Time & Distance

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Time and Distance Formulae: I)Speed = Distance/Time II)Time = Distance/speed III) Distance = speed*time IV) 1km/hr = 5/18 m/s V)1 m/s = 18/5 Km/hr VI)If the ratio of the speed of A and B is a:b,then the ratio of the time taken by them to cover the same distance is 1/a : 1/b or b:a VII) suppose a man covers a distance at x kmph and an equal distance at y kmph.then the average speed during the whole journey is (2xy/x+y)kmph Problems 1)A person covers a certain distance at 7kmph .How many meters does he cover in 2 minutes. Solution:: speed=72kmph=72*5/18 = 20m/s distance covered in 2min =20*2*60 = 2400m 2)If a man runs at 3m/s. How many km does he run in 1hr 40min Solution:: speed of the man = 3*18/5 kmph = 54/5kmph Distance covered in 5/3 hrs=54/5*5/3 = 18km 3)Walking at the rate of 4knph a man covers certain distance in 2hr 45 min. Running at a speed of 16.5 kmph the man will cover the same distance in. Solution:: Distance=Speed* time 4*11/4=11km New speed =16.5kmph therefore Time=D/S=11/16.5 = 40min

Top Complex Problems 1)A train covers a distance in 50 min ,if it runs at a speed of 48kmph on an average.The speed at which the train must run to reduce the time of journey to 40min will be. Solution:: Time=50/60 hr=5/6hr Speed=48mph distance=S*T=48*5/6=40km time=40/60hr=2/3hr New speed = 40* 3/2 kmph= 60kmph 2)Vikas can cover a distance in 1hr 24min by covering 2/3 of the distance at 4 kmph and the rest at 5kmph.the total

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distance is? Solution:: Let total distance be S total time=1hr24min A to T :: speed=4kmph diistance=2/3S T to S :: speed=5km distance=1-2/3S=1/3S 21/15 hr=2/3 S/4 + 1/3s /5 84=14/3S*3 S=84*3/14*3 = 6km 3)walking at ¾ of his usual speed ,a man is late by 2 ½ hr. the usual time is. Solution:: Usual speed = S Usual time = T Distance = D New Speed is ¾ S New time is 4/3 T 4/3 T – T = 5/2 T=15/2 = 7 ½ 4)A man covers a distance on scooter .had he moved 3kmph faster he would have taken 40 min less. If he had moved 2kmph slower he would have taken 40min more.the distance is. Solution:: Let distance = x m Usual rate = y kmph x/y – x/y+3 = 40/60 hr 2y(y+3) = 9x --------------1 x/y-2 – x/y = 40/60 hr y(y-2) = 3x -----------------2 divide 1 & 2 equations by solving we get x = 40 5)Excluding stoppages,the speed of the bus is 54kmph and including stoppages,it is 45kmph.for how many min does the bus stop per hr. Solution:: Due to stoppages,it covers 9km less. time taken to cover 9 km is [9/54 *60] min = 10min 6)Two boys starting from the same place walk at a rate of 5kmph and 5.5kmph respectively.wht time will they take to be 8.5km apart, if they walk in the same direction Solution:: The relative speed of the boys = 5.5kmph – 5kmph = 0.5 kmph Distance between them is 8.5 km Time= 8.5km / 0.5 kmph = 17 hrs 7)2 trains starting at the same time from 2 stations 200km apart and going in opposite direction cross each other ata distance of 110km from one of the stations.what is the ratio of their speeds. Solution:: In same time ,they cover 110km & 90 km respectively so ratio of their speed =110:90 = 11:9 8)Two trains start from A & B and travel towards each other at speed of 50kmph and 60kmph resp. At the time of the meeting the second train has traveled 120km more than the first.the distance between them. Solution:: Let the distance traveled by the first train be x km then distance covered by the second train is x + 120km x/50 = x+120 / 60 x= 600

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so the distance between A & B is x + x + 120 = 1320 km 9)A thief steals a ca r at 2.30pm and drives it at 60kmph.the theft is discovered at 3pm and the owner sets off in another car at 75kmph when will he overtake the thief Solution:: Let the thief is overtaken x hrs after 2.30pm distance covered by the thief in x hrs = distance covered by the owner in x-1/2 hr 60x = 75 ( x- ½) x= 5/2 hr thief is overtaken at 2.30 pm + 2 ½ hr = 5 pm 10)In covering distance,the speed of A & B are in the ratio of 3:4.A takes 30min more than B to reach the destion.The time taken by A to reach the destinstion is. Solution:: Ratio of speed = 3:4 Ratio of time = 4:3 let A takes 4x hrs,B takes 3x hrs then 4x-3x = 30/60 hr x = ½ hr Time taken by A to reach the destination is 4x = 4 * ½ = 2 hr 11)A motorist covers a distance of 39km in 45min by moving at a speed of xkmph for the first 15min.then moving at double the speed for the next 20 min and then again moving at his original speed for the rest of the journey .then x=? Solution:: Total distance = 39 km Total time = 45 min D = S*T x * 15/60 + 2x * 20/60 + x * 10/60 = 39 km x = 36 kmph 12)A & B are two towns.Mr.Fara covers the distance from A t0 B on cycle at 17kmph and returns to A by a tonga running at a uniform speed of 8kmph.his average speed during the whole journey is. Solution:: When same distance is covered with different speeds,then the average speed = 2xy / x+y =10.88kmph 13)A car covers 4 successive 3km stretches at speed of 10kmph,20kmph,30kmph&:60kmph resp. Its average speed is. Solution:: Average speed = total distance / total time total distance = 4 * 3 = 12 km total time = 3/10 + 3/20 + 3/30 + 3/60 = 36/60 hr speed =12/36 * 60 = 20 kmph

Top 14)A person walks at 5kmph for 6hr and at 4kmph for 12hr. The average speed is. Solution:: avg speed = total distance/total time = 5*6 + 4*12 / 18 =4 1/3 mph 15)A bullock cart has to cover a distance of 80km in 10hrs. If it covers half of the journeyin 3/5th time.wht should be its speed to cover the remaining distance in the time left. Solution:: Time left = 10 - 3/5*10 = 4 hr speed =40 km /4 hr =10 kmph

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16)The ratio between the speeds of the A& B is 2:3 an therefore A takes 10 min more than the time taken by B to reach the destination.If A had walked at double the speed ,he would have covered the distance in ? Solution:: Ratio of speed = 2:3 Ratio of time = 3:2 A takes 10 min more 3x-2x = 10 min A's time=30 min --->A covers the distance in 30 min ,if its speed is x -> He will cover the same distance in 15 min,if its speed doubles (i.e 2x) 17)A is twice as fast as B and B is thrice as fast as C is. The journey covered by B in? Solution:: speed's ratio a : b = 2: 1 b : c = 3:1 Time's ratio b : c = 1:3 b : c = 18:54 (if c covers in 54 min i..e twice to 18 min ) 18)A man performed 3/5 of the total journey by ratio 17/20 by bus and the remaining 65km on foot.wht is his total journey. Solution:: Let total distance is x x-(3/5x + 17/20 x) =6.5 x- 19x/20 = 6.5 x=20 * 6.5 =130 km 19)A train M leaves Meerat at 5 am and reaches Delhi at 9am . Another train N leaves Delhi at 7am and reaches Meerut at 1030am At what time do the 2 trains cross one another Solution:: Let the distance between Meerut & Delhi be x they meet after y hr after 7am M covers x in 4hr N covers x in 3 ½ i.e 7/2 hr speed of M =x/4 speed of N = 2x/7 Distance covered by M in y+2 hr + Distance covered by N in y hr is x x/4 (y+2) +2x/7(y)=x y=14/15hr or 56 min 20)A man takes 5hr 45min in walking to certain place and riding back. He would have gained 2hrs by riding both ways.The time he would take to walk both ways is? Solution:: Let x be the speed of walked Let y be the speed of ride Let D be the distance Then D/x + D/y = 23/4 hr -------1 D/y + D/y = 23/4 – 2 hr D/y = 15/8 --------2 substitute 2 in 1 D/x + 15/8 = 23/4 D/x = 23/4 -15/8 =46-15/8 =31/8 Time taken for walk one way is 31/8 hr time taken to walk to and fro is 2*31/8 = 31/4 hr =7 hr 45 min

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