Quant Practice Questions  Gaurav Kapoor

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!(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+ab8=0
x^210x+9=0
so x=1,9positive roots
so D>=0
q^24p^2>=0
(q2p)(q+2p)>=0
now sum of roots and product of roots wil be positive
so q/p so p > 0 , q < 0
so q2p < 0 is truetotal 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=42c2601=800let 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)let log_5a=k
so 6^k16^(1/k)=32
6^k * 2^(4/k)=32
so k=2
so so log_5a=2
so a=25Quick Approach 72=2^3*3^2
so 3a and 2b will be present in our ans
only 4th choice have that
so OA=4Quick 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 1n * (n+1)/2 x=756
approx value of n= 39
so 39 * 40/2=780
so missed value =780756=24f(x)=ax^2+bx+c
g(x)=dx^2+ex+m
now f(x)g(x)=x^2(ad)+x(be)+cm
let ad=M
be=N
cm=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^22x+2
so f(4)g(4)=10