Arun, Barun and Kiranmala start from the same place and travel in the same direction at speeds of 30, 40 and 60 km per hour respectively. Barun starts two hours after Arun. If Barun and Kiranmala overtake Arun at the same instant, how many hours after Arun did Kiranmala start?
Answer: C Let us assume that Arun started running at 10 AM and Barun started at 12 noon. So, in these two hours distance traveled by Arun is 60 km and the relative speed of Barun w.r.t Arun is 10 km/hr. So Barun will overtake Arun after =60/10 = 6hrs So, Barun reaches there at 6 PM. So, Kiranmala also overtakes Arun at 6 PM. Let us assume Kiranmala takes 't' time to overtake Arun and the relative speed of Kiranmala w.r.t Arun is 30 km/hr and Arun ran for 8 hrs. So, distance travelled by Arun is While Kiranmala's distance traveled is t = 4 hours So, after 4 hrs, Kiranmala will start running.
Q. No. 20:
Dishant and Prashant are competing with each other in a friendly community competition in a pool of 50m length and the race is for 1000m. Dishant crosses 50m in 2 min and Prashant in 3 min 15 sec. Each time they meet/cross each other, they do handshake's. How many such handshake's will happen if they start from the same end at the same time?
Answer: B When Dishant completes second round, they do handshake once. Now for every round which Dishant completes, there will be one hand shake as the ratio of speed is 13:8. D and P will meet at the pool end only after D completes 26 rounds. In the 20th round, D finish the race and the total handshake's will be 20 -1 = 19.
Q. No. 21:
A bus overtakes two boys who are walking in the direction of the bus at 2km/hr and 4km/hr in 9 sec and 10sec respectively. The length of the bus and its speed are
Answer: A a= length of the bus , b= speed of the bus in m/s Relative speed with respect to first one => b - 5/9 = (9b-5)/9 (As, 2kmph = 5/9 m/s) => a/((9b-5)/9) = 9 => a= 9b-5............(i) Similarly for second one => a/((9b-10)/9) = 10 => 90b-9a = 100................(ii) From Eq (i) and (ii) a = 50m and b= 50/9 m/s
Q. No. 22:
In a kilometer race, If Abhishek gives Bharti a 40m start, Abhishek wins by 19sec. But if Abhishek gives Bharti a 30 sec start, Bharti wins by 40 m. Find the time taken by Bharti to run 5,000m?
Answer: C If A takes x seconds and B takes y seconds to run 1km, then => x+19 = 960y/1000 and 960x/1000+30 =y => y= 150 sec and x =125 sec Answer = 150/1000 * 5000 = 750 sec
Q. No. 23:
A man jogging inside a railway tunnel at a constant speed hears a train approaching the tunnel from behind at a speed of 30km/h, when he is one third of the way inside the tunnel. Whether he keeps running forward or turns back, he will reach the end of the tunnel at the same time the train reaches that end. The speed at which the man is running is :
Answer: D Let the train is at distance y km from the tunnel and the length of the tunnel is x km. Man is at point C which is x/3 km away from B. A..............B.................C.......................D AB = y, BC = x/3 and CD = 2x/3. Let M km/h be the speed of man. Now, train is at A and man is at C and both will take same time for reaching from B. y/30 = x/3M M = 10x/y.......................(i) Also, train and man will take some time for reaching at D. (y+x)/30 = 2x/3M => M = 20x/(y+x)....................(ii) From both the equations we get, x = y. And on putting value in any equation we get, M = 10 km/hr.
Q. No. 24:
Two cities, A and B, at a distance of 50km, are connected by two separate roads. The speed of any vehicle traveling between the two cities on road 1 is 50km/hr, while the speed on road 2 is (80/n) km/hr, where n is the number of vehicles (including the concerned vehicle). If you travel in a vehicle from A to B on road 1 and come back from B to A on road 2 (where there are already three vehicles plying), your approximate average speed is :
Answer: B Time taken from road 1 = 50/50 = 1 hour. Time taken from road 2 is 50/ (80/4) = 2.5 hour. Total time taken = 3.5 hour. Total distance traveled = 100 km. Average speed = 100/3.5 km.hr = 29km/hr.