1 . | Answer : Option D | Explanation : | Let those three peoples as single group now 10 objects (9 people and a group) which can be arrange in (10 – 1)! ways = 9 ! ways But 3 person of group can be arranged in 3 ! ways Hence total possible no. of arrangements = 9 ! × 3 ! | | | 2 . | Answer : Option C | Explanation : | M and O are fixed at the start and end positions, Hence, we have to arrange 1, R, R, R among themselves ( i.e. in 4 places), and since R is 3 times. This can be done in $4!\over3!$ ways | | | 3 . | Answer : Option D | Explanation : | | | | 4 . | Answer : Option A | Explanation : | x ( x – 4) < 3x – 10 =${x^2}$– 7x + 10 < 0 = (x – 2) ( x – 5) < 0 ∴this possible when : Case I ; (x – 2 ) > 0 then (x – 5 ) < 0 ––– ∴x > 2 and x < 5 Case II ; (x – 2) < 0 then (x – 5) > 0 ∴ x < 2 and x > 5 | | | 5 . | Answer : Option D | Explanation : | 3 P 8 + 2 R 6 + 4 Q 5 = 989 For maximum value of R, the value of P and Q should be minimum (i.e. zero) ∴308 + 2 R 6 + 405 = 989 = 2 R 6 = 989 – 713 →2 R 6 = 276 ∴ R = 7 | | | 6 . | Answer : Option C | Explanation : | | | | 7 . | Answer : Option B | Explanation : | | | | 8 . | Answer : Option D | Explanation : | | | | 9 . | Answer : Option C | Explanation : | 4 king or 4 queen = 8 ∴ Req. probability = $8\over52$=$2\over13$ | | | 10 . | Answer : Option D | Explanation : | Sample space, S = ( HHH, HHT, HTH, HTT, THT, THH, TTH, TTT) Number of exhaustive cases = 8 Favourable cases is TTT ( no head probability to be found),BR. ∴ P(E)=$1\over8$ ∴ Probability of no heads =$Number of favourable cases\over Number of favourable cases$=$1\over8$ | | | |