Answer: B) 19/42 Explanation: A red ball can be drawn in two mutually exclusive ways  (i) Selecting bag I and then drawing a red ball from it.   (ii) Selecting bag II and then drawing a red ball from it.   Let E1, E2 and A denote the events defined as follows: E1 = selecting bag I, E2 = selecting bag II A = drawing a red ball Since one of the two bags is selected randomly, therefore  P(E1) = 1/2  and  P(E2) = 1/2 Now, PAE1 = Probability of drawing a red ball when the first bag has been selected = 4/7   PAE2  = Probability of drawing a red ball when the second bag has been selected = 2/6  Using the law of total probability, we have   P(red ball) = P(A) = PE1×PAE1+PE2×PAE2                              = 12×47+12×26=1942

Answer: C) 2/7 Explanation: P( only one of them will be selected) = p[(E and not F) or (F and not E)]   = PE∩F∪F∩E    = PEPF+PFPE    =17×45+15×67=27

Answer: B) 19/90 Explanation: ASSISTANT→AAINSSSTT STATISTICS→ACIISSSTTT Here N and C are not common and same letters can be A, I, S, T. Therefore  Probability of choosing A =  2C19C1×1C110C1 = 1/45   Probability of choosing I = 19C1×2C110C1 = 1/45 Probability of choosing S = 3C19C1×3C110C1 = 1/10 Probability of choosing T = 2C19C1×3C110C1 = 1/15 Hence, Required probability =   145+145+110+115= 1990

Answer: B) 15/39 Explanation: P( selecting atleast one couple) = 1 - P(selecting none of the couples for the prize)     = 1-16C1× 14C1×12C1×10C116C4=1539

Answer: B) 13 Explanation:  Let   Let x=333 = 333    Then, logx = 33 log3     = 27 x 0.47712 = 12.88224    Since the characteristic in the resultant value of log x is 12   ∴The number of digits in x is (12 + 1) = 13    Hence the required number of digits in 333is 13.

Answer: B) 3/2 Explanation:  = log 33 + log 23+ log 103log10×3×22        =log33 12+log 23+log 10312log(10×3×22)                 =12log 33+3 log 2+12 log103log10+log3+log22                 =32log 3 + 2 log 2 + log 10log 3 + 2 log 2 + log 10 = 32

Answer: C) p > q Explanation: kl>k+1l+1 for (k,l) > 0 and  k > l        Let     k = x+1    and   l = x       Therefore, x+1x>(x+1)+1(x)+1        (x + 1) > x       Therefore, log(x+1)log(x)>log(x+2)log(x+1)       ⇒logxx+1 >logx+1x+2

Answer: B) 0 Explanation: = log tan10+log tan890 + log tan20+ log tan880+⋯⋯+log tan450     = log [tan10 × tan890] + log [tan20 × tan880 ] +⋯⋯+log1     ∵ tan(90-θ)=cotθ and tan 450=1     = log 1 + log 1 +.....+log 1    = 0.