Question

a train covers 200 km in4 hrs 20m.if the speed is increased by 18 km/hr,the time taken to cover the same distance is​

Answer 1

Answer:

Given that distance(d) = 200km, time(t) = 4hrs 20min

First convert time into hours

t = 4 + 20/60 = 4 + 1/3 = (12 + 1)/3 = 13/3 hrs

Then calculate Speed(v1)

v1 = d/t = 200/(13/3) = (200*3)/13 = (600/13) km/hr

Now speed increased by 18km/hr

v2 = v1 + 18 = (600/13) + 18 = (600 + 234)/13 = (834/13) km/hr

So Time(t2) = d/v2

t2 = 200 / (834/13) = (200 * 13) / 834 = 2600/834 hr

t2 = 3.117 hrs or 3hr 7min 1 sec

Ans: Time = 3.117 hrs or 3hr 7min 1 sec

Answer 2

Answer:

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Step-by-step explanation:

First convert time into hours

First convert time into hourst = 4 + 20/60 = 4 + 1/3 = (12 + 1)/3 = 13/3 hrs

First convert time into hourst = 4 + 20/60 = 4 + 1/3 = (12 + 1)/3 = 13/3 hrsThen calculate Speed(v1)

First convert time into hourst = 4 + 20/60 = 4 + 1/3 = (12 + 1)/3 = 13/3 hrsThen calculate Speed(v1)v1 = d/t = 200/(13/3) = (200*3)/13 = (600/13) km/hr

First convert time into hourst = 4 + 20/60 = 4 + 1/3 = (12 + 1)/3 = 13/3 hrsThen calculate Speed(v1)v1 = d/t = 200/(13/3) = (200*3)/13 = (600/13) km/hrNow speed increased by 18km/hr

First convert time into hourst = 4 + 20/60 = 4 + 1/3 = (12 + 1)/3 = 13/3 hrsThen calculate Speed(v1)v1 = d/t = 200/(13/3) = (200*3)/13 = (600/13) km/hrNow speed increased by 18km/hrv2 = v1 + 18 = (600/13) + 18 = (600 + 234)/13 = (834/13) km/hr

First convert time into hourst = 4 + 20/60 = 4 + 1/3 = (12 + 1)/3 = 13/3 hrsThen calculate Speed(v1)v1 = d/t = 200/(13/3) = (200*3)/13 = (600/13) km/hrNow speed increased by 18km/hrv2 = v1 + 18 = (600/13) + 18 = (600 + 234)/13 = (834/13) km/hrSo Time(t2) = d/v2

First convert time into hourst = 4 + 20/60 = 4 + 1/3 = (12 + 1)/3 = 13/3 hrsThen calculate Speed(v1)v1 = d/t = 200/(13/3) = (200*3)/13 = (600/13) km/hrNow speed increased by 18km/hrv2 = v1 + 18 = (600/13) + 18 = (600 + 234)/13 = (834/13) km/hrSo Time(t2) = d/v2t2 = 200 / (834/13) = (200 * 13) / 834 = 2600/834 hr

First convert time into hourst = 4 + 20/60 = 4 + 1/3 = (12 + 1)/3 = 13/3 hrsThen calculate Speed(v1)v1 = d/t = 200/(13/3) = (200*3)/13 = (600/13) km/hrNow speed increased by 18km/hrv2 = v1 + 18 = (600/13) + 18 = (600 + 234)/13 = (834/13) km/hrSo Time(t2) = d/v2t2 = 200 / (834/13) = (200 * 13) / 834 = 2600/834 hrt2 = 3.117 hrs or 3hr 7min 1 sec

First convert time into hourst = 4 + 20/60 = 4 + 1/3 = (12 + 1)/3 = 13/3 hrsThen calculate Speed(v1)v1 = d/t = 200/(13/3) = (200*3)/13 = (600/13) km/hrNow speed increased by 18km/hrv2 = v1 + 18 = (600/13) + 18 = (600 + 234)/13 = (834/13) km/hrSo Time(t2) = d/v2t2 = 200 / (834/13) = (200 * 13) / 834 = 2600/834 hrt2 = 3.117 hrs or 3hr 7min 1 secAns: Time = 3.117 hrs or 3hr 7min 1 sec