Quant Boosters - Sibanand Pattnaik - Set 2


  • QA/DILR Mentor | Be Legend


    Method - 1 :
    If you are good in binomial then do it directly. Else you can also think this in these lines. Is this not same as 3x + 2y + z = 14 and is this not same as finding the positive distinct integral solutions of the equation a + b + c = 14? so (13c2 - 3 * 6)/6 = 10

    Method 2 :
    If you want to solve it by binomial then check below
    coefficient of x^14 in (x^3 + x^6 + x^9 + ...) * (x^2 + x^4 + x^6 + ...) * (x + x^2 + x^3 + ...)
    = coefficient of x^14 - 6 (1 + x^3 + x^6 + ...) * (1 + x^2 + x^4 + ...) * (1 + x + x^2 + x^3 + ...)
    = coefficient of x^8 in (1 + x^2 + x^3 + x^4 + ... + x^8 +...) * (1 + x^2 + x^3 + x^4 + ... + x^8 + ....)
    = 1 + 1 +1 + 1 + 1 + 2 + 1 + 2 = 10


  • QA/DILR Mentor | Be Legend


    Q9) In how many ways can an amount of rs. 100 be paid exactly using coins of denomination 1,2 and 5 such that at-least one coin of each denomination is used?


  • QA/DILR Mentor | Be Legend


    Basically the problem is to find +ve integer solutions to A + 2B + 5C = 100
    So A + 2B can take values 5 , 15, 25 ,......95
    The number of solutions corresponding to this will be 2 + 7 + 12 ... + 47 = 245
    Also, A + 2B can be 10 , 20 , 30 ... 90
    The number of solutions corresponding to this will be 4 + 9 + 14 ... + 44 = 216
    so total = 245 + 216 = 461.


  • QA/DILR Mentor | Be Legend


    Q10) In how many ways can 2310 be expressed as a product of 3 factors?


  • QA/DILR Mentor | Be Legend


    2310 = 2 * 3 * 5 * 7 * 11
    These 5 primes to be distributed into 3 places.
    Total : 3^5 ways , but this is ordered ( arranged) and we need unordered.
    We know that any triplet with distinct elements (a,b,c) is arranged in 3! ways, so all the distinct triplets in 3^5 = 243 are arranged in 3! Ways.
    Except for one triplet (1, 1, 2310) which is arranged in 3!/2= 3 ways
    So remove this from total and we are left with only distinct, is divide by 3! ,
    So (3^5 - 3)/3! , and just add that one case of 1, 1, 2310
    So (3^5-3)/3! + 1 = 41


  • QA/DILR Mentor | Be Legend


    Q11) How many pairs of positive integers (m, n) satisfy 1/m + 4/n = 1/12. Where n is an odd integer less than 60.


  • QA/DILR Mentor | Be Legend


    (m -12) * (n-48) = 12 * 48 = 2^6 * 3^2
    now for "n" to be odd , (n-48) has to be odd
    so n - 48 can be either 1 or 3 or 3^2 (because we need ODD)
    (m -12) * (n - 48) = 2^6 * 3^2 * 1
    OR (m -12) * (n - 48) = 2^6 * 3 * 3
    or (m -12) * (n - 48) = 2^6 * 3^2
    when n - 48 = 1, n = 49 (acceptable)
    when n - 48 = 3, n = 51 (acceptable)
    when n - 48 = 3^2, n = 57 (acceptable)
    only these 3 values are possible for n < 60 and odd


  • QA/DILR Mentor | Be Legend


    Q12) Find number of whole number solutions of a + b + c + d = 20 where a, b, c and d ≤ 8


  • QA/DILR Mentor | Be Legend


    Cofficient of x^20 in (1 + x + x^2 + x^3 + ... + x^8)^4
    = coff of x^20 in (1 - x^9)^4 * (1 - x)^-4
    = 23c3 - 4 * 14c3 + 4c2 * 5c3
    = 1771 - 1456 + 60
    = 375

    For more detailed explanation of the concept, refer the video.


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    Q13) If k is a factor of (n+5) and (6n+54) where n is a natural number, how many possible values of k are there?
    a) 6
    b) 7
    c) 8
    d) 9


  • QA/DILR Mentor | Be Legend


    As given k must be a factor of 6n+30 and is also a factor of 6n+54.
    So it must be a factor of (6n+54) – (6n+30) = 24 = 2^3 x 3
    Therefore k can take 8 values
    Hence (c)


  • QA/DILR Mentor | Be Legend


    Q14) The sum of all the factors of 480 including 1 and the number itself is 1512. What is the sum of the reciprocals of all the factors?
    a) 3.15
    b) 2.85
    c) 2.55
    d) 3.45


  • QA/DILR Mentor | Be Legend


    Sum of reciprocals = (sum of factors)/480
    1512/480= 3.15


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    Q15) Find the last two digits of 89^102
    a) 23
    b) 21
    c) 01
    d) 69


  • QA/DILR Mentor | Be Legend


    (89^2)^51
    89^2 same as 11^2 -> 21^51 = 21


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    Q16) The highest power of 5 in N! is 34. Find the highest power of 7 in N!.
    a) 23
    b) 26
    c) 22
    d) Cannot be determined


  • QA/DILR Mentor | Be Legend


    Highest power of 5 in 125!= 31.
    Highest power of 5 in 140!=34
    But highest power of 5 in 141!, 142!, 143! And 144! Is also 34.
    So N can be 140 to 144, any of these values
    However the highest power of 7 in any of these values is same i.e. 22


  • QA/DILR Mentor | Be Legend


    Q17) A! + B! = N^2, Where A, B and N are whole numbers. How many possible combinations are there?
    a) 1
    b) 2
    c) 3
    d) 4


  • QA/DILR Mentor | Be Legend


    0! + 4! = 5^2
    1! + 4! = 5^2
    0! + 5! = 11^2
    1! + 5! = 11^2
    Hence 4 combinations are possible


  • QA/DILR Mentor | Be Legend


    Q18) x, y and z are natural numbers such that x > y > z > 1 and PRODUCT= 3003. Let A and B be the maximum and minimum values of x + y + z. Find A-B.
    a) 152
    b) 45
    c) 107
    d) None of the above


 

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