Two men are walking towards each other alongside a railway track, A freight train overtakes one of them in 20 seconds and exactly 10 minutes later meets the other man coming from the opposite direction. The train passes this man is 18 seconds. Assume the velocities are constant throughout.
Q. No. 1:
How long after the train has passed the second man will the two men meet?
Answer: B Let ‘L’ be the length of train, ‘x’ be the speed of the first man, ‘y’ be the speed of the second man and ‘z’ be the speed of the train. 20 = 1/(z-x) and 18 = 1/(z+y). => z= 10x+9y Distance between the two men = 600(z + y) mt Time = {600(z + y) - 600(x + y)}/(x+y) = 600(9x+9y)/(x+y) = 90 minutes.
Q. No. 2:
The ratio of the velocities of the first man to the second man is
Answer: D Let ‘L’ be the length of train, ‘x’ be the speed of the first man, ‘y’
be the speed of the second man and ‘z’ be the speed of the train. 20 = 1/(z-x) and 18 = 1/(z+y). => z= 10x+9y Distance between the two men = 600(z + y) mt Time = {600(z + y) - 600(x + y)}/(x+y) = 600(9x+9y)/(x+y) = 90 minutes. x/y cannot be determined.
Q. No. 14:
Two boats, traveling at 5 and 10 kms per hour, head directly towards each other. They begin at a distance of 20 kms from each other. How far apart are they (in kms) one minute before they collide.
Answer: C The boats will be colliding after a time which is given by; 20/(5+10) = 4/3 hr = 80min. After this time of 80 minutes, boat (1) has covered 80* 5/60 = 20/3 kms whereas boat (2) has covered 80* 10/60 = 40/3 kms After 79 minutes, distance covered by the first boat (D1)= (20/3 - 5/60). After 79 minutes, distance covered by the second boat(D2) =(40/3 -10/60). So the separation between the two boats = 20 - (D1+D2) = 1/4 kms. Alternative method:- Relative speed of two boats = 5 + 10 = 15 km/hr i.e. in 60 min they cover (together) = 15 km in 1 min they will cover (together) = 15/60 = 1/4 kms.
Q. No. 15:
If a man cycles at 10 km/hr, then he arrives at a certain place at 1 p.m. If he cycles at 15 km/hr, he will arrive at the same place at 11 a.m. At what speed must he cycle to get there at noon?
Answer: B When speed of the man = 10 km/hr = d/t and, When speed of the man = 15 km/hr = d/(t-2). Equating the value of d: 10 × t = 15 × (t – 2) => t = 6 hours. Finally desired speed = d/(t-1) = 10t/(t-1) => (10*6)/5 = 12 km/hr
Q. No. 16:
Karan and Arjun run a 100-meter race, where Karan beats Arjun 10 metres. To do a favour to Arjun, Karan starts 10 metres behind the starting line in a second 100 metre race. They both run at their earlier speeds. Which of the following is true in connection with the second race?
A :
Karan and Arjun reach the finishing line simultaneously
Answer: D Situation (I): In whatever time Karan covers a distance of 100 m, Arjun covers 90 m in the same time. Situation (II): Now Karan is 10 m behind the starting point. Once again to cover 100 m from this new point Karan will be taking the same time as before. In this time Arjun will be covering 90 m eters only. This means that now both of them will be at the same point, which will be 10 meters away from the finish point. Since both of them are required to cover the same distance of 10 m now and Karan has a higher speed, he will beat Arjun. No need for calculations as option (4) is the only such option.
Q. No. 17:
Rajesh walks to and fro to a shopping mall. He spends 30 minutes shopping. If he walks at speed of 10 km an hour, he returns to home at 19.00 hours. If he walks at 15 km an hour, he returns to home at 18.30 hours. How fast must he walk in order to return at 18.15 hours?
Answer: D As per the question, let D be the total distance and ‘t’ is the time taken. So we have D = 10 t = 15(t – 0.5) =>t = 1.5 hrs =>D = 15 km Now, for the condition given we have 15 = S(t - 3/4) where ‘S’ is the required speed. => S(3/2 - 3/4) = 15 => S= 20km/hr
Q. No. 18:
A train left station X at A hour B minutes. It reached station Y at B hour C minutes on the same day, after travelling C hours A minutes (clock shows time from 0 hours to 24 hours). Number of possible value(s) of A is
Answer: A A hours + C hours = B hours … (i) A, C and B cannot have values greater than or equal to 24 B minutes + A minutes = C minutes … (ii) Looking at the two equations, we get that no value of A satisfies both the equations. Hence, option A.