Quant Practice Questions - Gaurav Kapoor



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    4 digit let abcd
    base 6 so 216a+36b+6c+d
    where each digit is less than 6
    now in base 10 , 1000a+100b+10c+d
    1000a+100b+10c+d=4*(216a+36b+6c+d)
    so 136a=44b+14c+3d
    when a=1 then b=2,c=3,d=2
    when b=2
    272=44b+14c+3d
    but max value of b=5 so 14c+3d=52
    which is not possible for b and c 10 which is not in option so no need to check further!

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    (x+1)(x+9)+8=0
    x^2+10x+17=0
    so a+b=-10
    and ab=17
    now roots of x^2+a+b)x+ab-8=0
    x^2-10x+9=0
    so x=1,9

    2_1492596580141_gk3.jpg

    positive roots
    so D>=0
    q^2-4p^2>=0
    (q-2p)(q+2p)>=0
    now sum of roots and product of roots wil be positive
    so -q/p so p > 0 , q < 0
    so q-2p < 0 is true

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    total people =42
    so number of handshake =42c2
    now subtract those cases when handshakes between 4 members of same family
    so 10 * 4c2=60
    now one handshake between Chris and his wife should be subtracted
    so OA=42c2-60-1=800

    0_1492596622704_gk5.jpg

    let 1 men in start so total work=M
    now in first 8 hours
    8(1+1/2+1/4+1/8+1/16)=M
    8+4+2+1+1/2=15+(1/2)

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    let log_5a=k
    so 6^k-16^(1/k)=32
    6^k * -2^(4/k)=32
    so k=2
    so so log_5a=2
    so a=25

    0_1492596661438_gk7.jpg

    Quick Approach- 72=2^3*3^2
    so 3a and 2b will be present in our ans
    only 4th choice have that
    so OA=4

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    Quick Approach = Product of roots [email protected]+1/@ minimum value of which is 2
    now one roots is 2 so other root will be greater than 1
    so only possible option 1

    0_1492596669966_gk9.jpg

    n * (n+1)/2 -x=756
    approx value of n= 39
    so 39 * 40/2=780
    so missed value =780-756=24

    0_1492596673670_gk10.jpg

    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


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