Quant Boosters - Gaurav Sharma - Set 1



  • S = 1/(3^2 – 1) + 1/(5^2 – 1) + 1/(7^2 – 1) … + 1/(99^2 – 1)
    = 1/(2 x 4) + 1/(4 x 6) + 1/(6 x 8 ) + … + 1/(98 x 100)
    = 1/2 [ 1/2 - 1/4 + 1/4 - 1/6 + 1/6 – 1/8 + 1/8 + … + 1/98 – 1/100]
    = 1/2 (1/2 – 1/100)
    = 1/2 ( 49 / 100)
    = 49/200



  • Q21) There are 11 numbers of the blackboard – six zeroes and five ones. You have to perform the following operation 10 times : cross out any two numbers. If they were equal write a zero on the blackboard and if they were not equal, write a one. This operation can be performed in any order as you wish. What is the number you have at the end.
    a) 11
    b) 00
    c) 1
    d) Cant say



  • The operation could be performed in a number of different ways, but 1 thing is always the same : After each operation the sum of the numbers on the blackboard is always odd.
    Also, we are losing 1 digit in each operation. So, finally we must have a single digit – either 0 or 1
    But it should be odd so our result must be 1.



  • Q22)A natural number is chosen at random from the first 100 natural numbers.
    The probability that (x + 100/x) > 50 is
    a) 1/2
    b) 7/18
    c) 5/12
    d) 11/20



  • We have x + 100/x > 50
    x^2 + 100 > 50x
    (x – 25)^2 > 525
    x – 25 < root(525) or x – 25 > root(525)
    x < 25 – root(525) or 25 + root(525)
    As x is positive and root(525) = 22.9
    We must have x ≤ 2 or x ≥ 48
    Thus 2 + 53 = 55 favorable cases.
    Probability = 55/100 = 11/20



  • Q23) Find the sum of first 100 terms of the series whose general term is given by T(k) = (K^2 + 1)k!



  • T(K) = (K^2 + 1)K!
    = (K(K+1) – (K -1))K!
    = K(K+1)! – (K-1)K!
    K(K+1)! – (K-1)K!

    T(1) = 1 x 2! – 0
    T(2) = 2 x 3! – 1 x 2!
    T(3) = 3 x 4! – 2 x 3!
    T(100) = 100 x 101! – 99 x 100!

    T(1) + T(2) + … T(100) = 100 x 101!



  • Q24) Odd integers are grouped as {1} {3,5} {7,9,11} and so on. What will be the sum of all the numbers in the twelfth group?
    a) 1000
    b) 1728
    c) 1536
    d) 1262



  • Sum of integers in group 1 = 1 = 1^3
    Sum of integers in group 2 = 8 = 2^3
    Sum of integers in group 3 = 27 = 3^3
    Observing the pattern, sum of integers in group 12 = 12^3 = 1728



  • Q25) An AP, GP and HP have the same first and last terms and the same odd number of terms. Then the middle terms of the three series are in
    a) AP
    b) GP
    c) HP
    d) None of these



  • Let the first and term be a & b respectively.
    The middle term of AP will be AM i.e, (a + b)/2
    The middle term of GP will be GM i.e root(ab)
    The middle term of HP will be HM i.e 2ab/(a+b)
    Now AM x HM = GM^2
    So the middle terms are in GP



  • Q26) Let x, y and n be positive integers, such that n > 1. How many different solutions are there to the equation x^2 – y^2 = 2^150
    a) 1
    b) 3
    c) 49
    d) 74



  • 2^150 is a perfect square divisible by 4
    So number of solutions of x^2 – y^2 = 2^150 are [(number of factors of 2^150/4) – 1] / 2
    = [number of factors of (2^148 – 1) / 2]
    = (149 – 1) / 2 = 74



  • Q27) Let a1, a2 … a11 be an arbitrary arrangement of the integers, 1,2 …. 11. Then the number (a1 – 1)(a2 – 2) … (a11 – 11) is
    a) Necessarily ≤ 0
    b) Necessarily 0
    c) Necessarily even
    d) None of the above



  • Necessarily ≤ 0 case : take values as (2-1)(3-2)(4-3) … (10-9)(1-10)(10-11)
    Now, all the terms except the last 2 terms are positive and there are 2 negative terms whose product will be even. Also, the product is greater than 0. Hence the statement is not necessarily negative or equal to 0

    Necessarily 0 case: not true. Proved above

    Necessarily even case: There are 6 odd and 5 even terms
    Let us try to make each term odd so that product is odd.
    (even – 1) (odd – 2)(even – 3)(odd – 4) … (even – 9)(odd – 10) after this there are no even terms left.
    Hence the last bracket will be (odd – 11) which is even number.
    So the product will be necessarily be even.
    Hence Option C.



  • Q28) In a chess tournament, each of the 5 players plays against every other player. No game results in a draw and the winner of each games gets one point and loser gets zero. Then which of the following sequences cannot represent the scores of the 5 players?
    a) 3, 3, 2, 1, 1
    b) 3, 2, 2, 2, 1
    c) 2, 2, 2, 2, 2
    d) 4, 4, 1, 1, 0



  • Total number of games played by 5 players will be 5C2 = 10
    Now in the 4th option it says that 2 of the players have won 4 games each.
    Let us consider the two players as A and B
    Now A has played 4 matches and won all of them = > A won when he played the match with B
    Also, B has played 4 matches and won all of them = > B won when he played the match with A
    Hence there is a contradiction.
    So Option 4 is not valid.



  • Q29) There are 11 vessels of capacity 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and 31 liters. All the vessels except one are filled with water, oil or alcohol. The quantity of alcohol is twice the quantity of water and the quantity of oil is thrice the quantity of alcohol. The capacity of the container that is empty is
    a) 13
    b) 23
    c) 7
    d) 9



  • Let the quantity of vessels containing water be x
    Quantity of alcohol = 2x
    Quantity of oil = 6x
    Total quantity will be 9x
    Hence the sum of the numbers must be divisible by 9 and
    2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 + 31 = 160
    Hence vessel kept empty would be 7 kgs as 160 – 7 = 153 is divisible by 9



  • Q30) One wishes to find positive integers x, y such that (x + y)/xy is also a positive integer, identify the correct alternative
    a) Not possible
    b) Possible and the pair (x,y) can be chosen in infinite ways
    c) Possible and there exist a finite number of ways of choosing the pair (x,y)
    d) Possible and the pair (x,y) is unique


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