A start walking from a place at a uniform speed of 2 kmph

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

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

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.

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

(B-2)(9/5)=1

B = 2.6

Hence B

Hope it helps
Cheers!
J

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

B's rate would be 23/9 not 23/5.

Your explanation is simple and neat. +1

B's rate/A's rate = A's time/B's time
B/2=2.3/1.8
B=2.6

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.

Given,
Speed of A - 2 KM/H
Time of A : 1 hour 48 Minutes = 108 Minutes

Time of B : 1 hour 8 minutes =78 Minutes (since B started walking after 30 minutes A started)
Distance for both A & B is same

We need to find the speed of B

Now, moving into formula:

2/60*108 = Speed of B*78
Speed of B =216/60*78
Speed of B = 0.05 KM/Minutes = 2.77 KM/ Hour (after multiplying by 60 ; since 60 minutes = 1 hour)

B is the answer

Why doesn't this work with a Gap approach?

The gap at the beginning is 1 km and after 1 h 48 min (1.80 h) it is zero
So he closes the 1km gap with a rate of 1 / 1.80

A travels with a rate of 2 / 1 per hour, which means he travels 3.60 / 1.80
Now given that the gap is shrinking B has to go with 4.60 / 1.8

Where is my mistake?

/edit: I just realized that I come up with 2.6 as well, I just not did the final division and left the 1.8 in the denominator.
Overall this approach doesn't flow well with this question though.. But I am so used to it that I wanted to fit it here as well

Its like this.
when B starts A is at 2KM distance, now in 1hr 48min = 2+2*(4/5)=3.6 hence total distance by A is 5.6
Now B has to cover 5.6 km in 1hr 48 min => a judgment is we need speed between 2.5 and 3 to get the job done.
only 1 satisfies

Answer B

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