MCQ Test: Height & Distance - 1 - SSC CGL MCQ

# MCQ Test: Height & Distance - 1 - SSC CGL MCQ

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## 15 Questions MCQ Test Quantitative Aptitude for SSC CGL - MCQ Test: Height & Distance - 1

MCQ Test: Height & Distance - 1 for SSC CGL 2024 is part of Quantitative Aptitude for SSC CGL preparation. The MCQ Test: Height & Distance - 1 questions and answers have been prepared according to the SSC CGL exam syllabus.The MCQ Test: Height & Distance - 1 MCQs are made for SSC CGL 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for MCQ Test: Height & Distance - 1 below.
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MCQ Test: Height & Distance - 1 - Question 1

### A vertical stick 12 cm long casts a shadow 8 cm long on the ground. At the same time, a tower casts a shadow 40 m long on the ground. The height of the tower is

Detailed Solution for MCQ Test: Height & Distance - 1 - Question 1

We can solve it through ratio proportion rule,
let the tower is x m long, then
12 cm stick casts → 8 cm shadow
x m tower casts → 40 m shadow
On cross multiplying, we get

Hence, option B is correct.

MCQ Test: Height & Distance - 1 - Question 2

### If the anlge of elevation of a balloon from two consecutive km stones along a road are 30 degree and 60 degree respectively, then the height of the balloon above the ground will be

Detailed Solution for MCQ Test: Height & Distance - 1 - Question 2

AB = Height of balloon = h km
BD = x km, CD = 1 km
From Δ ABD,

From Δ ABC,

Hence, option A is correct.

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MCQ Test: Height & Distance - 1 - Question 3

### A pole stands vertically, inside a saclene triangular park ABC, if the angle of elevation of the top of the pole from each corner of the park is same, then in ΔABC, the foot of the pole is at the

Detailed Solution for MCQ Test: Height & Distance - 1 - Question 3

AP = CP = BP
It is possible only when OA = OB = OC i.e. radii of circum circle.
Hence, option B is correct.

MCQ Test: Height & Distance - 1 - Question 4

Two poles of heights 6 m and 11 m stand vertically upright on a plane ground. If the distance between their feet is 12 m, what is the distance between their tops?

Detailed Solution for MCQ Test: Height & Distance - 1 - Question 4

Given that ther are two poles
AE = 11 m and, CD = 6 m
∴ BE = 6 m
[∵CD = BE]
∴ AB = AE – BE = 11 – 6 = 5m
distance between their feet
ED = 12 m
∴ BC = 12 m [∵ED = BC] Now, AC = ?
In ΔABC,
From Pythagorus theorum,
AC2 = AB2 + BC2
AC2 = 52 + 122
AC2 = 25 + 144 = 169
AC = √169
AC = 13
Hence, option C is correct.

MCQ Test: Height & Distance - 1 - Question 5

At an instant, the length of the shadow of a pole is square root of 3 times the height of the pole. Find the angle of elevation of the turn.

Detailed Solution for MCQ Test: Height & Distance - 1 - Question 5

Let AB = x
Then, BC = √ x
and θ = ?

tan θ = tan 30°
[∵ tan 30° = 1/√3]
∴ θ = 30°
Hence, option A is correct.

MCQ Test: Height & Distance - 1 - Question 6

The shadow of a tower is 15 m when the sun’s elevation is 30°. What is the length of the shadow when the sun’s elevation is 60°?

Detailed Solution for MCQ Test: Height & Distance - 1 - Question 6

Given, ∠ADB = 30° and ∠ ACB = 60°
When the sun's elevation is 30°, the shadow of tower is "BD = 15 m" and when the sun's elevation is 60°, the shadow of tower is "BC = ?"
Let, BC = x m In ΔABD, tan 30° = AB/BD

In ΔABC, tan 60° = AB/BC
√3 = AB/x
∴ AB = x √3 ...(ii)
From Eqs. (i) and (ii), we get
x √3 = 15/√3
x = 5 m
Hence, option C is correct.

MCQ Test: Height & Distance - 1 - Question 7

A telegraph post gets broken at a point against a storm and its top touches the ground at a distance 20 m from the base of the post making an angle 30° with the ground. What is the height of the post?

Detailed Solution for MCQ Test: Height & Distance - 1 - Question 7

Given, BC = 20 m
∠ACB = 30°
Total height of the telegraph post is (AB + CA) = ?
In Δ ABC, tan 30° = AB/BC

Hence, option B is correct.

MCQ Test: Height & Distance - 1 - Question 8

From the top of a cliff 90 m high, the angles of depression of the top and bottom of a tower are observed to be 30° and 60°, respectively. What is the height of the tower?

Detailed Solution for MCQ Test: Height & Distance - 1 - Question 8

Given, AB = 90 m
And ∠ACB = 60°
Then, DC = ?
Ratio of angles,

AE = 30 m
Now, DC = EB
= AB – AE
= 90 – 30 = 60 m
Hence, option C is correct.

MCQ Test: Height & Distance - 1 - Question 9

The angle of elevation of the top of an unfinished pillar at a point 150 m from its base is 30°. If the angle of elevation at the same point is to be 45°, then the pillar has to be raised to a height of how many metres?

Detailed Solution for MCQ Test: Height & Distance - 1 - Question 9

Given, BC = 150 m
∠ACB = 30°
and, ∠DCB = 45°

DB = 150
[∵ DB = AD + AB]
∴ AD = 150 – AB
= 150 – 86.6 = 63.4m
Hence, option D is correct.

MCQ Test: Height & Distance - 1 - Question 10

The angles of depression of two ships from the top of a light house are 45° and 30° towards east. If the ships are 200 m apart, find the height of the light house.

Detailed Solution for MCQ Test: Height & Distance - 1 - Question 10

Given, ∠ ACB = 45°
and distance between two ships, i.e.,
CD = 200 m
Then, AB = ?
Let BC = x m
In ΔABC, tan 45° = AB/BC
(∵ tan 45° = 1)
1 = AB/x
∴ AB = x m ....(i)
In ΔABD, tan30° = AB/BD

(∵ tan 30° = 1/√3 )
x = √ AB – 200 ...(ii)
From Eqs. (i) and (ii),
AB = √ AB – 200
√ AB – AB = 200
0.732 AB = 200
(∵√ = 1.732)

= 273 m
Hence, option D is correct.

MCQ Test: Height & Distance - 1 - Question 11

A telegraph post is bent at a point above the ground due to storm. Its top just meets the ground at a distance of 8√3 meters from its foot and makes an angle of 30°, then the height of the post is :

Detailed Solution for MCQ Test: Height & Distance - 1 - Question 11

Given, BC = 8 √ m
In ΔABC,

AB = 8 m
Again, sin 30° = AB/AC

AC = 16 m
∴ The height of the post = AC + AB = 16 + 8 = 24 m.
Hence, option C is correct.

MCQ Test: Height & Distance - 1 - Question 12

Two poles of equal height are standing opposite to each other on either side of a road which is 100 m wide. From a point between them on road, angle of elevation of their tops are 30° and 60°. The height of each pole (in metre) is

Detailed Solution for MCQ Test: Height & Distance - 1 - Question 12

Given, BC = 100 m,
Let, the height of each pole = h metre
And, BE = x metre
∴ CE = (100 – x) m
In ΔCDE,

h = x √3
h = (100 – h√ 3) √3 [From eq. (i)]
h = 100 √3 – 3h
4h = 100 √3
h = 25√3
∴ The height of each pole is 25 √3 meter.
Hence, option A is correct.

MCQ Test: Height & Distance - 1 - Question 13

The angle of elevation of the top of a tower from a point A on the ground is 30°. On moving a distance of 20 metres towards the foot of the tower to a point B, the angle of elevation increases to 60°. The height of the tower is

Detailed Solution for MCQ Test: Height & Distance - 1 - Question 13

Given, AB = 20 m
Let, the height of the tower = h metre
And, BC = x metre
∴ AC = AB + BC = (20 + x) m
In ΔACD,

h = (h √3 – 20) √3 [From eq. (i)]
h = 3h – 20 √3
2h = 20 √3
h = 10 √3
∴ The height of the tower is 10 √3 meter.
Hence, option C is correct.

MCQ Test: Height & Distance - 1 - Question 14

The angle of elevation of the top of a building from the top and bottom of a tree are x and y respectively. If the height of the tree is h metre, then (in metre) the height of the building is

Detailed Solution for MCQ Test: Height & Distance - 1 - Question 14

Given, the height of the tree, CD = h metre
Let, the height of the building, AB = H metre
And, BC = a metre
∴ AE = AB – EB = (H – h) metre [∵ CD = BE]
In ΔABC,

a = H cot y ...(i)

a = (H – h) cot x ...(ii)
From equations (i) and (ii),
H cot y = (H – h) cot x = H cot x – h cot x
H(cot x – cot y) = h cot x

∴ The height of the building

Hence, option C is correct.

MCQ Test: Height & Distance - 1 - Question 15

The distance between two pillars of length 16 metres and 9 metres is x metres. If two angles of elevation of their respective top from the bottom of the other are complementary to each other, then the value of x (in metres) is

Detailed Solution for MCQ Test: Height & Distance - 1 - Question 15

Given, AB = 16 m, CD = 9 m and BC = x metre
And, ∠ACB and ∠CBD are complementary.
∴ Let, ∠ACB = Θ and ∠CBD = (90° – Θ)
In ΔABC,

Now, In ΔBCD,

[∵ tan (90° – Θ) = cot Θ]
By multiplying eq. (i) & (ii),

x = 12 m
Hence, option C is correct.

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## Quantitative Aptitude for SSC CGL

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