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24 Dec 2025, 20:32
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M & D both leave at 09.00 AM
Dog reaches at the park at 09.30 AM, i.e. distance of park = 30 mins speed of dog
Dog plays 20 mins and comes back to meet M at 10.00 AM, i.e. it started at 09.50 AM from the park and reached the M in 10 minutes, i.e. 1/3rd of the distance covered by dog and 2/3 covered by M till 10.00 AM
Distance covered by M till 10.00 AM, i.e. in 1 hr = 2/3 of total
Time required to cover full distance = 3 / 2 hrs i.e. he'll reach the park at 10.30 AM
Hence, Answer Option D
Dog reaches at the park at 09.30 AM, i.e. distance of park = 30 mins speed of dog
Dog plays 20 mins and comes back to meet M at 10.00 AM, i.e. it started at 09.50 AM from the park and reached the M in 10 minutes, i.e. 1/3rd of the distance covered by dog and 2/3 covered by M till 10.00 AM
Distance covered by M till 10.00 AM, i.e. in 1 hr = 2/3 of total
Time required to cover full distance = 3 / 2 hrs i.e. he'll reach the park at 10.30 AM
Hence, Answer Option D
24 Dec 2025, 20:57
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Time taken by dog tor each park = 30min
Let Vd be the velocity of dog
Then distance to park D = Vd*time = 0.5Vd
Time travelled by man and dog to meet at 10am is 10 min = 1/6hr
Distance Dog travelled in 1/6 hr = Vd/6
Since the distance dog travelled back + distance man travlled towards the park should sum to D, man travelled Vd/2 - Vd/6
Man travelled Vd/2 - Vd/6 in 1hour. So Velocity of man Vm= Vd/3
The man has to travel Vd/6 more to reach the park.
Time taken to travel Vd/6 distance = (Vd/6)/(Vd/3)
= 0.5 hour. So he will reach park at 10:30am
Let Vd be the velocity of dog
Then distance to park D = Vd*time = 0.5Vd
Time travelled by man and dog to meet at 10am is 10 min = 1/6hr
Distance Dog travelled in 1/6 hr = Vd/6
Since the distance dog travelled back + distance man travlled towards the park should sum to D, man travelled Vd/2 - Vd/6
Man travelled Vd/2 - Vd/6 in 1hour. So Velocity of man Vm= Vd/3
The man has to travel Vd/6 more to reach the park.
Time taken to travel Vd/6 distance = (Vd/6)/(Vd/3)
= 0.5 hour. So he will reach park at 10:30am
Bunuel
24 Dec 2025, 21:51
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Let, man speed= x & dog speed=y
Distance home to park= z
Dog reach park in 1/2 hour, z=y/2
Dog play for 20 minutes then runs back in 10 minutes or 1/6 hour
Distance dog run back is y/6
At 10 am meeting point is z-y/6
Man walks for 9-10am, 1 hour
Let that distance is x
Distance since they met: x=z-y/6 =(y/2)-(y/6) =y/3
y=3x....
z=y/2=3x/2
Time taken by man= z/x = (3x/2)/x =3/2 hour =1 hour 30 minutes
Till 10 he already walked 1 hour, he will need 30 minutes more to reach park. He will reach at 10:30 am.
D
Distance home to park= z
Dog reach park in 1/2 hour, z=y/2
Dog play for 20 minutes then runs back in 10 minutes or 1/6 hour
Distance dog run back is y/6
At 10 am meeting point is z-y/6
Man walks for 9-10am, 1 hour
Let that distance is x
Distance since they met: x=z-y/6 =(y/2)-(y/6) =y/3
y=3x....
z=y/2=3x/2
Time taken by man= z/x = (3x/2)/x =3/2 hour =1 hour 30 minutes
Till 10 he already walked 1 hour, he will need 30 minutes more to reach park. He will reach at 10:30 am.
D
