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

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

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

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

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

External radius = 4 cm,
Internal radius = 3 cm.
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

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

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

l = 10 m
h = 8 m

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

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

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

(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

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

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

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

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

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.
∴ Correct answer is (c).

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.
∴ Correct answer is (d).

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.
∴ Correct answer is (c).

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.
∴ Correct answer is (d).

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.
∴ Correct answer is (c).