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: B) 30 kmph Explanation: Let the speed of the faster train be 'S' kmph   Then speed of the slower train will be '(S-5)' kmph   Time taken by faster train = 350/S hrs   Time taken by slower train = 350/(S-5) hrs  350s-5 - 350s = 2 hrs 20 min = 213 = 73   => S = 30 km/hr.

Answer: A) 3 : 2 Explanation: % of milk in first bottle = 64% % of milk in second bottle = 100 - 26 = 74% Now, ATQ   64%            74%                   68%   6                    4   Hence, by using allegation method,   Required ratio = 3 : 2

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: B) 15/17 Explanation: The original fraction is  2551 × 350+100150+100  = 15/17.

Answer: C) 66.6 mts Explanation: From above diagramAC represents the hill and DE represents the tower Given that AC = 100 m angleXAD = angleADB = 30° (∵ AX || BD ) angleXAE = angleAEC = 60° (∵ AX || CE) Let DE = h Then, BC = DE = h, AB = (100-h) (∵ AC=100 and BC = h), BD = CE tan 60°=AC/CE => √3 = 100/CE =>CE = 100/√3 ----- (1) tan 30° = AB/BD => 1/√3 = 100−h/BD => BD = 100−h(√3)∵ BD = CE and Substitute the value of CE from equation 1 100/√3 = 100−h(√3) => h = 66.66 mts The height of the tower = 66.66 mts.

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: B) Option B Explanation: Man walked 6 km at 1.5 kmph, again he walked 8 km at speed of 2 kmph and 40 km at a speed of 8kmphtime taken indivisually: => 6/1.5 = 4 m=> 8/2 = 4 m=> 40/8 = 5 m   Average speed of man= total distance/ total time=> 54/13 = 4213 kmph