Olympiad Test: Elementary Mensuration-II - 1 - Class 7 MCQ

# Olympiad Test: Elementary Mensuration-II - 1 - Class 7 MCQ

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## 20 Questions MCQ Test Mathematics Olympiad Class 7 - Olympiad Test: Elementary Mensuration-II - 1

Olympiad Test: Elementary Mensuration-II - 1 for Class 7 2024 is part of Mathematics Olympiad Class 7 preparation. The Olympiad Test: Elementary Mensuration-II - 1 questions and answers have been prepared according to the Class 7 exam syllabus.The Olympiad Test: Elementary Mensuration-II - 1 MCQs are made for Class 7 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Olympiad Test: Elementary Mensuration-II - 1 below.
Olympiad Test: Elementary Mensuration-II - 1 - Question 1

### A right triangle with sides 3 cm, 4 cm and 5 cm is rotated the side of 3 cm to form a cone. The volume of the cone so formed is:

Detailed Solution for Olympiad Test: Elementary Mensuration-II - 1 - Question 1

Clearly, we have r = 3 cm and h = 4 cm.
∴ Volume = 1/3 πr2h
= (1/3 × π × 32 × 4) cm3
= 12π cm3

Olympiad Test: Elementary Mensuration-II - 1 - Question 2

### In a shower, 5 cm of rain falls. The volume of water that falls on 1.5 hectares of ground is:

Detailed Solution for Olympiad Test: Elementary Mensuration-II - 1 - Question 2

1 hectare = 10,000 m2
So, Area = (1.5 × 10000) m2 = 15000 m2
Depth     = (5/100) m = (1/20) m
∴ Volume = (Area × Depth)
= (15000 × 1/20) m3 = 750 m3

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Olympiad Test: Elementary Mensuration-II - 1 - Question 3

### A hall is 15 m long and 12 m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls, the volume of the hall is:

Detailed Solution for Olympiad Test: Elementary Mensuration-II - 1 - Question 3

Given 2(15 + 12) × h = 2(15 × 12)
⇒ h = (180/27) m = (20/3) m
∴ Volume = (15 × 12 × 20/3) m3
= 1200 m3

Olympiad Test: Elementary Mensuration-II - 1 - Question 4

66 cubic centimetres of silver is drawn into a wire 1 mm in diameter. The length of the wire in metres will be:

Detailed Solution for Olympiad Test: Elementary Mensuration-II - 1 - Question 4

Let the length of the wire be h.
Radius = 1/2 mm = 1/20 cm. Then,
⇒ (22/7 × 1/20 × 1/20 × h) = 66
∴ h = (60 × 20 × 20 × 7)/22 = 8400
cm = 84 m

Olympiad Test: Elementary Mensuration-II - 1 - Question 5

A hollow iron pipe is 21 cm long and its external diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weighs 8 g/cm3, then the weight of the pipe is:

Detailed Solution for Olympiad Test: Elementary Mensuration-II - 1 - Question 5

Volume of iron
= (22/7 × [(4)2 – (3)2] × 21) cm3
= (22/7 × 7 × 1 × 21) cm3
= 462 cm3
∴ Weight of iron= (462 × 8) gm
= 3696 gm
= 3.696 kg

Olympiad Test: Elementary Mensuration-II - 1 - Question 6

A boat having a length 3 m and breadth 2 m is floating on a lake. The boat sinks by 1 cm when a man gets on it. The mass of the man is:

Detailed Solution for Olympiad Test: Elementary Mensuration-II - 1 - Question 6

Volume of water displaced
= (3 × 2 × 0.01) m3
= 0.06 m3
∴ Mass of man = Volume of water displaced × Density of water
= (0.06 × 1000) kg
= 60 kg

Olympiad Test: Elementary Mensuration-II - 1 - Question 7

50 men took a dip in a water tank 40 m long and 20 m broad on a religious day. If the average displacement of water by a man is 4 m3, then the rise in the water level in the tank will be:

Detailed Solution for Olympiad Test: Elementary Mensuration-II - 1 - Question 7

Total volume of water displaced
= (4 × 50) m3 = 200 m3
∴ Rise in water level = [200/( 40 × 20)] m
= 0.25 m =25 cm

Olympiad Test: Elementary Mensuration-II - 1 - Question 8

The slant height of a right circular cone is 10 m and its height is 8 m. Find the area of its curved surface.

Detailed Solution for Olympiad Test: Elementary Mensuration-II - 1 - Question 8

l = 10 m
h = 8 m

∴ Curved surface area = πrl
= (π × 6 × 10) m2 = 60 π m2

Olympiad Test: Elementary Mensuration-II - 1 - Question 9

A cistern 6m long and 4 m wide contains water up to a depth of 1 m 25 cm. The total area of the wet surface is:

Detailed Solution for Olympiad Test: Elementary Mensuration-II - 1 - Question 9

Area of the wet surface
= [2(lb + bh + lh) – lb]
= 2(bh + lh) + lb
= [2 (4 × 1.25 + 6 × 1.25) + 6 × 4] m2
= 49 m2

Olympiad Test: Elementary Mensuration-II - 1 - Question 10

A metallic sheet is of rectangular shape with dimensions 48 m x 36 m. From each of its corners, a square is cut off so as to make an open box. If the length of the square is 8 m, the volume of the box (in m3) is:

Detailed Solution for Olympiad Test: Elementary Mensuration-II - 1 - Question 10

Clearly, l = (48 – 16)m = 32 m,
b = (36 – 16)m = 20 m
h = 8 m.
∴ Volume of the box = (32 × 20 × 8) m3
= 5120 m3

Olympiad Test: Elementary Mensuration-II - 1 - Question 11

The curved surface area of a cylindrical pillar is 264 m2 and its volume is 924 m3. Find the ratio of its diameter to its height.

Detailed Solution for Olympiad Test: Elementary Mensuration-II - 1 - Question 11

(πr2h)/ (2πrh) = 924/264
r = [(924 /264) × 2] = 7 m and, 2πrh = 264
h = (264 × 7/22 × 1/2 × 1/7) = 6 m
∴ Required ratio = 2r/h =14/6 = 7 : 3

Olympiad Test: Elementary Mensuration-II - 1 - Question 12

A cistern of capacity 8000 litres measures externally 3.3 m by 2.6 m by 1.1 m and its walls are 5 cm thick. The thickness of the bottom is:

Detailed Solution for Olympiad Test: Elementary Mensuration-II - 1 - Question 12

Let the thickness of the bottom be x cm.
Then, [(330 – 10) × (260 – 10) × (110 × x)]
= 8000 × 1000
⇒ 320 × 250 × (110 – x) = 8000 × 1000
⇒ (110 – x) = (8000 × 1000)/ (320 × 250) = 100
x = 10 cm = 1 dm

Olympiad Test: Elementary Mensuration-II - 1 - Question 13

What is the total surface area of a right circular cone of height 14 cm and base radius 7 cm?

Detailed Solution for Olympiad Test: Elementary Mensuration-II - 1 - Question 13

h = 14 cm, r = 7 cm.

∴ Total surface area = πrl + πr2
= (22/7 × 7 × 75 + 22/7 × 72) cm2
= [154(5 + 1)] cm2 = (154 × 3.236) cm2
= 498.35 cm2

Olympiad Test: Elementary Mensuration-II - 1 - Question 14

A large cube is formed from the material obtained by melting three smaller cubes of 3, 4 and 5 cm side. What is the ratio of the total surface areas of the smaller cubes and the large cube?

Detailed Solution for Olympiad Test: Elementary Mensuration-II - 1 - Question 14

Volume of the large cube = (33 + 43 + 53)
= 216 cm3
Let the edge of the large cube be a.
So, a3 = 216 a = 6 cm
∴ Required ratio
= [6 × (32 + 42 + 52)]/ (6 × 62
= 50/36 = 25 : 18

Olympiad Test: Elementary Mensuration-II - 1 - Question 15

How many bricks, each measuring 25 cm × 11.25 cm × 6 cm, will be needed to build a wall of 8 m × 6 m × 22.5 cm?

Detailed Solution for Olympiad Test: Elementary Mensuration-II - 1 - Question 15

Number of bricks = Volume of the wall/ Volume of 1 brick = (800 × 600 × 22.5)/ (25 × 11.25 × 6)
= 6400

Olympiad Test: Elementary Mensuration-II - 1 - Question 16

What is the volume of 32 metre high cylindrical tank?
I. The area of its base is 154 m2.
​II. The diameter of the base is 14 m.

Detailed Solution for Olympiad Test: Elementary Mensuration-II - 1 - Question 16

Given, height = 32 m
From I, the area of the base = 154 m2
∴ Volume = (Area of the base × Height)
= (154 × 32) m3
Thus, I alone gives the answer.
From II, the radius of the base = 7 m
∴ Volume = π r2h = (22/7 × 7 × 7 × 32) m3
= 4928 m3
Thus, II alone gives the answer.

Olympiad Test: Elementary Mensuration-II - 1 - Question 17

Is a given rectangular block a cube?
I. At least 2 faces of the rectangular block are squares.
II. The volume of the block is 64.

Detailed Solution for Olympiad Test: Elementary Mensuration-II - 1 - Question 17

From I, any two of l, b, h are equal.
From II, lbh = 64.
From I and II, the values of l, b, h may be (1, 1, 64), (2, 2, 16), (4, 4, 4).
Thus, the block may be a cube or cuboid.

Olympiad Test: Elementary Mensuration-II - 1 - Question 18

What is the capacity of a cylindrical tank?
I. Radius of the base is half of its height which is 28 metres.
II. Area of the base is 616 sq. metres and its height is 28 metres.

Detailed Solution for Olympiad Test: Elementary Mensuration-II - 1 - Question 18

From I, h = 28 m and r = 14.
∴ Capacity = π r2h, which can be obtained.
Thus, I alone gives the answer.
From II, π r2 = 616 m2 and h = 28 m
∴ Capacity = (π r2 × h) = (616 × 28) m3
Thus, II alone gives the answer.

Olympiad Test: Elementary Mensuration-II - 1 - Question 19

What is the height of a circular cone?
I. The area of that cone is equal to the area of a rectangle whose length is 33 cm.
II. The area of the base of that cone is 154 sq. cm.

Detailed Solution for Olympiad Test: Elementary Mensuration-II - 1 - Question 19

II gives the value of r.
But, in I, the breadth of rectangle is not given.
So, we cannot find the surface area of the cone.
Hence, the height of the cone cannot be determined.

Olympiad Test: Elementary Mensuration-II - 1 - Question 20

What is the volume of a cube?
I. The area of each face of the cube is 64 square metres.
II. The length of one side of the cube is 8 metres.

Detailed Solution for Olympiad Test: Elementary Mensuration-II - 1 - Question 20

Let each edge be a metre.
Then, I. a2 = 64
a = 8 m
Volume = (8 × 8 × 8) m3 = 512 m3
Thus, I alone gives the answer.
II. a = 8 m
Volume = (8 × 8 × 8) m3 = 512 m3
Thus, II alone gives the answer.

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