Let the certain distance be d and time t.
Now, by given condition,
d/3 = ( t + 15 ) min ( t +15 )/60 h and
d/4 = (t - 15) min = ( t - 15 )/60 h

Correct Option: B

Let the certain distance be d and time t.
Now, by given condition,
d/3 = ( t + 15 ) min ( t +15 )/60 h
⇒ 20d = t + 15
⇒ t = 20d - 15 ......(i)

And from another condition,
d/4 = (t - 15) min = ( t - 15 )/60 h
⇒ 15d = t - 15
⇒ t = 15d + 15 ...(ii)

From Eqs. (i) and (ii), we get
20d - 15 = 15d + 15
⇒ 5d = 30
∴ d = 6 km.

Relative speed = 21 - 15 = 6 m per minute.
Relative distance = 114 m

Correct Option: A

Relative speed = 21 - 15 = 6 m per minute.
Relative distance = 114 m
Time taken to catche = 144/6 min = 19 min

Required average speed = Total distance covered/Total time taken
= (3 x 39)/[39/26 + 39/39 +39/52]

Correct Option: D

Required average speed = Total distance covered/Total time taken
= (3 x 39)/[39/26 + 39/39 +39/52]
= 3 x 13/(1/2 + 1/3 + 1/4)
= 3 x 13 x 12/13 = 36 km/h

Let they meet after T h.
then, according to the question, 3T + 4T = 17.5

Correct Option: A

Let they meet after T h.
then, according to the question, 3T + 4T = 17.5
⇒ 7T = 17.5
∴ T = 17.5/7 = 2.5 h
So, they meet 2.5 h after 8 : 00 am.
It means they meet at 10 : 30 am.

Let usual speed of a man is V and time is t h.
By the formula, V₁t₁ = V₂t₂
Vt = 1¹/₂ V(t - 20/60)
⇒ t = 3/2(t - 1/3)

Correct Option: C

Let usual speed of a man is V and time is t h.
By the formula, V₁t₁ = V₂t₂
Vt = 1¹/₂ V(t - 20/60)
⇒ t = 3/2(t - 1/3)
⇒ 3/2 t - t = 1/2
∴ t = 1 h
Hence, the time taken by the man to reach his office at his usual speed is 1 h.