Quantitative Aptitude

General Questions

Quantitative Aptitude Exercise Mode

General Questions

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Question 1
From a point P on a level ground, the angle of elevation of the top tower is 30º. If the tower is 100 m high, the distance of point P from the foot of the tower is:

A
149 m
B
156 m
C
173 m
D
200 m
Correct Answer: Option C

Let AB be the tower.

Then, APB = 30º and AB = 100 m.

AB = tan 30º = 1
AP 3

AP = (AB x 3) m
= 1003 m
= (100 x 1.73) m
= 173 m.

Question 2
The angle of elevation of the sun, when the length of the shadow of a tree 3 times the height of the tree, is:

A
30º
B
45º
C
60º
D
90º
Correct Answer: Option A

Let AB be the tree and AC be its shadow.

Let ACB = .

Then, AC = 3         cot = 3
AB

= 30º.

Question 3
A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30º with the man's eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 60º. What is the distance between the base of the tower and the point P?

A
(4 * 1.732)units
B
8 units
C
12 units
D
Data inadequate
E
None of these
Correct Answer: Option D

One of AB, AD and CD must have given.

So, the data is inadequate.

Question 4
Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is:

A
173 m
B
200 m
C
273 m
D
300 m
Correct Answer: Option C

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 3

      AC = AB x 3 = 1003 m.

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

CD = (AC + AD) = (1003 + 100) m
= 100(3 + 1)
= (100 x 2.73) m
= 273 m.

Question 5
The angle of elevation of a ladder leaning against a wall is 60º and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is:

A
2.3 m
B
4.6 m
C
7.8 m
D
9.2 m
Correct Answer: Option D

Let AB be the wall and BC be the ladder.

Then, ACB = 60º and AC = 4.6 m.

AC = cos 60º = 1
BC 2

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

Question 6
An observer 1.6 m tall is 203 away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The heights of the tower is:

A
21.6 m
B
23.2 m
C
24.72 m
D
None of these
Correct Answer: Option A

Let AB be the observer and CD be the tower.

Draw BE CD.

Then, CE = AB = 1.6 m,

      BE = AC = 203 m.

DE = tan 30º = 1
BE 3

DE = 203 m = 20 m.
3

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

Question 7
On the two sides of a road are two tall buildings exactly opposite to each other. The height of the taller building is 60 m. If the angle of elevation from the top of the smaller building to the top of the taller one is 30° and the angle of depression from top of the taller building to the foot of the smaller one is 30°, then find the height of the smaller building.

A
45
B
30
C
20
D
40
Correct Answer: Option D


Let AB be the taller building of height 60 m and CD be the smaller one of height h m.

 DB / AB = tan 30º = 1/ 3

 DB = AB x tan 30º

= 34.64 m.

tan 30º = AE/CE = AE/DB = 1/ = AE/34.64

AE = 60 – h = 20

h = 40 m

Question 8
A ship is approaching an observation tower. If the time taken by the ship to change the angle of elevation from 30° to 45° is 10 minutes, then find the time the ship will take to cover the remaining distance and reach the observation tower assuming the ship to be travelling at a uniform speed.

A
15 minutes 20 seconds
B
13 minutes 40 seconds
C
16 minutes 40 seconds
D
Cannot be determined
Correct Answer: Option


Let AB be the observation tower and h be its height.

Also, let the ship be at C when the angle of elevation is 30° and at D when the angle of elevation is 45°.

The time taken by the ship to travel from C to D is 10 minutes and we need to find out the time the ship will take to reach B from D.

 tan 30º = AB/CB = h / CB = 1/ 3

 CB = 3 x h

tan 45º = AB / DB = h / DB = 1

 DB = h

 CD = CB – DB = (3h – h) = h(3– 1)

Now, as h(3 – 1) distance is covered in 10 minutes, a distance of h is covered in = 13.66 minutes = 13 minutes 40 seconds

Question 9
A man standing on the terrace of a building watches a car speeding towards him. If at that particular instant the car is 200 m away from the building makes an angle of depression of 60° with the man’s eye and after 8 seconds the angle of depression is 30°, what is the speed of the car?

A
15 m/s
B
25 m/s
C
16.67 m/s
D
Cannot be determined
Correct Answer: Option C

Let AB be the building and the man is standing at A.

When the car is 200m away from the building, the angle of depression is 60°

tan 60° = BD/AB = 3

 200/AB = 3

 AB = 115.47 m

 tan30° =BC/AB = x/115.47 = 1/ 3

 x = 66.67

Now, the car travels distance CD in 8 seconds

CD = BD – BC = 200 – 66.67 = 133.33 m

Speed = 133.33/8 = 16.67 m/s

Question 10
Two boats are spotted on the two sides of a light house. If the angle of depression made by both the boats from top of the lighthouse is 30° and 45° and the height of the light house is 125 m then find the distance between the two boats.

A
188.56 m
B
1 m
C
197.17 m
D
250 m
Correct Answer: Option C

Let AB be the light house and the two boats be at C and D

AB = 125 m

tan30° = BC/AB = x/125 = 1/3

=> x = 72.17 m


tan45° = BD/AB = y/125 = 1

=> y = 125 m


 The distance between the two boats is = x + y

= 72.17 + 125

= 197.17 m

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