A man riding a bicycle form his house at 10km/h and reaches his office late by 6 min. He increases his speed by 2km/h and reaches 6 min before. How far is the office form his house?

Let distance and original speed of the man be d km and s km/h.
Then,
d/s - d/s + 3 = 2/3
? [d(s + 3 - s)]/[s(s + 3)] = 2/3
? 9d = 2s(s + 3) ...(i)

And d/(s - 2) - d/s = 2/3
? [d(s - s + 2)]/[s(s - 2)] = 2/3
? 3d = s(s - 2) ...(ii)

From Eqs.(i) and (ii), we get
3s(s - 2) = 2s(s + 3)
? 3s² - 6s = 2s² + 6s
? s² = 12s
? s = 12

From Eq.(ii), we get
3d = 12(12 - 2)
? d = 40 km

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
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]
= 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
? 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 B catches A after T h.
Then, distance travelled by A in (T + 2) h = Distance travelled by B in T h

According to the question,
2 (T + 2 ) = 5T
? 2T + 4 = 5T
? 3T = 4
? T = 4/3 h

Distance travelled by B in 4/3 h
= (4/3) x 5 = 20/3 = 6²/₃ km

Let total distance be 100 km.
Then, Average = 100/[ (50/40) + (30/60) + (20/30)]
= 100/( 5/4 + 1/2 + 2/3) = 100/[(15 + 6 + 8)/12]
= (100 x 12)/29
= 42.35

Distance of his office from his house = {ab/(b - a)} x (t₁ + t₂)
Here, a = 20 km/h, t₁ = 1 h, b = 20 + 5 = 25 km/h
and t₂ = 30 min = 0.5 h
? Distance = {20 x 25/(25 - 20)} x (1 + 0.5) = 500(1.5)/5
= 100 x 1.5 = 150 km

Let the distance = x
Here, the difference in time
= 20 - 5 = 15 min
= 15/60 = 1/4h
Speed during next journey
= 15 + 5 = 20 km/h
According to the question,
x/15 - x/20 = 1/4
? 4x - 3x/60 = 1/4
? x = 60/4 = 15
? x = 15 km

Let distance covered by dog in 1 leap is x and
Distance covered by cat in 1 leap is y

Then, 3x = 4y
? x= 4/3y

Now, Ratio of speed of dog and cat = Ratio of distance covered by them in the same time = 4x : 5y
= 16/3y : 5y
= 16 : 15