Quant Boosters - Sibanand Pattnaik - Set 2


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    Q26) N is a natural number such that N/9 is a perfect square and N/4 is a perfect cube. Find how many factors does the smallest value of N have?
    a) 18
    b) 24
    c) 14
    d) 21


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    N = 2^2 * 3^6
    factors = 21


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    Q27) How many ways can 1500 be expressed as product of 3 integers ?


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    1500 = 2^2 *3 * 5^3
    ordered solutions = 4c2 * 3c2 * 5c2 = 180
    integral : here we can have any 2 out 3 elements as negative ( for example , xyz = -x * -y * z)
    so 3c2 = 3 ways so total 180 * 3 = 540 negative solutions
    so finally 180 + 540 = 720 ordered integral solutions
    total perfect squares in 1500 = 4
    Total cubes = 1
    So (a,a,b) cases = 3 and (a,a,a) cases = 1
    now for integral (a,a,b), each of it can be (-a,-a,b) and (a,-a,-b)
    finally 2 * 4 -1 = 7 aab types and only 1 aaa type
    unordered = [720 - 3 * 7 -1]/6 + 8 = 124
    so its 124 ways


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    Q28) How many 2 digit numbers are there such that the sum of their digits is a prime number?
    a) 31
    b) 33
    c) 25
    d) 27


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    Sum of the digits is prime means the sum of the digits can be
    2: 11,20
    3:12,21,30
    5:41,14,23,32,50
    7:61,16,25,52,34,43,70
    11:92,29,83,38,74,47,65,56
    13:94,49,85,58,76,67
    17:98,89
    There are 33 such numbers. Hence (b)


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    Q29) N has 12 factors. Which of the following cannot be the number of factors N^2?
    a) 33
    b) 35
    c) 45
    d) 54


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    N can be of any of the following form:
    P^5x Q^1 therefore N^2 = P^10x Q^2; therefore N^2 has 33 factors
    P^3x Q^2 therefore N^2 = P^6x Q^4; therefore N^2 has 35 factors
    P^2x Q^1 x R^1 therefore N^2 = P^4x Q^2 x R^2; therefore N^2 has 45 factors
    Therefore 54 is the answer


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    Q30) How many factors (x, y, z) of 1000 exist such that HCF (x,y,z) = 1 ?


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    1000 = 2^3 * 5^

    When we say HCF of factors (x,y,z) = 1 ; then we mostly talk of 3 cases ..

    Case -1 ; ( 1, 2^3, 5^3) ..
    now here ordered arrangements = 3^2*3! = 54

    Case - 2 ( 1, 1, 2^3) ; (1, 1, 5^3) ; (1, 1, 2^3 * 5^3)
    now total arrangement of these cases is 3!/2! ( 15) = 45

    case -3 when (1 , 1, 1) ; as we are taking of factors of 1000 with HCF 1 , so this has to be a case ... And it can be arranged in only n only 1 way .. so 1

    So finally 54 + 45 + 1 = 100


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