Quant Boosters - Kamal Lohia - Set 2


  • Faculty and Content Developer at Tathagat | Delhi College of Engineering


    Number of Questions - 30
    Topic - Quant Mixed Bag
    Solved ? : Yes
    Source :


  • Faculty and Content Developer at Tathagat | Delhi College of Engineering


    Q1) First 20 whole numbers are written on black board. A student comes, picks two numbers randomly say a and b. And replace them with a + b - 1. Then next student comes and does the same thing and this process goes on till there is a single number left on the board. What is that number?


  • Faculty and Content Developer at Tathagat | Delhi College of Engineering


    Important point to keep in mind here is that the process is being repeated exactly 19 times as in each step we are decreasing one number and finally we are to decrease total 19 numbers. So the answer is 190 - 19 = 171.


  • Faculty and Content Developer at Tathagat | Delhi College of Engineering


    Q2) First 20 whole numbers are written on black board. A student comes, picks two numbers randomly say a and b. And replace them with |a - b|. Then next student comes and does the same thing and this process goes on till there is a single number left on the board. Will the number be Even or Odd ?


  • Faculty and Content Developer at Tathagat | Delhi College of Engineering


    If we add all numbers from 0 to 19, then sum is even as calculated above. Now when we replace a, b with |a - b|, we are decreasing the total sum by a + b - |a - b| i.e. an even quantity. And EVEN - EVEN is EVEN always. Hence the final number will be EVEN certainly


  • Faculty and Content Developer at Tathagat | Delhi College of Engineering


    Q3) First 20 whole numbers are written on black board. A student comes, picks two numbers randomly say a and b. And replace them with a - b + 1. Then next student comes and does the same thing and this process goes on till there is a single number left on the board. Will the number be Even or Odd?


  • Faculty and Content Developer at Tathagat | Delhi College of Engineering


    Now when we replace a+ b by a - b + 1, we are changing the sum by (a + b) - (a - b + 1) = 2b - 1 i.e. an odd quantity. So effectively in the total sum i.e. 190, when we add/subtract an ODD quantity (2b - 1), an ODD (19) number of times, resultant number will be ODD certainly.


  • Faculty and Content Developer at Tathagat | Delhi College of Engineering


    Q4) Two diagonally opposite corners of a 8 by 8 chess board are removed. In how many ways can we cover the entire chess board (remaining part only) with 31 pieces of 1 by 2 rectangular dominoes without overlapping?


  • Faculty and Content Developer at Tathagat | Delhi College of Engineering


    Important point to understand is that the two diagonally opposite corner squares are of same colour i.e. both WHITE or both BLACK. That means the remaining 62 squares have 32 squares of one colour and 30 of other. Now each domino will cover exactly two squares; one of each colour. So it is not possible to cover remaining chessboard completely with 31 dominoes.


  • Faculty and Content Developer at Tathagat | Delhi College of Engineering


    Q5) In a class there are 5 rows of 5 desks each. Each desk is occupied by one student. Teacher announce them that they will shift to one step left or right or forward or backward of their respective positions simultaneously. In how many ways the students may sit finally so that again each desk is being occupied by one student?


  • Faculty and Content Developer at Tathagat | Delhi College of Engineering


    Assume the 5 by 5 desk arrangement as 5 by 5 square grid which has alternate black and white colored squares. Let's assume that corner squares are White, so we have 13 White and 12 Black squares. Now when they shift, so White one will certainly move to a Black one in either of the at most four possible positions and Black one will shift to White one. But we do not have 13 Black squares where 13 White square ones can shift to. Hence NOT POSSIBLE


  • Faculty and Content Developer at Tathagat | Delhi College of Engineering


    Q6) In a bag there are 10 red balls and 10 blue balls. How many least number of balls should be taken out of bag blindfolded so that I have taken out at least two balls of same color?


  • Faculty and Content Developer at Tathagat | Delhi College of Engineering


    Worst case is the key word here. WORST CASE means you are picking balls one by one but the required configuration is not achieved....so to the extent it is not achieved is the worst scenario and after that next ball makes your cause certain. We have 10R and 10B balls and we are to ensure that we've picked 2 balls of same color. So first ball can be any of the two colors. Right? Now second ball, if of same color then problem is solved. But look for the worst possible. That second ball is of different color. Now worse phase is over, so whatever be the third ball, you CERTAINLY have at least two balls of same color.


  • Faculty and Content Developer at Tathagat | Delhi College of Engineering


    Q7) In a bag there are 10 red balls and 10 blue balls. How many least number of balls should be taken out of bag blindfolded so that I have taken out at least two red balls?


  • Faculty and Content Developer at Tathagat | Delhi College of Engineering


    We want to have 2R balls picked, so worst case is that in first 10 chances not a single R ball is picked. But next two draws will ensure that we have taken at least 2R balls out of the bag.


  • Faculty and Content Developer at Tathagat | Delhi College of Engineering


    Q8) In a bag there are 10 red balls and 10 blue balls. How many least number of balls should be taken out of bag blindfolded so that I have taken out at least five balls of same colour?


  • Faculty and Content Developer at Tathagat | Delhi College of Engineering


    we want to pick five balls of same colour. So worst case is that you have been able to pick at most 4 balls of each colour i.e. 8 balls in all. After that 9th ball will bring good fortune for sure


  • Faculty and Content Developer at Tathagat | Delhi College of Engineering


    Q9) In a bag there are 10 red balls and 10 blue balls. How many least number of balls should be taken out of bag blindfolded so that I have taken out at least five red balls?


  • Faculty and Content Developer at Tathagat | Delhi College of Engineering


    Here, where we want 5R, worst is not a single R ball is picked in first 10 draws, but next 5 draws ensure that we have picked the desired


  • Faculty and Content Developer at Tathagat | Delhi College of Engineering


    Q10) In a bag there are 10 red balls and 10 blue balls. How many least number of balls should be taken out of bag blindfolded so that I have taken out at least five red balls OR seven blue balls?


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