A car starts running with the initial speed of 40 km/h with its speed increasing every hour by 5 km/h. How many hours will it taken to cover a distance of 385 km?

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.

x/20 + y/70 = (x + y)/50
? (70x + 20y)/1400 = (x + y)/50
? 42x = 8y
? x/y = 4/21

Time taken by express train to cover 75 km including stoppage
= (60 / 100 x 75) min + = 48 min
Time taken by express train to cover 600 km.
= Time taken by it to cover 75 km
= (48 / 75 x 525 ) min + (60 / 100 x 75 ) min
= (336 + 45) min = 381
Time taken by local train to9 cover 25 km including stoppage
= (60 / 50 x 25) min + 1 min = 31 min
In 31 min, distance covered = 25 km
In(31 x 12) min. distance covered
= (25 / 31 x 31 x 12 ) = 300 km
In last 9 min, distance covered
= (25 / 31 x 9 ) = 7.25 km
? Required total distance
= 300 + 7.25 = 307.25 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