Quant Boosters - Kamal Lohia - Set 2



  • 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.



  • 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?



  • 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



  • 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?



  • 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.



  • 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?



  • 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.



  • 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?



  • 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



  • 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?



  • 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



  • 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?



  • we want 5R or 7B, so worst case is that so far you have been not to achieve it i.e. you have taken out 4R and 6B so far. Now 11th ball will ensure the desired configuration



  • Q11) 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 AND seven blue balls?



  • We want 5R and 7B, so worst case is that we have taken all 10R balls, so next 7 draws will ensure the required number.



  • Q12) Set S is formed by positive integers from 1 to 100 so that sum of no two elements of S is 103. What is the maximum possible number of subsets of S?



  • Basically we need to maximize the number of elements in S to maximize the number of subsets. So we can take exactly one element from each of brackets i.e. {1}, {2}, {3, 100}, {4, 99}, {5, 98}... {51, 52} i.e. 51 in all and thus number of subsets become 2^51.



  • Q13) In a community of 2029 inhabitants, at least how many have same initials for their first name and last name?



  • For example it is possible that at most all of 2029 inhabitants have same initials, AB (say)
    Total different initials possible = 26 * 26 = 676. So considering WORST CASE, first layer of 676 inhabitants have all different initials, then second layer also, then third one. And finally one person remains that confirms that at least 4 persons are there which have same English initials for their first name and last name



  • Q14) The number 3 can be written as 3, 2 + 1, 1 + 2, 1 + 1 + 1 in four ways. In how many ways the number "n" can be written?


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