Quant Boosters by Shubham Ranka, IIM Calcutta


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    Let x^2 + 3xy + 2y^2 = k
    Given that 1 = 2x^2 + 3y^2
    So, x^2 + 3xy + 2y^2 = k = k*1 = k(2x^2 + 3y^2)
    == > x^2 (1-2k) + 3xy + y^2 (2-3k) = 0
    Dividing by y^2:-
    (x/y)^2 (1-2k) + (x/y)*3 + (2-3k) = 0
    For the above quadratic eqn in (x/y) to have a real solution:-
    D > = 0 thus
    9 - 4(1-2k)(2-3k) > = 0'
    so 24k^2 - 28k - 1 < = 0. Thus the interval is [N,M] so N and M are roots of the equation, thus M + N = 7/6 and MN = -1/24
    so M + N - MN= 7/6 - 1/24 = 27/24 = 9/8


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    Q 21) Find the value of the series 3/1! + 7/2! + 15/3! + 27/4! + 43/5! + ...


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    Taylor series expansion for e^x = 1 + x + x^2/2! + x^3/3! +...
    e-1 = 1+1/2!+1/3!+1/4!+....
    3,7,15,27,43,
    Numerator:-
    Tn = Tn-1 + 4(n-1), n > =2
    T1=3
    Tn = 3+ 4 (n(n-1)/2)
    == > Generic Tn = [3+2(n(n-1))]/n!
    = 3/n! + 2/(n-2)! (n > =2)
    = 3(e-1) + 2e
    = 5e-3


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    Q 22) In the year 1980, the age of a person is 1/89th of the year of his birth. What was the age of person in 2012?


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    Let x be the man's year of birth,
    Then x/89 = 1980 - x
    == > x = 1958
    == > In 2012 Age was: 54 years


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    Q23) Let a, b, c, d belongs to [1/2, 2]. Suppose abcd=1. Find the maximum and minimum value of (a + 1/b)(b + 1/c)(c + 1/d)(d + 1/a).


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    (ab+1)(bc+1)(cd+1)(ad+1)
    (ab+1)(cd+1) = (ab+1)(1/ab + 1)
    Now: (x+1)(1/x +1) = 1+x+1/x+1
    We know, that: x+1/x > =2 (2 @ x=1)
    == > (ab+1)(1/ab + 1) > 2+2 =4
    Similarly:- (bc+1)(ad+1) > =4
    Since, in the given domain: a=b=c=d=1 satisfies:
    == > Min. values= 4*4 = 16
    Put a=b=c=d=2 to get the max. value


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    Q24) In the series 3,6,9,12.. how many of the first 100 terms can be expressed as a difference of 2 perfect squares ?


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    (a+b)(a-b)=3k {k=1 to 100}
    Sum of factors and difference of factors =even . So, both even or both odd.
    For odd values of k, a-b=1, and take a=3k+1/2, b=3k-1/2. So, 50 values of k
    For even values of k, should be a multiple of 4. So, 25 values of k.
    Hence, a total of 75 terms can be expressed.


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    Q25) In the game of cricket , Runs scored in one ball can be 0(dot ball) , 1, 2, 3 , 4 or 6. What is the probability that there is at least one dot ball in an over in which 7 runs was scored. No extra deliveries were bowled.


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    a+b+c+d+e+f = 7 ----- (A)
    a,b,c,d,e,f > 0
    ==> 6c5
    ==> p = 1-6c5/N
    N = no. of solns of (A)
    = 12c5 - x
    x = Ways in which he scores 5 or 7
    For x = 7 ----> 6c1 = 6
    When any of the value is 5
    a'+b'+c'+d'+e'+f' = 7-5 = 2
    7c5
    ==> 1 - 6/ (12c5-6c1-7c5) = 769/775


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    Q 26) P and Q are the two opposite ends of a pond and the distance between them is 420 metres. Messi and Ronaldo start swimming towards each other at the same time from P and Q, with speeds in the ratio 5 : 9 respectively. As soon as any of them reaches an end, he turns back and starts swimming towards the other end. At what distance (in metres) from P will they meet when Ronaldo is in his 13th round?


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    For the same distance, Ratio of time will be inverse of speeds = 9 : 5
    When Ronaldo covers 9 rounds, Messi covers 5.
    ==> When Ronaldo covers 13 , Messi will cover 13 x 5 / 9 = 65/9 rounds.
    ==> Distance = 420 x 65/9 = 9100/3 m
    FOr P, nearest multiple of the track length = 420 x 7 (closest to 65/9 ) = 2960m.
    ==> Distance travelled from P = 9100/3 - 2960 = 43.33m


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    Q27) No of negative integral solution of 7a + 17b + 1000 = 0 is ?


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    a0
    7x + 17y = 1000
    mod 3y/7 = 6
    x y
    138 2
    Tn = 138 + (n-1) * -17 > 0
    n < 9. xx
    ===> 9 solns


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    Q 28) 20% of notes are fake.Probability of accepting a fake for an original is .1 & original never rejected.The probability of accepting original is


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    p(F) = 0.2 | p(O) = 0.8
    p(Acceptance) = 0.8 + 0.2*.1 = .82
    p(Acceptance of Original) = .8/.82 = .97


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    Q 29) There are three classes X, Y and Z. The average age of the students of classes X and Y together is 20 years. The average age of the students of classes Y and Z together is 35 years. The average age of the students of classes X and Z together is 15 years. Which of the following can be the average age of all the three classes together?
    a) 16
    b) 38
    c) 30
    d) 25


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    X+Y = 20(a+b) ----------------------------(1)
    Y+Z=35(b+c) ------------------------------(2)
    X+Z = 15(a+c) ----------------------------(3)
    Let, X+Y+Z/a+b+c = M
    Adding (1), (2) and (3) :
    X+Y+Z = 5(7a+11b+10c)
    = 35(a+b+c) + 5 (4b+3c)
    M = 35 + 5 (4b+3c)/(a+b+c)
    which is greater than 35 ===> (B)


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    Q 30) A hospital nursery has 3 boys and some girls. Overnight a child is born and is added to the group. The next morning, one child is chosen from the group at random, with uniform probability, and turns out to be a boy.Determine the probability p that the overnight addition to the group was also a boy


 

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