Quant Boosters by Hemant Malhotra - Set 14


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


    Find numerically the greatest term in the expansion of (2+3x)^9 when x=3/2

    (n-r+1)/(r) *(a/x) >= 1
    (10-r)/r * (3x/2) >= 1
    (10-r)/r *(9/4)>= 1
    90-9r> = 4r
    so 13r< = 90
    r < = 6. 9 so r=6
    so 7th term

    If a quadrilateral with sides 2, 3, 4 and 5 is inscribed in a circle and circumscribed to another circle then
    find area of Quadrilateral

    Direct formula : If any quadrilateral with sides a, b, c and d inscribed in a circle and circumscribed to another circle then area of quadrilateral is sqrt(abcd). So here answer is sqrt(120)

    Find remainder when 1^99+2^99+3^99+....99^99 is divided by 100

    1 to 99 there are 99 terms
    1^99+99^99 will be div by 100
    2^99+98^99 will be div by 100
    now 49^99+51^99 mod 100=0
    but there will be one more term
    50^99 mod 100
    which is also divisible by 100
    so OA=0

    How many integral solutions (a,b) does the equation a^b=a * b have
    (1) 1
    (2) 2
    (3) 3
    (4) 4
    (5) More than four

    a^b = a * b
    let b=1
    then a=a
    we can put any integral value of a here so More than 4
    OA= 5

    What is the minimum value of 2x^2+3y^2-4x-12y+18=0 for real x and y

    Method1-
    partial differentiation
    with respect to x
    4x-4=0 so x=1
    now with respect to y
    6y-12=0 so y=2
    now put x=1 and y=2
    2+12-4-24+18=4
    so OA=4

    Method2-
    try to make perfect square form
    2*(x-1)^2+3*(y-2)^2+4
    so x=1 and y=2 will give us min value

    F(x) and G(x) are two quadratic Functions such that
    f(1)-G(1)=1
    f(2)-G(2)=2
    F(3)-G(3)=5
    Find value of F(4)-G(4)
    (1) 8
    (2) 9
    (3)10
    (4)12

    Method1- f(x)=ax^2+bx+c
    g(x)=dx^2+ex+m
    now f(x)-g(x)=x^2(a-d)+x(b-e)+c-m
    let a-d=M
    b-e=N
    c-m=T
    so f(x)-g(x)= Mx^2+Nx+T
    so M+N+T=1
    4M+2N+T=2
    9M+3N+T=5
    now find M,N,T= M=1 N=-2 and T=2
    so F(x)-g(x)=x^2-2x+2
    so f(4)-g(4)=10

    Quick approach = this is double difference AP if u are aware of this... it will take less than 20 seconds to solve. 1, 2, 5 so next will be 10

    Centroid of triangle is at (1,-1) while its orthocenter is at (5,3) then circumcentre of triangle could be
    (1) (-1,-3)
    (2) (8/3 , 0)
    (3) (0,8/3)
    (4) (7/3,1/3)

    Many Methods to tackle this, centroid bisect ortho and circum in 2:1 ratio
    distance between (1,-1) and (5,3) is sqrt(16+16)=sqrt32
    so distance between centroid and circum should be sqrt32/2
    check
    sqrt(4+4)=sqrt8
    so OA= A

    There is a clock that has a special way of telling the time. It does not have any hands or numbers on it, but it has a chimer. If the time is 1 o'clock, it chimes once. If the time is 2 o'clock, it chimes twice, and so forth. The time gap between any two chimes is 4 seconds. How many seconds would it take you to know the time, after the first chime is heard, if it is 11 o'clock?
    a) 40 seconds
    b) 44 seconds
    c) 42 seconds
    d) none of these

    OA=44 , 40+ extra 4 seconds u have to wait to confirm chimer rung no more

    In triangle ABC, AB = 3000
    BC = 875, CA = 3125
    If I is the in center of the triangle, what is the ratio of AI:ID where AD is angle bisector

    Method1- AD is angle bisector
    so AB/AC=BD/CD
    so 3000/3125=BD/CD
    let BD=x then CD=875-x
    so we can find x from here
    now in triangle ABD AI is angle bisector
    so BA/BD=AI/ID
    3000/x=AI/ID

    Alternate -
    AI/ID = (AB+AC)/BC, valid for every triangle
    so ((6125/875)

    The average of three consecutive multiples of 3 is a.
    The average of four consecutive multiples of 4 is a + 27.
    The average of the smallest and largest of these seven integers is 42.
    Determine the value of a.

    3(k+1) = a
    4p+6 = a+27
    4p+3k = 72
    3 equations 3 variables.
    k = 8 ; p = 12 and a = 27.


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