Quant Boosters by VP - Set 1



  • Q26) If x + (1/x) = 1, find x^127 + (1/x)^127



  • (x + 1/x) = 1
    x^2 + 1 = x
    x^2 + 1 - x = 0
    Now try to observe
    Where have you seen such terms
    a^3 + b^3 = (a+b) (a^2 + b^2 - ab)
    So here we have to multiply by
    (x+1) on both sides
    (x+1) (x^2 -x +1 ) = 0
    => (x^3 +1) = 0
    x^3 = -1
    Now x^127 => (x^3)^42 * x
    => (-1)^42 * x
    => x
    So x + (1/x) which is equal to 1



  • Q27) A natural number equals 75 times the average of its digits. How many three digit numbers satisfy this condition



  • If average is 1
    Then number is 75 (which can't be possible )

    If average is 2
    Then number is 150 ( as yes it's average is 2 also )

    If average is 3
    Number is 75*3 => 225 ( yes average of the digits is also 3)

    Average is 4
    Number is 300 ( but average of digits is not 4)

    Average is 5
    Number is 375 ( average of digits is 5 ,so possibile )

    Average is 6
    75 * 6 => 450 ( average of digits is not 6)

    Average is 7
    75 * 7 => 575 ( average is not 7)

    Average is 8
    75 × 8 => 600 ( again not possible )

    Average is 9
    75 * 9 => 675 ( but average can't be 9)

    So.here Basically two concepts

    1. Table of 75
    2. Average of digits can't be more than 9

    As it is a three digit number
    If you take average as 10 then sum.of digits would be more than 27 which can't be possible
    So three numbers are possible - 150, 225 , 375



  • Q28) The regional passport office has a waiting list of 6665 applicants for passport. The list shows that there at least 6 males between any two females. The largest possible number of females in the waiting list is



  • First person is female
    Then eight person is female
    So one combination of
    1F MMMMMM
    Repeats
    6665 /7 => 952(1/7)
    So 952 combinations will have 952 female
    And one more female in the last
    So 952 + 1 = 953



  • Q29) In an arithmetic progression, the sum of the first N terms is T and the sum of the first 2N terms is 6T. If the sum of the first 3N terms is kT, then find k



  • Take N = 1
    Sum of first term is T
    Sum of first two terms is 6T
    I.e first term = T
    Second term = 5T
    So third term has to be 9T
    Sum of first three terms = T + 5T + 9T
    => 15T
    So k = 15



  • Q30) A clock was correct at 2 p.m, but then it began to lose 30 minutes each hour. It now shows 6 pm, but it stopped 3 hours ago. What is the correct time now?



  • The clock loses 30 minutes per hour.
    i.e 30 minutes of this faulty clock = 60 minutes of the correct clock
    From 2 p.m to 6 p.m ,
    total number of hours = 4 hours
    4 hours of this faulty clock => 4 x 60/30= 8 hours of original clock
    So, The correct time when the clock show 6 p.m = 6 p.m + 4 = 10 p.m
    But The clock stopped 3 hours ago ,
    So present time is 10 p.m + 3 hours = 1 a.m


 

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