Ram and Shyam run a race between points A and B, 5 km apart. Ram start

But the question says overtake, not meet. Therefore, it should be after 10.15

Ram reaches B at 10 am (Trip time 1 hour: 5 km at 5 kph). At this point, Shyam has traveled for 15 minutes (0.25 hrs) covering 2.5 kms (10 kph*0.25 hrs)
so the distance separating them is 2.5 kms (5-2.5). Thereafter, Ram turns back and runs towards Shyam so their relative speed is 15 kph (5+10). Thus the time they will take to meet is:
2.5/15 = (1/6) hr = 10 mins. So they will meet at 10:10 am. ANS:B

Explanation:

Ram reached B by 10 am
By then shyam reach 2.5 km (10*15/60) From A towards B
Distance between them is now 2.5 km
Relative speed 15km /hr (10+5 =15, As coming towards each other)
So, They Will first Meet at = [(2.5/15)*60 ] i.e, 10 mins i.e. 10.10 am
IMO-B

Since it takes Ram 1 hour to make it to B, Ram starts running back from B to A at 10 am.

By 10 am, Shyram has run for 1/4 hour and has run 10 x 1/4 = 10/4 = 5/2 = 2.5 km from A toward B.

So, the distance between them is 5 - 2.5 = 2.5 km.

Let’s create an equation to show the time it takes (counting from 10 am) for them to meet each other:

5t + 10t = 2.5

15t = 2.5

t = 25/150

t = 1/6 hour = 10 minutes. So, they meet at 10:10 AM.

Answer: B

Ram leaves A at 9am.
Since his speed = 5kph, and the distance between A an B = 5k, the time for Ram to travel between A and B = 1 hour.
Implication:
Ram reaches B at 10am and.returns to A at 11am.
The answer choices are all earlier than 11am, indicating that Ram meets Shyam on his return trip to B.

In the figure below, the blue arrows represent Ram, the red arrow represents Shyam, and X represents where the two travelers meet:
A ----------------------->B
A ------------->X<----- B

Let t = Shyam's time.
Since Shyam's rate = 10kph, the red distance traveled by Shyam = 10t
Since Ram leaves 45 minutes earlier than Shyam, his time = t + 3/4
Since Ram's rate = 5kph, the blue distance traveled by Ram = 5(t + 3/4)
Since the sum of the two colored distances = 10k, we get:
10t + 5(t + 3/4) = 10
15t + 15/4 = 10
60t + 15 = 40
60t = 25
t = 25/60 hour = 25 minutes

Since Shyam leaves at 9:45am and travels for 25 minutes, the time at which she meets Ram = 9:45 am + 25 minutes = 10:10am

r=d/t;
To find time where they meet same direction, subtract rates;
To find time where they meet opposite or towards direction, add rates.

R reached B at 5km/5=1hr so at 10:00
S began at 09:45 so at 10:00 traveled 1/4*10=2.5km
Distance between S and R: 5-2.5=2.5km
Time they first meet towards direction: 2.5km/(10+5)=2.5/15*60=10mins
10:00 + 00:10 = 10:10

Ans (B)

­my approach is like the distance travel by ram untill shyam starts is 5*45/60 that is 15/4 so so to cover the distance it would take shyam 45 min to cover as distance between them is 15/4 and to cover that distance i used relative speed formula that is (15/4)/(10-5) answer is 3/4 that is 45 min and then they would meet on 10:30 is which is wrong so where i went wrong could anyone help me fix my approach ?

The relative speed approach is appropriate when elements are traveling in the same direction.
But in this problem, Ram and Shyam do not travel in the same direction for the entire race.
After an hour of travel, Ram arrives at B and begins his return to A.
Implication:
When they meet, Ram is traveling TOWARD A while Shyam is traveling TOWARD B.
Since they are not traveling in the same direction, the relative speed approach is not viable.­