Answer: B Due to stoppage the train travels (45-36) = 9km less in 1 hour and time taken to travel 9 km is the time taken at stoppages. Therefore, time taken to cover 9 km = 9/45 * 60 = 12.
Q. No. 8:
A train leaves station X at 5am and reaches station Y at 9am. Another train leaves station Y at 7am and reaches station X at 10:30 am. At what time do the two trains cross each other?
Answer: B Let the distance between X and Y is d km Then, speed of A is d/4 km/hr and that of B is 2d/7 km/hr. X.............................................Y (XY = d km) Relative speed = (d/4 + 2d/7) =15d/28 km/hr Now distance between these trains at 7 am = d- d/2 = d/2 km Hence, time = (d/2)/(15d/28) = 14/15 * 60 = 56 min Hence both of them meet at 7:56 am.
Q. No. 9:
A ship 77 km from the shore, springs a leak which admits to 9/2 tonnes of water in 11/2 minutes. 92 tonnes of water would sink it. But the pumps can throw out 12 tonnes of water per hour. Find the average rate of sailing so that the ship may just reach the shore as it begins to sink.
Answer: A Given leak admits 9/4 tonnes of water in 11/2 min. Leak admits one tank of water = 11/2 * 4/9 = 22/9 min Leak admits 9/22 tonne of water in one min Now, pump throws 12 tonne of water in 60 min. Pump throws 1 tonne of water in 5 min. In 1 minutes it throws 1/5 tonne of water. Water accumulated in the ship in 1 min = 9/22 - 1/5 = 23/110 tonnes or 92 tonnes of water which is sufficient to get the ship sinked can be accumulated in 92/(23/110) = 440 min = 22/3 hours. Rate of sailing in order that the ship may just reach the shore = (77/22) / (22/3) => 10.5 km/hr.
Q. No. 10:
X and Y start walking towards each other at 10am at speeds of 3km/hr and 4km/hr respectively. They were initially 17.5 km apart. At what time do they meet?
Answer: D Let after T hours they meet Then, 3T +4T = 17.5 T= 2.5 Time = 10:00 am + 2.5 hour = 12:30 pm
Q. No. 11:
A train covered a certain distance at a uniform speed. If the train had been 6km/hr faster, it would have taken 4 hour less than the scheduled time. And, if the train were slower by 6km/hr, the train would have taken 6hr more than the scheduled time. The length of the journey is
Answer: C Let the length of the journey be d km and the speed of train be S km/hr. Then, d/(S+6) = t-4 ...........(i) and d/(S-6) = t+6..........(ii) Substracting the 1 equation from another we get d/(S-6) - d/(S+6) =10........(iii) Now , t= d/S....substitute in equation (i) and solve for d and S We get S=30 d= 720 km.
Q. No. 12:
A tower is 61.25m high. A rigid body is dropped from its top and at the same instant another body is thrown upwards from the bottom of the tower with such a velocity that they meet in the middle of the tower. The velocity of projection of the second body is
Answer: A Let the body moving downwards take 't' sec to reach half the height. => 245/2 = 1/2 * 9.8 * t*t => t = 5/2 sec Again, assume that the second body is projected minimum velocity 'v' upwards => 245/8 = v*5/2 - 245/8 => v = 49/2 = 24.5 m/sec