Quant Boosters - Sibanand Pattnaik - Set 2


  • QA/DILR Mentor | Be Legend


    You will finally end up with a prime number which will have 2 divisors. 2 has 2 divisors. Hence (b)


  • QA/DILR Mentor | Be Legend


    Q22) In how many ways can 3660 be written as the sum of 2 or more consecutive positive integers?
    a) 2
    b) 3
    c) 4
    d) More than 4


  • QA/DILR Mentor | Be Legend


    Direct formula for natural numbers = number of odd factors - 1
    So 3660 = 2^2 * 3 * 5 * 61
    so odd factors = 8
    Answer = 8 - 1 = 7
    as in case of "integers" its 2 * (number of odd factors) - 1
    had it been asked "integers then ur answer would have been 16 - 1 = 15

    Else you can think this way :

    3660 = 61 x 3 x4 x5
    3660= a + (a+1) + (a+2)...+ (a+(n-1)
    Then 3660 = n x Average of the above series
    If the number of terms in the series is odd, then the middle term of the series will be the average
    So say the number of terms is 61, then the middle term will be 3x4x5= 60. So the series will be: (30+31+32...60+61+62...90)
    If the number of terms is 3, the average will be 61x4x5= 1220. So the series is 1219+1220+1221.
    Therefore if the number of terms is odd, its easy to construct a series like this
    3660 has 7 odd factors other than 1


  • QA/DILR Mentor | Be Legend


    Q23) N has 6 factors and lies between 250 and 350. How many values of N are possible?
    a) 12
    b) 13
    c) 14
    d) None of the above


  • QA/DILR Mentor | Be Legend


    If a and b are prime numbers then N= a^5 or N= a^2 x b
    N= a^5 :No values
    N= a^2 x b
    If a= 2, then b can take prime values of 67, 71, 73, 79, 83: 5 values
    If a= 3, then b can take prime values of 29, 31, 37: 3 values
    If a= 5, then b can take prime values of 11, 13: 2 values
    Total values: 10


  • QA/DILR Mentor | Be Legend


    Q24) N^2 has 15 factors. M^2 has 9 factors. Which of the following cannot be the number of factors that MxN can have, if M and N are co prime to each other?
    a) 24
    b) 32
    c) 20
    d) 40


  • QA/DILR Mentor | Be Legend


    N= a^7 OR N= a^2xb^1
    M= p^4 OR M= p^1xq^1
    Where a,b, p and q are prime numbers
    Therefore M x N=
    i) a^7 x p^4 : 40 factors
    ii) a^2 x b x p^4 : 30 factors
    iii) a^7 x p x q : 32 factors
    iv) a^2 x b x p x q= 24 factors


  • QA/DILR Mentor | Be Legend


    Q25) Sum of HCF and LCM of 2 numbers is 30. How many such pairs of numbers exist?
    a) 6
    b) 4
    c) 7
    d) 8


  • QA/DILR Mentor | Be Legend


    h(1+xy) = 30
    h=1 xy=29 so 2^0=1
    h=2 xy= 14 = 2*7 so 2^1= 2
    h=3 xy=9 = 3^2 so 2^0= 1
    h=5 xy=5 so 2^0=1
    h=6 xy=4 so 2^0=1
    h=10 xy=2 so 2^0=1
    h=15 xy=1 so 1
    -> 1 + 2 + 5 * 1= 8


  • QA/DILR Mentor | Be Legend


    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


  • QA/DILR Mentor | Be Legend


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


  • QA/DILR Mentor | Be Legend


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


  • QA/DILR Mentor | Be Legend


    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


  • QA/DILR Mentor | Be Legend


    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


  • QA/DILR Mentor | Be Legend


    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)


  • QA/DILR Mentor | Be Legend


    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


  • QA/DILR Mentor | Be Legend


    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


  • QA/DILR Mentor | Be Legend


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


  • QA/DILR Mentor | Be Legend


    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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