Speed Distance Time — Set #2 | Practice

{"setId":"f3da06e3-c2ff-41eb-8aa6-0fabc3bbd3ac","setTitle":"Speed, Distance and Time 2 – UPSC CSAT","topicName":"Speed Distance Time","topicSlug":"speed-distance-time","setNumber":2,"totalQuestions":10,"questions":[{"id":"c537b21e-0ee7-430d-8b95-76000f85815d","order":1,"statement":"A person travels a certain distance at a uniform speed in 5 hours. In return journey, he increases his speed by 5 kmph and covers the same distance in 4 hours. What was his initial speed?","options":{"A":"30 kmph.","B":"25 kmph.","C":"20 kmph.","D":"36 kmph."},"correctOption":"C","solution":"Ratio of initial:final time = 5:4, so ratio of speeds = 4:5. Let speeds be 4x and 5x. Given difference in speeds = 5x-4x=5 kmph, so x=5. Initial speed = 4x=4×5=20 kmph."},{"id":"14f6ef3d-eb8c-4c04-a619-53cac4b5fe7e","order":2,"statement":"In a 1000 m race 'A' wins over 'B' by 150 m and in 1000 m race 'B' can give a start of 250 m to 'C'. By how much distance 'A' gives a start to 'C', so that 'A' beats 'C' by 200 metres in a race of 1000 m?","options":{"A":"162.5 m","B":"160.5 m","C":"155 m","D":"200 m"},"correctOption":"A","solution":"Ratio of distances run by A:B=1000:850=80:68. Ratio of B:C=1000:750=68:51. So A:B:C=80:68:51. When A runs 1000m, C runs (51/80)×1000=637.5m. So C runs 362.5m less than A. For A to beat C by only 200m, A should give C a start of 362.5-200=162.5m."},{"id":"dc4ad2ff-c69a-4ebc-a731-b42dd5bb9c68","order":3,"statement":"A man goes 50 km by Rickshaw, 60 km by Bike and 60 km by Bus, he takes 15 hours to cover the whole distance. If speed of bus is 4 times that of bike and 3 times that of Rickshaw. Find the speed of Rickshaw.","options":{"A":"6 kmph","B":"8 kmph","C":"10 kmph","D":"12 kmph"},"correctOption":"C","solution":"Ratio of speeds: Bus:Bike=4:1, Bus:Rickshaw=3:1, so Bus:Rickshaw:Bike=12:4:3. Let speeds be 12x,4x,3x. Time taken=Distance/Speed: Rickshaw=50/4x=12.5/x, Bus=60/12x=5/x, Bike=60/3x=20/x. Total time=12.5/x+5/x+20/x=37.5/x=15hours, giving x=2.5. Speed of Rickshaw=4x=4×2.5=10 kmph."},{"id":"4fb24bce-b325-44e5-ba89-f9955f209273","order":4,"statement":"A person can cover a journey in 4 hours. If he decreases his speed by 1/12, he covers 15 km less. Find his original speed.","options":{"A":"45 km/hr","B":"65 km/hr","C":"50 km/hr","D":"64 km/hr"},"correctOption":"A","solution":"Ratio of original:decreased speed=12:11, and since speed is proportional to distance covered in same time, ratio of distances=12:11 or 12x:11x. Given difference=12x-11x=x=15km. Original distance=12×15=180km. Original speed=Distance/time=180/4=45 km/hr."},{"id":"f2f90961-3fd8-4fc7-b900-94dbd887955d","order":5,"statement":"Two trains start at the same time from two stations and proceed towards each other at the rates of 25 km/hr and 40 km/hr respectively. When they meet, it is found that one train has travelled 90 km more than the other. Find the distance between the two stations.","options":{"A":"490 km","B":"390 km","C":"540 km","D":"None of the above"},"correctOption":"B","solution":"Let distance travelled by slower train=d km, so faster train travels (d+90)km in the same time. Equation: d/25=(d+90)/40. Solving: 40d=25d+2250, giving 15d=2250, d=150. Total distance = 150+150+90=390 km."},{"id":"a0983fdc-1a33-450c-90a4-a60e5e8778c5","order":6,"statement":"A boat covered a certain distance travelling downstream in 40 minutes, while it came back to the starting point in one hour 20 minutes. The speed of the stream is 3 kmph. What is the speed of the boat in still water?","options":{"A":"6 km/hr","B":"7 km/hr","C":"8 km/hr","D":"9 km/hr"},"correctOption":"D","solution":"Let boat speed in still water=x km/hr, one-way distance=d km. Downstream: d/(x+3)=40/60, so d=2(x+3)/3. Upstream: d/(x-3)=80/60, so d=4(x-3)/3. Equating: 2(x+3)/3=4(x-3)/3, giving x+3=2(x-3)=2x-6, so x=9 km/hr."},{"id":"5a99cc01-26a8-4cc0-ab30-5a672b0c7c55","order":7,"statement":"A boat travels upstream from Q to P and downstream from P to Q in 8 hours. If the speed of the boat in still water is 12 kmph. and the speed of the current is 3 km/hr, the distance between P and Q (in km) is","options":{"A":"30km","B":"42km","C":"45km","D":"36km"},"correctOption":"C","solution":"Downstream speed=12+3=15kmph, upstream speed=12-3=9kmph. Let distance=D km. Total time: D/15+D/9=8. Solving: (3D+5D)/45=8, giving 8D=360, D=45 km."},{"id":"5657da4f-fd26-4906-be16-683c15fa6182","order":8,"statement":"Walking at 3/4 of his normal speed, a person is 10 minutes late in reaching his office. What time does he usually take to cover the distance?","options":{"A":"40 min","B":"45 min","C":"30 min","D":"Data is insufficient"},"correctOption":"C","solution":"Let usual time=x min. Since speed decreases to 3/4, time increases proportionally by a factor of 4/3, meaning extra time taken = (1/3) of usual time = 10 minutes. So x/3=10, giving x=30 min."},{"id":"1649fd8c-c218-42f6-84f1-bccf5c57a95f","order":9,"statement":"The distance between the two stations Delhi and Lucknow is 850 km. A train travels from Delhi to Lucknow at a speed of 75 km per hour and then returns from Lucknow to Delhi at a constant speed of 80 km per hour. What is the average speed of the train for the entire round trip?","options":{"A":"77.42 km/h","B":"74.72 km/h","C":"72.47 km/h","D":"72.77 km/h"},"correctOption":"A","solution":"Average speed for a round trip with equal distances = 2xy/(x+y), where x and y are the two speeds. Here x=75, y=80. Average speed = (2×75×80)/(75+80) = 12000/155 = 77.42 km/h."},{"id":"0ee8eafc-2a45-4dff-955e-72a9bd1e86cd","order":10,"statement":"The speed of the boat in still water is 8 km/hr. If it can travel 22 km downstream and 10 km upstream in the same time, then what is the speed of the stream?","options":{"A":"3 km/hr","B":"2 km/hr","C":"4 km/hr","D":"None of the above"},"correctOption":"A","solution":"Let speed of current=x km/hr. Given 22/(8+x)=10/(8-x). Cross multiplying: 22(8-x)=10(8+x), giving 176-22x=80+10x, so 96=32x, x=3 km/hr."}]}