Two train of length 250 m and 140 m are running on parallel lines in the same direction at 58 km/h and 32 km/h, respectively. Find the time taken by the slower train to pass the driver of the faster one?

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

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

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 .

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

Since, the speed of X and Y are 60 km/h each. The distance between A and B is 700 km. Distance traveled by X upto 11 am is 60 km. Since, X stops at 10 am and distance traveled by up to 11 am is 120 km.
Now, the distance between them = 700 - 180 = 520 km
Now, let they will meet at a distance of x km from X's position.
? Distance traveled by X = x km
and distance traveled by Y = (520 - x) km
? x/60 = (520 - x)/60
? 2x = 520
? x = 260
? T = 260/60 = 4¹/₃ h
Thus, they will cross each other after, 4¹/₃ h after 11 am i.e., at 3 : 20 pm.

We know that, if two equal distances are covered at two different speeds A and B, then
Average speed = 2AB/(A + B) = (2 x 58 x 52)/(58 + 52)
= 54.8 km/h
= 55 km/h (Approx)