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A start walking from a place at a uniform speed of 2 kmph
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24 Aug 2013, 22:54
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A start walking from a place at a uniform speed of 2 kmph in a particular direction. After half an hour, B starts from the same place and walks in the same direction as A at a uniform speed and overtakes A after 1 hour 48 minutes. Find the speed of B. A) 4.7 kmph B) 2.6 kmph C) 4 kmph D) 7 kmph E) 5.3 kmph
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Re: A start walking from a place at a uniform speed of 2 kmph
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24 Aug 2013, 23:41
rrsnathan wrote: A start walking from a place at a uniform speed of 2 kmph in a particular direction. After half an hour, B starts from the same place and walks in the same direction as A at a uniform speed and overtakes A after 1 hour 48 minutes. Find the speed of B.
A) 4.7 kmph B) 2.6 kmph C) 4 kmph D) 7 kmph E) 5.3 kmph can u please let me know the calc  i got 2(t+30) = 108x. can u correct me here



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Re: A start walking from a place at a uniform speed of 2 kmph
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24 Aug 2013, 23:53
rrsnathan wrote: A start walking from a place at a uniform speed of 2 kmph in a particular direction. After half an hour, B starts from the same place and walks in the same direction as A at a uniform speed and overtakes A after 1 hour 48 minutes. Find the speed of B. A) 4.7 kmph B) 2.6 kmph C) 4 kmph D) 7 kmph E) 5.3 kmph Distance covered by A in 30 min = 1 Km B covers extra 1km in 1 hour 48 minutes (9/5 hr) i.e. Relative speed of B over A = 1/(9/5) = 5/9 So the speed of B = Speed of A + 5/9 = 2 + 5/9 = 2.55 Answer B
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Re: A start walking from a place at a uniform speed of 2 kmph
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25 Aug 2013, 00:27
clematischestnut wrote: rrsnathan wrote: A start walking from a place at a uniform speed of 2 kmph in a particular direction. After half an hour, B starts from the same place and walks in the same direction as A at a uniform speed and overtakes A after 1 hour 48 minutes. Find the speed of B.
A) 4.7 kmph B) 2.6 kmph C) 4 kmph D) 7 kmph E) 5.3 kmph can u please let me know the calc  i got 2(t+30) = 108x. can u correct me here 1 Distance traveled by both is the same ie., 2(t+0.5) = tx 2. we know t=1.8 hrs, therefore t +0.5= 2.3 3 substituting the above values in (1), 2*2.3= 1.8*x 4. x=2.6 km/h
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Re: A start walking from a place at a uniform speed of 2 kmph
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25 Aug 2013, 05:35
A overcame 1km+2km+0.8*2km=4.6km (if its speed is 2kmph, then in half an hour A overcame 1km, in an hour and 48 mins (2+1.6)km) so, B's speed =4.6 km/1.8h=2.56 kmph
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Re: A start walking from a place at a uniform speed of 2 kmph
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09 Sep 2013, 22:14
A start walking from a place at a uniform speed of 2 kmph in a particular direction. After half an hour, B starts from the same place and walks in the same direction as A at a uniform speed and overtakes A after 1 hour 48 minutes. Find the speed of B. A) 4.7 kmph B) 2.6 kmph C) 4 kmph D) 7 kmph E) 5.3 kmph
In half an hour, A will travel 1 km. B catches A in 1 hour 48 minutes, which means A traveled for another 1 hr 48 min in addition to 30 min that he had traveled before B started his journey. So, A travels for a total of 2 hours 18 min. For A: 1km (30 min) + 2km (1hour) + {(4/5)*2}km in 48 min (48 min are 4/5th of an hour) = 1+2+(8/5) = 23/5 km
B travels for 1 hr 48 min. And he covers 23/5 km in that time. You have the time and the distance. Let's use Rxt=D formula for a quick calculation. 1 hr 48 min can be written as 1 4/5 hours or 9/5. So B's Rate = (23/5)/(9/5) = 23/5 = 2. something. Only option B has a 2. something. So B is the correct answer.
This can be done fairly quickly of course, but had to be a long post to explain properly.
Another Method: Quick guessing  A travels for roughly 2 hours at a rate of 2 km per hour for a total of 4 km. B covers 4 km in roughly 2 hours, so his rate has to be close to 2 kmh. None of the answers except B are close to 2. If some were, we could have looked a little deeper to see that A traveled for slightly more than 2 hours, therefore slightly more than 4km. B traveled a little less than 2 hours. So the ratio will be slightly more than 2. Hence B.



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Re: A start walking from a place at a uniform speed of 2 kmph
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13 Jan 2014, 11:57
rrsnathan wrote: A start walking from a place at a uniform speed of 2 kmph in a particular direction. After half an hour, B starts from the same place and walks in the same direction as A at a uniform speed and overtakes A after 1 hour 48 minutes. Find the speed of B.
A) 4.7 kmph B) 2.6 kmph C) 4 kmph D) 7 kmph E) 5.3 kmph I did it in a similar way. Let A and B for the speeds of each body A traveled 1 km in 2 hours as per its rate of 2kmph For B to reach A in 1hr48m or (9/5hr) then (B2)(9/5)=1 B = 2.6 Hence B Hope it helps Cheers! J



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Re: A start walking from a place at a uniform speed of 2 kmph
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14 Jan 2014, 00:03
Half hour= 1/2 hour. 1 hour 48 minutes=9/5 Since b overtake A by 1 hour 48 minutes after A has been waalking in 30 minutes. Therefore, Distance of A : (1/2+9/5)2=4.6 Speed of B would be 4.6/1.8=2.55555=2.6



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Re: A start walking from a place at a uniform speed of 2 kmph
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07 Feb 2014, 20:59
taransambi wrote: A start walking from a place at a uniform speed of 2 kmph in a particular direction. After half an hour, B starts from the same place and walks in the same direction as A at a uniform speed and overtakes A after 1 hour 48 minutes. Find the speed of B. A) 4.7 kmph B) 2.6 kmph C) 4 kmph D) 7 kmph E) 5.3 kmph
In half an hour, A will travel 1 km. B catches A in 1 hour 48 minutes, which means A traveled for another 1 hr 48 min in addition to 30 min that he had traveled before B started his journey. So, A travels for a total of 2 hours 18 min. For A: 1km (30 min) + 2km (1hour) + {(4/5)*2}km in 48 min (48 min are 4/5th of an hour) = 1+2+(8/5) = 23/5 km
B travels for 1 hr 48 min. And he covers 23/5 km in that time. You have the time and the distance. Let's use Rxt=D formula for a quick calculation. 1 hr 48 min can be written as 1 4/5 hours or 9/5. So B's Rate = (23/5)/(9/5) = 23/5 = 2. something. Only option B has a 2. something. So B is the correct answer.
This can be done fairly quickly of course, but had to be a long post to explain properly.
Another Method: Quick guessing  A travels for roughly 2 hours at a rate of 2 km per hour for a total of 4 km. B covers 4 km in roughly 2 hours, so his rate has to be close to 2 kmh. None of the answers except B are close to 2. If some were, we could have looked a little deeper to see that A traveled for slightly more than 2 hours, therefore slightly more than 4km. B traveled a little less than 2 hours. So the ratio will be slightly more than 2. Hence B. B's rate would be 23/9 not 23/5. Your explanation is simple and neat. +1



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A start walking from a place at a uniform speed of 2 kmph
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29 Aug 2015, 16:31
B's rate/A's rate = A's time/B's time B/2=2.3/1.8 B=2.6



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Re: A start walking from a place at a uniform speed of 2 kmph
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04 Aug 2018, 06:46
A has walked for around 2hr 20mins when B overtakes A (1hr 48mins + 30mins ~ 2hr 20 mins) *we can round 1hr 48mins to 1hr 50mins. 2hr 20min * 2kmph = 4.67km B's speed = 4.67/1hr 50mins = ~2.5kmph, only possible answer is option B, the rest are eliminated.
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Re: A start walking from a place at a uniform speed of 2 kmph &nbs
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