An express train travelled at an average speed of 100 km/h, stopping for 3 min after every 75 km. A local train travelled at a speed of 50 km/h, stopping for 1 min after every 25 km. If the trains began travelling at the same time, how many kilometres did the local train travel in the time in which the express train travel 600 km?

Required number of hours is the number of terms of the series 40 + 45 + 50 +... as speed increases every hour.
Given, sum of the series is 385
? a = 40, d=5, s = 385 and n= ?
Using S = n/2[2a + (n - 1)d]
? 385 = n/2[80 + 5n - 5]
? 770 = 5n² + 75n
? n² + 15n - 154 = 0
? n² + 22n - 7n - 154 = 0
? n(n + 22) - 7(n + 22) = 0
? (n + 22) (n - 7) = 0
? n = 7
Remaining part = 1/12 - 1/20 = (5 - 3)/60 = 2/60 = 1/30 i.e., 1/30 th part is filled by B in 1 min
Hence, required time to fill the whole tank = (165 + 1 + 1) min = 167 min

Let speed of A = 5x
Then, spend of B = 6x
Given that,speed of B = 6x = 90
? x = 90/6 = 15
? Speed of A = 5x = 5x = 5 x 15 = 75 km/h
Let A and B meet after T h. Then,
75 x T + 90 x T =88
? T = 88/165 h = (88 x 60)/165 = 32 min

Let two trains meet at a km from 'X'
[Time taken by M to cover (450 - a)km] - [Time taken by the L to cover a km ] = 40/60
? (450 - a)/80 - a/60 = 40/60
? (450 - a)/80 = 40/60 + a/60
? (450 - a)/8 - (a + 40) / 6 = 0
? 3(450 - a) - 4(a + 40) = 0
? 7a = 1190
? a = 1190 / 7 = 170
Time taken by L to cover 170 km = 170/60 h = 2 h 50 min
So, the two trains will meet 2 h 50 min after 6:00 pm. it means that the two trains will meet at 8 : 50 pm .

Required average speed = 4/(1/60 + 1/40 + 1/120 +1/20)
= (4 x 120)/12 = 40 km/h

Here, S₁ = 8 km/h
and S₂ = 16 km/h
? d₁ =s₁ x t₁ = 8t₁ ...(i)
and d₂ = s₂ x t₂ = 16t₂ ...(ii)
We know that,
t₁ + t₂ = 8 ...(iii)
and
d₁ + d₂ = 80 (given) ...(iv)
From Eqs.(i) and (ii) put the value of d₁ and d₂ in Eq. (iv), we get
d₁ + d₂ = 80
8t₁ + 16t₂ = 80
? 8t₁ + 8t₂ + 8t₂ = 80
8(t₁ + t₂) + 8t₂ = 80
From Eq (iii),
8 x 8 + 8t₂ = 80

8t₂ = 80 - 64 = 16
? t₂ = 16/8 = 2h
? t₁ = 8 - 2 = 6 h
? Distance travelled by foot = d₁ = 8 x 6 = 48 km.

Let the distance be x km.
? Time = Distance/speed
1st day time taken = x/4 h
2nd day time taken = x/6 h
Difference in time = (x/4 - x/6) h
Actual difference between these two times = 15 min = 1/4 h
? x/4 - x/6 = 1/4 ? (3x - 2x)/12 = 1/4
? x = 3 km

Relative speed = (58 - 32) = 26 km/h
= (26 x 5/18) m/s = 65/9 m/s
Time taken to cross each other = (250 + 140)/(65/9)
= (390 x 9)/65 = 54 s

Let T h, he travels on foot.
? 4T + 10(8 - T) = 50
? 80 - 6T = 50
? 6T = 30,
? T = 5 h
? The distance travelled on foot = 4 x 5 = 20 km

Walking time + Cycling time = 5 h 45 min = 345 min ....(i)
If he had cycled both way he would have gained 2 h (120 min).
? 2 x Cycling time = 345 - 120 = 225 min ...(ii)
Walking time = 2 x 345 - 225 = 690 - 225 = 465 min
Time taken by him to walk both ways = 7 h 45 min

Let w be the time taken in one way by walking and c be the time taken in one way by car.
Now, according to the question,
In first case, w + c = 6 h 45 min
or
2w + 2c = 13 h 30 min ..(i)
In second case, 2c = 4 h 45 min ....(ii)
From Eqs. (i) and (ii), we get
2w + 2c = 13 h 30 min
? 2w + 4h 45 min = 13 h 30 min
? 2w = 13 h 30 min - 4 h 45 min
? 2w = 8 h 45 min
? If he walks both ways, then time taken = 8 h 45 min