Answer: B) 961 x 961 Explanation: We know that am×an=am+n  ∴3131-27 = 314 = 961 x 961.

Answer: B) 392 Explanation: The given number series 11, 13, 30, 96, ? follows a pattern that, 11 11 x 1 + 2 = 13 13 x 2 + 4 = 30 30 x 3 + 6 = 96 96 x 4 + 8 = 392   Hence, the next number in the given series is 392.

Answer: A) 99 Explanation: Let, Distance between A and B = d   Distance travelled by P while it meets Q = d + 11   Distance travelled by Q while it meets P = d – 11   Distance travelled by Q while it meets R = d + 9   Distance travelled by R while it meets Q = d – 9   Here the ratio of speeds of P & Q => SP : SQ = d + 11 : d – 11   The ratio of speeds of Q & R => SQ : SR = d + 9 : d – 9   But given Ratio of speeds of P & R => P : R = 3 : 2  SPSR = SPSQxSQSR = d+11d+9d-11d-9   => d+11d+9d-11d-9 = 3/2   =>  d = 1, 99   => d = 99 satisfies.   Therefore, Distance between A and B = 99

Answer: A) 200, 600, 400 Explanation: Ratio of efficiencies of P, Q and R = 2 : 3 : 4 From the given data, Number of working days of P, Q, R = 5 : 10 : 5 Hence, ratio of amount of p, Q, R = 2x5 : 3x10 : 4x5 = 10 : 30 : 20 Amounts of P, Q, R = 200, 600 and 400.

Answer: B) 3.5 years Explanation: Given that Rs. 1860 will become Rs. 2641.20 at 12% => Simple Interest = 2641.20 - 1860 = Rs. 781.20 We know I = PTR/100 => 781.20 x 100 = 1860 x T x 12 => T = 78120/1860x12 => T = 78120/22320 => T = 3.5 years.

Answer: D) 150 lit Explanation: To cover a distance of 800 kms using a 800 cc engine, the amount of diesel required = 80 litres. => the amount of diesel required to cover a distance of 1000 kms = 1000 x 80/800 = 100 litres However, the vehicle uses a 1200 cc engine and it is given that the amount of diesel required varies directly as the engine capacity. i.e., for instance, if the capacity of engine doubles, the diesel requirement will double too. Therefore, with a 1200 cc engine, quantity of diesel required = 1200/800 x 100 = 150 litres.

Answer: A) Sunday Explanation: The year 2004 is a leap year. So, it has 2 odd days. But, Feb 2004 not included because we are calculating from March 2004 to March 2005. So it has 1 odd day only. The day on 5th March, 2005 will be 1 day beyond the day on 5th March, 2004. Given that, 5th March, 2005 is Monday. 5th March, 2004 is Sunday (1 day before to 5th March, 2005).