Arithmetic Aptitude :: Height and Distance

Home > Arithmetic Aptitude > Height and Distance > General Questions

NA
SHSTTON
2
Solv. Corr.
6
Solv. In. Corr.
8
Attempted
0 M:0 S
Avg. Time

1 / 17

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 ?

A43 units

B8 units

C12 units

DNone of these

Explanation:

Workspace

NA
SHSTTON
12
Solv. Corr.
11
Solv. In. Corr.
23
Attempted
0 M:44 S
Avg. Time

2 / 17

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

A2.3 m

B4.6 m

C7.8 m

D9.2 m

| | | Asked In nagarro |

Explanation:

Workspace

NA
SHSTTON
7
Solv. Corr.
13
Solv. In. Corr.
20
Attempted
0 M:58 S
Avg. Time

3 / 17

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

A30â€ŽÂ°

B45â€ŽÂ°

C60â€ŽÂ°

D90â€ŽÂ°

| | | Asked In nagarro |

Explanation:

Let AB be the tree and AC be its shadow, So as per question, the shadow of a tree is âˆš3 times the height of the tree.

Lets assume h be the height of the tree.

AB/AC = tanÎ¸

=> h/âˆš3h = tanÎ¸

=> tanÎ¸ = 1/âˆš3

So, Angle = 30âˆ˜

Workspace

NA
SHSTTON
2
Solv. Corr.
5
Solv. In. Corr.
7
Attempted
0 M:0 S
Avg. Time

4 / 17

Choose the correct option.

From height of 8 mts a ball fell down and each time it bounces half the distnace back. What will be the distance travelled

A24

B23

C19

DNone of these

Explanation:

8+4+4+2+2+1+1+0.5+0.5+ and etc .. = 24

Workspace

NA
SHSTTON
6
Solv. Corr.
8
Solv. In. Corr.
14
Attempted
0 M:0 S
Avg. Time

5 / 17

Choose the correct option.

The base of a triangular field is three times its altitude. If the cost of cultivating the field at Rs. 24.68 per hectare be Rs. 333.18, find its base and height.

ABase = 100 m & Height = 200m.

BBase = 900 m & Height = 300m.

CBase = 1900 m & Height = 300m.

DBase = 900 m & Height = 1300m.

Explanation:

Area of the field = Total cost/Rate = (333.18/24.68) hectares =13.5 hectares.
= (13.5*10000) m^2 =135000m^2.
Let altitude = x meters and base = 3x meters.
Then, 1/2 *3x* x= 135000 or x^2 = 9000 or x= 300.
Therefore, base =900 m & altitude = 300m.

Workspace

NA
SHSTTON
2
Solv. Corr.
3
Solv. In. Corr.
5
Attempted
0 M:0 S
Avg. Time

6 / 17

Choose the correct option.

Find the area of a rhombus one side of which measures 20 cm and one diagonal

A32 cm.

B22 cm.

C52 cm.

DNone of these

Explanation:

Lets assume other diagonal = 2x cm,
Since halves of diagonals and one side of rhombus form a right angled triangle with side as hypotenuse, we have:
(20)^2 =(12)^2+x^2 or x=?(20)^2-(12)^2 =?256=16 cm.
So, other diagonal = 32 cm.

Workspace

NA
SHSTTON
4
Solv. Corr.
10
Solv. In. Corr.
14
Attempted
0 M:0 S
Avg. Time

7 / 17

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 degree and 45 degree respectively. If the lighthouse is 100 m high, the distance between the two ships is:

A276 metre

B273 metre

C270 metre

D263 metre

| | | Asked In Infosys |

Explanation:

Workspace

NA
SHSTTON
3
Solv. Corr.
11
Solv. In. Corr.
14
Attempted
0 M:0 S
Avg. Time

8 / 17

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 30Â° and the angle of depression from top of the taller building to the foot of the smaller one is 30Â°, then find the height of the smaller building.

A45 m.

B30 m.

C40 m.

D20 m.

| | | Asked In Infosys |

Explanation:

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

tan30= DB/AB = DB/60 = 1/v3

DB = 34.64 m

tan30= AE/CE = AE/DB = 1/v3 = AE/34.64

AE = 60 - h = 20

h = 40 m

Workspace

NA
SHSTTON
2
Solv. Corr.
6
Solv. In. Corr.
8
Attempted
0 M:0 S
Avg. Time

9 / 17

Satish was looking at a new building Kohli Towers constructed having height of 200 m. The angle of elevation of the top of building from a point on ground is 30Â°. What is the distance of the point from the foot of the building?

A166 m

B254 m

C346 m

D273 m

| | | Asked In Infosys |

Explanation:

Given towers Angle APB = 30Â°. And AB = 200 m
AB/AP = tan 30Â° = 1/âˆš3
AP = (AB x âˆš3) m
200âˆš3 m = 200 x 1.73 = 346m

Workspace

NA
SHSTTON
4
Solv. Corr.
4
Solv. In. Corr.
8
Attempted
0 M:0 S
Avg. Time

10 / 17

Jaya Towers and Karuna towers are two tall buildings standing 400 metres apart. Two fast flying birds were sitting on top of these two towers and were aiming at catching the grains put up in one place. Jaya Tower's height is 300 metres and the grains 160 metres apart from Jaya Towers. Both the birds were flying down at the same speed and reached the grains spot at the same time. What is the approximate height of Karuna Towers?

A200 m

B240 m

C160 m

D180 m

| | | Asked In Infosys |

Explanation:

Jaya Tower is 300 metre tall
Lets assume Karuna Tower's height be x
Distance of grains spot from Jaya Towers -- 160 metres.
Both the birds were flying down at the same speed and reached the grains spot at the same time
i.e Time taken for bird on Jaya tower to reach grain = Time taken for bird on Karuna tower to reach grain
Or Distance travelled by bird on Jaya tower / Speed of bird on Jaya tower = Distance travelled by bird on Karuna tower / Speed of bird on Karuna tower ->(1)
But as it is given that both the birds travelled with same speed, Speed of bird on Jaya tower = Speed of bird on Karuna tower. Applying this to equation (1) we get,
Distance travelled by bird on Jaya tower = Distance travelled by bird on Karuna tower -> (2)
Apply Pythagoras theorem, we know that (Distance travelled by bird on Jaya tower)2 = (300)2 + (160)2 -> (3)
Again by applying Pythagoras theorem to the other triangle. (Distance travelled by bird on Karuna tower)2 = (400-160)2 + x2 -> (4)
From equations 2,3 and 4 we get
(300)2 + (160)2 = (400-160)2 + x2
115600 = 57600 + x2
58000 = x2
240.83 m = x
Therefore x is approximately 240 m

Workspace