From a point P, on the surface of radius 3cm, two cockroaches A and B started moving along two different circular paths, each having the maximum possible radius, on the surface of the sphere, that lie in the two different planes which are inclined at an angle of 45 degree to each other. If A and B takes 18 sec and 6 sec respectively, to complete one revolution along their respective circular paths, then after how much time will they meet again, after they start from P?

Answer: C Both the circular paths have the maximum possible radius hence, both have a radius of 3cm each. Irrespective of the angle between the planes of their circular paths, the two cockroaches will meet again , at the point Q only, which is diametrically opposite end of P. A will takes 9 seconds to reach point Q, completing half a revolution. On the other hand, B would have completed 3/2 of her revolution and it will also reach. point Q simultaneously.

Q. No. 2:

P and Q travels from D to A and break journey at M in between. Somewhere between D and M, P asks "how far have we travelled?" Q replies,"Half as far as the distance from here to M".Somewhere between M and A, exactly 300km from the point where P asks the first question, "How far have we to go?" Q replies, "Half as far as the distance from M to here". The distance between D and A is

Answer: B Point Z when Q asked the first question. Point Y when Q asked the second question. Therefore, distance is DA. => DZ+DY+YA => x/2 + 300+ (300-x)/2 => (x+600+300-x)/2 = 900/2 = 450 km

Q. No. 3:

Two motorist A and S are practicing with tow different speed cars: Ferrari and Maclarun, on the circular racing track, for the car racing tournament to be held next month. Both A and S start from the same point on the circular track. A completes one round of the track in 1 min and S takes 2 min to complete a round. While A maintains same speed for all the rounds, S halves his speed after the completion of each round. How many times A and S will meet between the 6th round and the 9th round of S (6th and the 9th round is excluded)? Assume that the speed of S remains steady throughout each round and changes only after the completion of that round.

Answer: D According to the question, S's time to cover a round keeps getting halved ater completion of each round while Anil maintains his initial speed or the time. Till 6th round, S's time to cover a round = 2^{6} During 7th and 8th round, time takes by S will be 2^{7} and 2^{8} minutes Now since A maintains his speed which comes 1 min, so they will meet => (128+256-2)= 382 times between the 6th round and 9th round. Note:- Subtraction of 2 indicates the exclusion of 6th and 9th round.

Q. No. 4:

Laila drives to the station each day to pick up her husband Manju, who usually arrives on a train at 6o’clock. Last Monday, Manju finished work earlier, caught an earlier train and arrived at the station at 5 ‘o clock. He started to walk home and eventually met Laila who drove him the rest of the way, getting home 20 minutes earlier than usual. On Tuesday, he again finished early and found himself at the station at 5 : 30. Again he began to walk home, again he met Laila on the way, and she drove him home the rest of the way, Assume constant speed throughout with no wasted time for waiting, backing of the car etc. How earlier than the usual time were theory home on Tuesday?

Answer: C Everyday, Laila leaves from A (the time of her departure is same everyday) to reach point D (the station) exactly at 6 ‘O clock where he finds Majnu waiting for him. On Monday:- A-------------------C-------------------------D Laila leaves from A at her usual time but Majnu reaches the station(D )at 5’O clock (i.e. an hour earlier than his normal time ) and he starts walking towards home (i.e. towards point A). Laila meets him on the way at C and from there both of them head towards A. A total of 20 minutes are saved this way. Where have they saved these 20 minutes’ time? These 20 minutes are saved just because of the fact that today Laila did not have to travel the distance C→D +D→C . So we can deduce that Laila must be taking a total of 20 minutes time in a to-and-fro travel between C and D daily. As Laila (and for that matter Majnu also) have a constant speed throughout the journey, we can also deduce that for Laila, the time taken in going from C to D is the same as the time taken in going from D to C. So we can say, had Majnu been at D at 6’O clock on Monday as well, then Laila must have traveled the C to D distance and she must have passed point C at exactly 10 minutes before 6 ‘ O clock i.e. at 5 : 50 PM. This only means that on Monday when they meet at point C, the time of their meeting was exactly 5:50 pm. Now what should this tell us about their speeds? We know that Laila takes 10minute’s time to move from C to D. And now we also know that Majnu had been walking till 5:50 PM (he reaches the station and starts walking till he reaches the point C). So Majnu takes exactly 50 minutes in traveling the same distance. If V_{L} and V_{M} be the speeds of Laila and Majnu respectively, then we must have:
V_{M} / V_{L} = 1/5 On Tuesday:- A----------------------E--------------------D Majnu reaches D at 5:30 PM and starts walking towards A. Lets assume that he walks for x minutes and reaches the point E, where he meets Laila. So that Laila has saved time in moving from E to D and from D to E. Today, Majnu reaches at D at 5:30 + x PM. From Laila’s point, we can say, Laila must have reached the point E at exactly 30 − x minutes before 6:00 PM. (for example if 5:30 + x had been something like 5:45 PM, then she must have reached 15 minutes before 6:00 PM). In other words, today both of them have saved a total of 2×(30 – x)minutes in reaching back to home (point A). To find the value of x: Majnu Moved from D to E in x minutes, and Laila (would have moved) from E to D in 30 – x minutes. Now, => x/(30-x) = V_{L} / V_{M }= 5 Which gives x = 25 minutes.So, today they have saved a total of 2×(30 – 25) = 10minutes Hence (C) option is the correct one.

Q. No. 5:

In a 3600 m race around a circular track of length 400m, the faster runner and the slowest runner meet at the end of the fourth minute, for the first time after the start of the race. All the runner maintains a uniform speed throughout the race. If the faster runner runs at thrice the speed of the slowest runner. Find the time taken by the faster runner to finish the race.

Answer: B As, the faster runner is thrice as fast as the slowest runner, the faster runner would have completed three rounds by the time the slowest runner completes one round. And that is their second meeting.Their first meeting takes place after the fastest runner takes 4 min to complete one and the half round. i.e 400 * 3/2 = 600m He takes 3600/600 * 4 = 24 minutes to finish the race.

Q. No. 6:

Three athletes A,B and C run a race, B finished 24 meters ahead of C and 36 m ahead of A, while C finished 16 m ahead of A. If each athlete runs the entire distance at their respective constant speeds, what is the length of the race?

Answer: D Let the length of the race be 'd'. When B finished the race. A and C would have run (d-36) and (d-24) meters respectively. When C finishes the race. A would have run (d-16) meters. The ratio of speeds of C and A is => (d-24)/(d-36) = d/(d-16) => d = 96 m