Quant Boosters - Sibanand Pattnaik - Set 2


  • QA/DILR Mentor | Be Legend


    Max power of "a" in the expression = 1+2+3+4+.....+20 = (20 * 21 )/2 = 210
    To get co-efficient of a^204 we have to find the constant term of expression having power a^6 or remove factors of a^6" , which are

    1. (x^1-1) * (x^5-5)" product of constant term = 5
    2. (x^2-2) * (x^4-4)" product of constant term = 8
    3. (x^1-1)" (x^2-2)" (x^3-3)" product of constant term = -6
    4. (x^6 - 6) * 1 .. product of constant terms = -6

    so your required coefficient = 5 + 8 - 6 - 6 = 1


  • QA/DILR Mentor | Be Legend


    Q5) How many 5 digit numbers are there whose sum of digits is 12 ?


  • QA/DILR Mentor | Be Legend


    a + b + c + d + e = 12
    now a should be at-least 1
    (x+1) + b + c + d + e = 12
    => x + b + c + d + e = 11 so 15c4 cases

    now remove the cases when x > = 9
    so, (x + 9) + b + c + d + e = 11
    => x + b + c + d + e = 2 i.e 6c4 cases

    and when b, c , d , e > = 10 , we have to remove those cases as well .. isn't it ?
    so , x + (b+10) + c + d + e = 11
    => x + b + c + d + e = 1 i.e 5c4 cases

    Required numbers = 15c4 - 4 * 5c4 - 6c4
    = 1365 - 20 - 15
    = 1365 - 35
    = 1330


  • QA/DILR Mentor | Be Legend


    Q6) There are exactly 2005 ordered pairs (x, y) of positive integers satisfying 1/x + 1/y = 1/N. Find the number of digits in smallest possible value of N if N is a natural number.


  • QA/DILR Mentor | Be Legend


    2005 = 5 * 401
    N=a^p * b^q
    N^2 = a^(2p) * b^(2q)
    So (2p + 1)(2q + 1)= 5 * 401
    p = 200, q = 2
    N = 2^200 * 3^2
    [LogN] + 1 = [200log2 + 2log3] + 1 = [61.16] + 1 = 62 digits


  • QA/DILR Mentor | Be Legend


    Q7) Total no of integral solution for x + y + z = 15 where 1≤ x, y, z ≤ 6


  • QA/DILR Mentor | Be Legend


    (6 - a ) + (6 - b) + (6 - c) = 15
    => a + b + c = 3
    5c2
    Total no . of integral solutions = 10

    Alternate method : Coefficient of x^15 in (x + x^2 + x^3 + ... + x^6)^3
    = x^12 in (1 - x^6)^3 * (1-x)^-3
    = 14c2 - 3c1 * 8c2 + 3c2
    = 10


  • QA/DILR Mentor | Be Legend


    Q8) Find the coefficient of x^14 in (x^3 + x^6 + x^9 +...) * (x^2 + x^4 + x^6 + ...) * (x + x^2 + x^3 + ...)


  • 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


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    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)


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    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


 

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