Two trains A and B starts from Howrah and Patna towards Patna and Howrah respectively at the same time. After passing each other, they take 4, h, 48 min and 3 h, 20 min to reach Patna and Howrah, respectively. If the train from Howrah is moving at 45 km/h, then the speed of the other train is?

Speed of the person = 4 km/h = 4.5 x (5/18) = m/s = 5/4 m/s = 1.25 m/s
Speed of 2nd person = 5.4 km/h = 5.4 x (5/18) m/s = 3/2 m/s = 1.5 m/s

Let speed of train be x m/s.
Then, (x - 1.25) x 8.4 = (x - 1.5) x 8.5
? 8.4x - 10.5 = 8.5x - 12.75
? 0.1x = 2.25
? x = 22.5
? Speed of the train = 225 x (18/5 ) = 81 km/h .

Let speed in the return journey = V
? Speed in onward journey = (125/100) x V = 5V/4 km/h
Average speed = (2x 5V/4 x V) / (5V/4 + V) = 10V/9 km/h
? 1600 x (9/10V) = 30 [? halts for 2 h ]
? V = 1600 x 9/30 x 10 = 48 km/h
? Speed in onward journey = 5V/4 = 5/4 x 48 = 60 km/h

Let the distance between Mokama and Hazipur be D km
Time taken by train P to cover D km = 1 h
Time taken by Q to cover D km = 3/2 h
? Speed of train P = D km/h
Speed of train Q = 2D/3 km/h .
Let they cross each other T h after 5 : 00 pm.
Then, DT + 2DT/ 3 = D
? T + 2T/3 = 1
? T(1 + 2/3) = 1
? T = 3/5h = (3/5) x 60 = 3 x 12 = 36 min.

Hence, the two trains meet 36 min after 5 : 00 pm.
? Meeting time = 5:36 pm

Basically they will exchange their speeds just after half of the time required for the whole journey. It means after covering 210 km distance they will exchange their speeds. check it out graphically for more clarification.

A's speed / B's speed = ?(Time taken by B to reach X) / (Time taken by A to reach Y)
? 45/B's speed = ?(10/3) x (5/24) = 5/6
? B's speed = 45 x (6/5) km/hr = 54 km/hr.

Let the length of slower train be L meters and the
length of faster train be (L/2) meters
Their relative speed = (36 + 54) km/hr
= 90 x (5/18)
= 25 m/sec

? 3L / (2 x 25) = 12
? 3L = 600
? L = 200
? Length of slower train = 200 meters
Let the length of platform by M meters
Then, 200 + M/[36 x (5/18)] = 90 sec.
? 200 + M = 900
? M = 700 meters
Length of platform = 700 meters.

Consider the diagram is shown above where PR represents the ladder and RQ represent the wall.
Cos 60° = PQ / PR
? 1 / 2 = 12.4 / PR
? PR = 2 × 12.4 = 24.8 m

Let us draw a figure below as per given question.
Let A be the position of a person on the bank of a river and OP the tree on the opposite bank and ?OAP = 60°. When the person retires to the position B, then AB = 40 meter and ?OBP = 30°
Let us assume OA(Breadth of the river) = x meter and height of tree OP = h meter
In ?OAP, Use the trigonometry formula
Tan60° = P/B = Perpendicular distance / Base distance
? Tan60° = OP / OA
? OP = OA Tan60°
Put the value of OP and OA, We will get
? h = x?3 ..............(1)
Now in the triangle ?OBP
Tan30° = OP / OB
? OP = OB Tan30°
? OP = (x + 40)/?3
? h = (x + 40)/?3 ...................(2)
From Equation (1) and (2), We will get
? (x + 40)/?3 = x?3
? (x + 40) = x?3 X ?3
? (x + 40) = 3x
? 3x - x = 40
? x = 20 m

Let us draw the figure from the given question.
Let AB = h meter be the height of the tower; C and D are the two points on the ground such that BC = 60 m; ?ACB = 45° and ?ADB = 30°
Now from right triangle ABC,
tan 45° = h/60
? 1 = h/60
? h = 60 m;
Again from right triangle ABD;
tan 30° = h/(x + 60)
? 1/?3 = 60/(x + 60)
? x + 60 = 60?3
? x = 60(1.73 - 1) = 43.8 meter
Hence, speed of boat = 43.8/5 m/s = 43.8/5 x 18/5 = 31.5 km/hr.

Let us draw a figure from given question.
Let, AB = 100 m be the height of a tower, P is a point on the ground such that ∠APB = 30°
From right triangle ABP,
BP = 100 cot 30°
⇒ BP = 100 √3
⇒ BP = 100 X 1.73
⇒ BP = 173 meter