Answer: D) 120 litres Explanation: wine(left)wine(added) = 343169   It means  wine(left)wine(initial amount) = 343512    (since 343 + 169 = 512)    Thus,  343x = 512x1 - 15k3    343512 = 783 = 1 - 15k3    1-15k=78=1-18    Thus the initial amount of wine was 120 liters.

Answer: A) 2880 Explanation: There are total 9 places out of which 4 are even and rest 5 places are odd.   4 women can be arranged at 4 even places in 4! ways.   and 5 men can be placed in remaining 5 places in 5! ways.   Hence, the required number of permutations  = 4! x 5! = 24 x 120 = 2880

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: C) 246 Explanation: A Committee of 5 persons is to be formed from 6 gentlemen and 4 ladies by taking.    (i) 1 lady out of 4 and 4 gentlemen out of 6  (ii) 2 ladies out of 4 and 3 gentlemen out of 6  (iii) 3 ladies out of 4 and 2 gentlemen out of 6  (iv) 4 ladies out of 4 and 1 gentlemen out of 6    In case I the number of ways = C14×C46 = 4 x 15 = 60  In case II the number of ways = C24×C36 = 6 x 20 = 120  In case III the number of ways = C34×C26 = 4 x 15 = 60 In case IV the number of ways = C44×C16 = 1 x 6 = 6    Hence, the required number of ways = 60 + 120 + 60 + 6 = 246

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) 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