A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30° with the man's eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 60°. What is the distance between the base of the tower and the point P?

Correct Answer: 81 km/hr

Explanation:

4.5 km/hr =	❨	4.5 x	5	❩	m/sec =	5	m/sec = 1.25 m/sec, and
			18			4

5.4 km/hr =	❨	5.4 x	5	❩	m/sec =	3	m/sec = 1.5 m/sec.
			18			2

Let the speed of the train be x m/sec.

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 =	❨	22.5 x	18	❩	km/hr = 81 km/hr.
			5

Correct Answer: 82 km/hr

Explanation:

Let the speed of the second train be x km/hr.

Relative speed	= (x + 50) km/hr
	= [ (x + 50) x 5 ] m/sec 18
	= [ 250 + 5x ] m/sec. 18

Distance covered = (108 + 112) = 220 m.

∴	220	= 6
	❨ 250 + 5x ❩ 18

⟹ 250 + 5x = 660

⟹ x = 82 km/hr.

Correct Answer: 60 km/hr

Explanation:

Let the speed of the slower train be x m/sec.

Then, speed of the faster train = 2x m/sec.

Relative speed = (x + 2x) m/sec = 3x m/sec.

∴	(100 + 100)	= 3x
	8

⟹ 24x = 200

⟹ x =	25	.
	3

So, speed of the faster train =	50	m/sec
	3

=	❨	50	x	18	❩km/hr
		3		5

   = 60 km/hr.

Correct Answer: 24 sec

Explanation:

Relative speed =	= (45 + 30) km/hr
	= ❨ 75 x 5 ❩ m/sec 18
	= ❨ 125 ❩ m/sec. 6

We have to find the time taken by the slower train to pass the DRIVER of the faster train and not the complete train.

So, distance covered = Length of the slower train.

Therefore, Distance covered = 500 m.

∴ Required time =	❨	500 x	6	❩	= 24 sec.
			125

Correct Answer: 50 m

Explanation:

2 kmph =	❨	2 x	5	❩	m/sec =	5	m/sec.
			18			9

4 kmph =	❨	4 x	5	❩	m/sec =	10	m/sec.
			18			9

Let the length of the train be x metres and its speed by y m/sec.

Then,	❨	x	❩	= 9 and	❨	x	❩	= 10.
		y - 5 9				y - 10 9

∴ 9y - 5 = x and 10(9y - 10) = 9x

⟹ 9y - x = 5 and 90y - 9x = 100.

On solving, we get: x = 50.

∴ Length of the train is 50 m.

Correct Answer: 9.2 m

Explanation:

Let AB be the wall and BC be the ladder.

Then, ∠ACB = 60° and AC = 4.6 m.

AC	= cos 60° =	1
BC		2

⟹ BC	= 2 x AC
	= (2 x 4.6) m
	= 9.2 m.

Correct Answer: 21.6 m

Explanation:

Let AB be the observer and CD be the tower.

Draw BE ⊥ CD.

Then, CE = AB = 1.6 m,

      BE = AC = 20√3 m.

DE	= tan 30° =	1
BE		√3

⟹ DE =	20√3	m = 20 m.
	√3

∴ CD = CE + DE = (1.6 + 20) m = 21.6 m.

Correct Answer: 173 m

Explanation:

Let AB be the tower.

Then, ∠APB = 30° and AB = 100 m.

AB	= tan 30° =	1
AP		√3

⟹ AP	= (AB x √3) m
	= 100√3 m
	= (100 x 1.73) m
	= 173 m.

Correct Answer: 30°

Explanation:

Let AB be the tree and AC be its shadow.

Let ∠ACB = Θ.

Then,	AC	=	√3     ⟹     cot Θ = √3
	AB

∴ Θ = 30°.

Correct Answer: 273 m

Explanation:

Let AB be the lighthouse and C and D be the positions of the ships.

Then, AB = 100 m, ∠ACB = 30° and ∠ADB = 45°.

AB	= tan 30° =	1	⟹     AC = AB x √3 = 100√3 m.
AC		√3

AB	= tan 45° = 1     ⟹     AD = AB = 100 m.
AD

∴ CD = (AC + AD)	= (100√3 + 100) m
	= 100(√3 + 1)
	= (100 x 2.73) m
	= 273 m.