Mohan walking at a speed of 20 km/h reaches his office 1 h late. Next time he increases his speed by 5 km/h, then he reaches his office 30 min early. What is the distance of his office from his house?

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

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

Here, a = 10 km/h
b = ( 10 + 2 ) = 12 km/h,
t₁ = 6 min = 1/10 h
and t₂= 1/10 h.
? Required distance = ab( t₁ + t₂ )/(b - a)
= 10 x 12 x ( 1/10 + 1/10 ) /(12 - 10) = (10 x 12 x 1)/(5 x 2) = 12km

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.

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

Let the total distance be y km.
Then, 2y / 5 = 1200
? y = (1200 x 5) /2 = 3000 km.

Distance traveled by car
= (1/3 X 3000) = 1000 km.

Distance traveled by train
= [3000- (1200 + 1000) ] km.
= 800 km.

Relative velocity = u + v = 18 + 20 = 38 km
They are 38 km. apart in 1 hr.
? They will be 95 km. apart in ( 95 / 38) hrs.
= 2 hrs . 30 min

Distance left = (1/2 x 80 ) km. = 40km.
Time left - [(1-3/5) x 10 ] hrs.
= 4 hours

Required speed = (40 / 4 ) km/hr
= 10 km/hr.