Quant Boosters - Hemant Malhotra - Set 11


  • Director at ElitesGrid | CAT 2017 - QA 100 Percentile / CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    abc = 540
    540 = 2^2 * 3^3 * 5^1
    a^x * b^y * c^z = 2^2 * 3^3 * 5^1
    a will some number which will contain 2 ,3 and 5
    same for b
    and same for
    and there will be some power of 2 ,3 and 5 in them whatever will be the power 0,1,2 or 3
    but sum of the power of 2 will be 2 , sum of the power of 3 will be 3 and sum of the power of 5 will be 1
    so x + y + z = 2 (where x,y,z could vary from 0 to2)
    so number of ways=2+3-1c3-1=4c2
    x+y+z=3
    3+3-1c3-1=5c2
    x+y+z=1
    so 1+3-1c3-1=3c2
    so total number of positive solutions will be
    =4c2 * 5c2 * 3c2=180
    so ordered number of positive solutions will be 180

    now if we want total number of integral solutions then we will find negative solutions also
    for negative out of a,b,c ,2 numbers will be negative
    so choose 2 numbers out of those 3 numbers
    so 3c2 * positive solutions
    so 3c2 * 180
    so total ordered solutions =180 + 3c2 * 180


  • Director at ElitesGrid | CAT 2017 - QA 100 Percentile / CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    Q23) If a * b * c * d = 648, then how many ordered integral values of a,b,c,d are possible ?


  • Director at ElitesGrid | CAT 2017 - QA 100 Percentile / CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    648=2^3 * 3^4
    so x+y+z+k=3 so 6c3=20
    and x+y+z+k=4 =7c3=35
    so total positive =700
    now cases can be
    all positive=700
    2 positive 2 negative =4c2 * 700
    or all negative =700
    so 700+700+6 * 700
    4200+1400=5600
    a^x * b^y * c^z * d^k = 2^3 * 3^4
    now a ,b,c,d all will contain 0,1,2,or 3 power of 2 and sum of all those will be equal to 3
    so x+y+z+k=3
    and same for 3
    so x+y+z+k=4


  • Director at ElitesGrid | CAT 2017 - QA 100 Percentile / CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    Q24) In how many ways can 5^17 be written as a product of three positive numbers


  • Director at ElitesGrid | CAT 2017 - QA 100 Percentile / CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    a + b + c = 17
    19C2 = 171
    when two are same :
    2a + c = 17
    total 9 solutions
    (171 - 3 * 9)/ 3! + 9 = 33


  • Director at ElitesGrid | CAT 2017 - QA 100 Percentile / CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    Q25) |x * y * z| = 210. Find number of Integral Solutions


  • Director at ElitesGrid | CAT 2017 - QA 100 Percentile / CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    210= 2 * 3 * 5 * 7
    So positive = 3 * 3 * 3 * 3 = 3^4
    Now each of three can be either positive or negative
    So 2 * 2 * 2 * 3^4 = 8 * 3^4
    Unordered: 81 - 3 / 6 + 1 = 14


  • Director at ElitesGrid | CAT 2017 - QA 100 Percentile / CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    Q26) In how many ways 1001^2 can be written as product of 3 integers (Ordered and unordered)


  • Director at ElitesGrid | CAT 2017 - QA 100 Percentile / CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    a * b * c = 7^2 * 11^2 * 13^2
    x+y+z=2=4c2=6
    x+y+z=2=4c2=6
    x+y+z=2=4c2=6
    so 6^3+3 * 6^3
    4 * 6^3=864
    now unordered
    2 same
    a^2 * b=1^2 * x
    7^2 * x
    11^2 * x
    13^2 * x
    (7 * 11)^2 * x
    (7 * 13)^2 * x
    (11 * 13)^2 * x
    (7 * 11 * 13)^2
    same 8 for negative so 16
    so 864-16 * 3/6 +16


  • Director at ElitesGrid | CAT 2017 - QA 100 Percentile / CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    Q27) In how many ways 1000 can b expressed as the product of 3 integers? (ordered and unordered)


  • Director at ElitesGrid | CAT 2017 - QA 100 Percentile / CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    1000=2^3 * 5^3
    so x+y+z=3
    so 5c2=10
    and x+y+z=3
    so 10
    so 10 * 10=100
    and negative 3 * 100=300
    so total =400

    now unordered
    same a^2 * b=2^3 * 5^3
    1^2 * x
    2^2 * x
    5^2 * x
    10^2 * x
    (-1)^2 * x
    (-2)^2 * x
    (-5)^2 * x
    (-10)^2 * x
    so total y=8
    but one case will be all same here so y=7
    and z=1
    so 6x + 3 * 7 + 1=400
    so x=400-22/6
    so unordered=x+y+z


  • Director at ElitesGrid | CAT 2017 - QA 100 Percentile / CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    Q28) Find coefficient of x^3 * y^4 * z^3 in expansion of (2x + y + 4z)^10


  • Director at ElitesGrid | CAT 2017 - QA 100 Percentile / CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    general term of this equation is = 10!/(r1! * r2! * r3!) * (2x)^r1 * (y)^r2 * (4z)^r3
    = 10!/(r1! * r2! * r3!) * (2)^r1 * (4)^r3 * x^r1 * y^r2 * z^r3
    now we have to find x^3 * y^4 * z^3 so r1=3 ,r2=4 and r3=3
    put tvalue of r1, r2, r3 here 10!/(r1! * r2! * r3!) * (2)^r1 * (4)^r3 * x^r1 * y^r2 * z^r3
    so coefficient is 10!/(3! * 4! * 3!) * (2)^3 * (4)^3


  • Director at ElitesGrid | CAT 2017 - QA 100 Percentile / CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    Q29) Find the largest coefficient in (x + y + z + k)^15


  • Director at ElitesGrid | CAT 2017 - QA 100 Percentile / CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    general term = 15!/(r1! * r2! * r3! * r4!) * x^r1 * y^r2 * z^r3 * k^r4
    so basically we have to maximize 15!/(r1! * r2! * r3! * r4!) term
    so we have to minimize (r1! * r2! * r3! * r4!) so for this symmetry is useful so put values as close as possible here
    r1 = r2 = r3 = 4 and r4 = 3
    so largest coefficient is 15!/(3! * (4!)^3)


  • Director at ElitesGrid | CAT 2017 - QA 100 Percentile / CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    Q30) Find the coefficient of x^6 in (1 + x^2 - x^3)^8


  • Director at ElitesGrid | CAT 2017 - QA 100 Percentile / CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    general term = 8!/r1! * r2! * r3! * (1)^r1 * (x^2)^r2 * (-x^3)^r3
    = 8!/r1! * r2! * r3! * x^(2r1) * (-1)^r3 * (x)^3r3
    = 8!/r1! * r2! * r3! * (-1)^r3 * x^(2r2+3r3)
    now we have to find coefficient of x^6 here (for any coefficient same procedure )
    so r1 + r2 + r3 = 8
    2r2 + 3r3 = 6
    find r1, r2 and r3 here
    let r3=0 so r2=3 so r1=2 and let r3=2,so r2=0 so r1=6 (r1,r2,r3 should be integer because term can't be negative)
    now coefficient will be 8!/r1! * r2! * r3! *(-1)^r3
    put both values of r1,r2 and r3 and add
    so 8!/5! * 3!+8!/2! * 6!=84



  • @hemant_malhotra as the no. of real roots is asked so it will always be 3


 

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