does not reach 6174 are repdigits such as 1111, which give the result 0000 after a single iteration.
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  Gyan Room - Geometry - Concepts & Shortcuts
  mission iimpossible • • Mission IIMpossible

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  Mass point Geometry lecture - @gaurav_sharma
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  Gyan Room : Arithmetic - Concepts & Shortcuts
  mission iimpossible • • Mission IIMpossible

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  Multiplication factor This is a very important understanding of percentages from calculations point of view. What this basically does is: it converts a percentage question to a ratio question. And dealing with ratios are always easier than to deal with percentages, as ratios are easier to calculate. So, if a value increases by 20%, which is 1/5th: the multiplication factor must be (1+1/5) = 6/5, as the value is increasingIf the value reduces by 30%, which is 3/10: the multiplication factor must be (1 - 3/10) = 7/10, as the value is reducing Try to use multiplication factors in the basic questions to ease up your calculations
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  Time, Speed & Distance Concepts & Solved Examples for CAT - Nitin Gupta, AlphaNumeric (Part 2/2)
  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
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  Sequence & Series Concepts & Solved Examples for CAT - Nitin Gupta, AlphaNumeric
  sequence & series • • sibanand_pattnaik

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

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  Time, Speed & Distance Concepts & Solved Examples for CAT - Nitin Gupta, AlphaNumeric (Part 1/2)
  time speed & distance • • sibanand_pattnaik

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

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

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  Area Of The Region Bounded By The Curves - Concepts & Shortcuts
  • zabeer

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  Finding Number of Positive & Non-Negative Integral Solutions For a + b + c + ... = N & Its Applications
  theory of equations permutation & combination • • zabeer

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  Permutations & Combinations - Nikhil Goyal
  permutation & combination • • Nikhil Goyal

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  Simple Equations - Nikhil Goyal
  theory of equations • • Nikhil Goyal

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