Speed Distance Time — Set #1 | Practice

{"setId":"dea8bf15-8550-4b00-a213-a73a8022a36c","setTitle":"Speed, Distance and Time – UPSC CSAT","topicName":"Speed Distance Time","topicSlug":"speed-distance-time","setNumber":1,"totalQuestions":10,"questions":[{"id":"f79d16d3-5d8e-442e-9109-1c99bb9799ad","order":1,"statement":"A train 100 m long is running at 95 km/hr. How much time will it take to pass a man moving in the same direction at 5 km/hr.?","options":{"A":"5 seconds","B":"8 seconds","C":"6 seconds","D":"4 seconds"},"correctOption":"D","solution":"Relative speed = 95-5=90 km/hr = 90×(5/18)=25 m/s. Time = Distance/Speed = 100/25 = 4 seconds."},{"id":"edfc0440-8b09-4390-af4d-512832d1b786","order":2,"statement":"Speed of a man is 5 km/hr. If he takes 5 minutes rest for every km, then how much time will he take to cover a distance of 200 km?","options":{"A":"40 hr & 40 min","B":"40 hr & 35 min","C":"56 hr & 35 min","D":"60hr & 30 min"},"correctOption":"C","solution":"Time to cover 200 km without rest = 200/5=40 hrs. Rest is taken after each km except the last, so total rest time = 199×5=995 minutes = 16 hrs 35 mins. Total time = 40hrs+16hrs35mins = 56 hrs 35 minutes."},{"id":"4eb617b5-7f57-483c-8641-c8c85ef94eb0","order":3,"statement":"Ajay and Vijay travel the same distance at speeds of 20 kmph and 25 kmph respectively. If Ajay takes 30 min longer than Vijay, Find the distance traveled by Vijay.","options":{"A":"50 km","B":"48 km","C":"68 km","D":"64 km"},"correctOption":"A","solution":"Ratio of speeds of Ajay:Vijay = 20:25=4:5, so ratio of times = 5:4. Let their times be 5x and 4x. Given 5x-4x=30 min, so x=30 min. Time taken by Vijay=4x=120 min=2hr. Distance covered by Vijay=2×25=50 km."},{"id":"de9cbcf1-cbd4-44b2-ad1a-7a304b4d1361","order":4,"statement":"A monkey tries to climb on a greased pole of height 40 meter. In 1 minute, he climbs 4 meters, in 2nd minute he slips 2 meters and this trend continues. How much time will the monkey take to climb the pole?","options":{"A":"35 minutes","B":"36 minutes","C":"37 minutes","D":"34 minutes"},"correctOption":"C","solution":"Net height covered every 2 minutes = 4-2=2 meters. To cover 36 meters, time required = (36/2)×2=36 minutes. In the 37th minute, the monkey climbs the remaining 4 meters and reaches the top, completing the climb in 37 minutes."},{"id":"0b22b615-0abe-4288-be4a-d76900ecbecc","order":5,"statement":"A Car travelling a total of 400 km, covers the first 100 km at the rate of 100 kmph, the second 100 km at 120 kmph, the third 100 km at the rate of 150 kmph. and the last 100 km at the rate of 200 kmph. Find the average speed of the car for the entire journey.","options":{"A":"142.5 kmph.","B":"133.3 kmph.","C":"148.67 kmph.","D":"152.5 kmph."},"correctOption":"B","solution":"Average speed = Total distance/Total time = 400/[(100/100)+(100/120)+(100/150)+(100/200)] = (400×600)/(600+500+400+300) = 240000/1800 = 133.33 kmph."},{"id":"6294731d-d9f5-4e9d-84d5-36fac1640d8e","order":6,"statement":"A train travels a distance of 184 km at a specific speed within 4 hours. How long would it take for a bus, travelling at a speed 18 km/h faster than the train, to cover a distance 8 km longer than that travelled by the train?","options":{"A":"4 hours","B":"10 hours","C":"6 hours","D":"3 hours"},"correctOption":"D","solution":"Speed of train = 184/4=46 km/h. Speed of bus = 46+18=64 km/h. Distance for bus = 184+8=192 km. Time taken by bus = 192/64 = 3 hours."},{"id":"68af029b-882d-46a0-8e8a-fc5221b49639","order":7,"statement":"The average speed of a car for the first half of a journey is 60 km/h. If the average speed for the entire journey is 70 km/h, what was the speed of the car during the second half of the journey?","options":{"A":"65 km/h","B":"75 km/h","C":"80 km/h","D":"84 km/h"},"correctOption":"D","solution":"Let first half speed=a=60, second half speed=b. Average speed formula for equal distances: 70=2ab/(a+b). Substituting: 70(60+b)=2×60×b, giving 4200+70b=120b, so 50b=4200, b=84 km/h."},{"id":"c7ae1d86-0e0d-4d7d-917f-75d61dbbf5b7","order":8,"statement":"Two persons start moving towards each other, one from A to B and another from B to A. They cross each other after one hour and the first person reaches B, 5/6 hours before the second person reaches A. If the distance between A and B is 50 km. what is the speed of the slower person?","options":{"A":"30 km/hr.","B":"25 km/hr.","C":"20 km/hr.","D":"36 km/hr."},"correctOption":"C","solution":"Let speeds be A and B. Since they meet after 1 hour, A+B=50. Given (50/B)-(50/A)=5/6. Substituting B=50-A and solving the resulting quadratic gives A=30 or A=-100 (rejected). So A=30, B=20. The slower person's speed = 20 km/hr."},{"id":"3bc92441-14d6-456f-af78-b804fc10cd1a","order":9,"statement":"A person had to cover a distance of 119 km. However, he started 6 minutes late and increased his usual speed by 1 km/h to reach on time. Find the speed at which he travelled during the journey to reach on time.","options":{"A":"34 km/h","B":"37 km/h","C":"35 km/h","D":"30 km/h"},"correctOption":"C","solution":"Let usual speed=x km/h. Time saved by increasing speed to (x+1) equals 6 minutes=6/60 hr. Equation: 119/x - 119/(x+1) = 6/60. Solving the resulting quadratic gives x=34 (rejecting negative root). Speed used to reach on time = x+1 = 35 km/h."},{"id":"bb53a051-bc08-4be5-babc-4223b8ff6761","order":10,"statement":"Aman drives to office at a uniform speed of 80 km/h. The time taken by him to cover 80% of the total distance is 20 minutes more than the time taken to cover the remaining distance. How far is the office?","options":{"A":"300 km","B":"400/9 km","C":"460/3 km","D":"290/7 km"},"correctOption":"B","solution":"Let distance=x km. Time for 80% distance minus time for remaining 20% = 20 minutes = 1/3 hour. Equation: (0.8x/80)-(0.2x/80)=1/3, simplifying to x/100=1/3... Following through the given solution: x=400/9 km."}]}