Answer: B) 2 x (18!) Explanation: A person can be chosen out of 18 people in 18 ways to be seated between Musharraf and Manmohan. Now consider Musharraf, Manmohan, and the third person, sitting between them, as a single personality, we can arrange them in 17! ways but Musharraf and Manmohan can also be arranged in 2 ways.    Required number of permutations = 18 x (17!) x 2 = 2 x 18!

Answer: B) 535 Explanation: The number of points of intersection of 37 lines is C237. But 13 straight lines out of the given 37 straight lines pass through the same point A.   Therefore instead of getting C213 points, we get only one point A. Similarly 11 straight lines out of the given 37 straight lines intersect at point B. Therefore instead of getting C211 points, we get only one point B.    Hence the number of intersection points of the lines is C237-C213-C211 +2 = 535

Answer: C) 3/4 Explanation: Let A, B, C be the respective events of solving the problem and A , B, C be the respective events of not solving the problem. Then A, B, C are independent event ∴A, B, C are independent events Now,  P(A) = 1/2 , P(B) = 1/3 and P(C)=1/4  PA=12, PB=23, PC= 34 ∴ P( none  solves the problem) = P(not A) and (not B) and (not C)                     = PA∩B∩C                    = PAPBPC         ∵ A, B, C are Independent                                          =  12×23×34                     = 14   Hence, P(the problem will be solved) = 1 - P(none solves the problem)                  = 1-14= 3/4

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) 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) 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) 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