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Swaraj India Public School: Class: 8 Subject: Maths TOPIC: Mensuration EXERCISE: 18.3 Solution of Classwork Sums

This document contains the solutions to various maths problems involving the calculation of volumes and ratios from mensuration questions. It includes calculating the volume of bricks required to fill an excavated pit, finding the ratio of volumes when an edge of a cube is tripled, and determining the volume of wood required to make a closed box from its dimensions. Various geometry problems involving cylinders are also solved, such as calculating the ratio of heights of two cylinders given the ratio of their radii, and finding the radius of a single cylinder required to hold the total volume of milk from two other cylinders.

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ANUSHKA SINGH
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113 views5 pages

Swaraj India Public School: Class: 8 Subject: Maths TOPIC: Mensuration EXERCISE: 18.3 Solution of Classwork Sums

This document contains the solutions to various maths problems involving the calculation of volumes and ratios from mensuration questions. It includes calculating the volume of bricks required to fill an excavated pit, finding the ratio of volumes when an edge of a cube is tripled, and determining the volume of wood required to make a closed box from its dimensions. Various geometry problems involving cylinders are also solved, such as calculating the ratio of heights of two cylinders given the ratio of their radii, and finding the radius of a single cylinder required to hold the total volume of milk from two other cylinders.

Uploaded by

ANUSHKA SINGH
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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SWARAJ INDIA PUBLIC SCHOOL

CLASS: 8 SUBJECT: MATHS


TOPIC: Mensuration EXERCISE: 18.3
SOLUTION OF CLASSWORK SUMS

Q5. Dimensions of rectangular pit = 1.4 m x 90 cm x 70 cm


= 140 cm x 90 cm x 70 cm
Volume of the pit = 𝒍 x b x h
= (140 x 90 x 70) cm3
Volume of earth dug out = Volume of the pit
= (140 x 90 x 70) cm3
Earth dug out is used to make 1000 bricks
∴ Volume of 1000 bricks = (140 x 90 x 70) cm3
𝟏𝟒𝟎 𝐱 𝟗𝟎 𝐱 𝟕𝟎
∴ Volume of each brick =
𝟏𝟎𝟎𝟎
= (14 x 9 x 7) cm3
Length & breadth of each brick = 21 and 10.5 cm respectively
Volume of each brick = 14 x 9 x 7
𝒍 x b x h = 14 x 9 x 7
21 x 10.5 x h = 14 x 9 x 7
𝟏𝟒 𝐱 𝟗 𝐱 𝟕
h=
𝟐𝟏 𝐱 𝟏𝟎.𝟓
h = 4 cm
Q6. Let the edge of the cube = 𝒙 cm
Volume of the cube = (𝒙)3 cm3
On tripling the edge
New edge = 3𝒙 cm
New volume = (3𝒙)3
= 27 𝒙3 cm3
𝟐𝟕 𝒙𝟑
Ratio of the new volume to the original volume = 𝟑 = 27:1
𝒙
∴ If the edge of the cube is tripled, the volume becomes
27 times.
Q8. External dimensions of the closed wooden box
= 84 cm x 75 cm x 64 cm
External volume of the closed wooden box = 𝒍 x b x h
= (84 x 75 x 64) cm3
= 403200 cm3
Width of the wood all around = 2 cm
Internal dimensions of the box 𝒍 = 84 – (2 + 2) = 80 cm
b = 75 – (2 + 2) = 71 cm
h = 64 – (2 + 2) = 60 cm
Internal volume of the box = 𝒍 x b x h
= 80 x 71 x 60
= 340800 cm3
Volume of the wood required to make the box
= External volume – Internal volume
= 403200 – 340800
= 62400 cm3
Ans. The volume of the wood required to make the box is 62400 cm3.

Q9. Ratio of the diameter of two cylindrical vessels = 3:4


Ratio of the radius of two cylindrical vessels = 3:4
Let the radius of 2 cylindrical vessels be 3𝒙 & 4𝒙 respectively.
Let the height of two cylindrical vessels be h1 & h2 respectively.
Acc. to the ques.
𝝅r2h1 = 𝝅r2h2
𝟐𝟐 𝟐𝟐
x 3𝒙 x 3𝒙 x h1 = x 4𝒙 x 4𝒙 x h2
𝟕 𝟕
𝟐 𝟐
9𝒙 h1 = 16𝒙 h2
𝒉𝟏 𝟏𝟔 𝒙𝟐
= ⇒ h1:h2 = 16: 9
𝒉𝟐 𝟗 𝒙𝟐
Ans. The ratio of their heights is 16:9.
Q12. Internal diameter of cylindrical pipe = 7 cm
𝟕 𝟑.𝟓
Internal radius of cylindrical pipe = = 3.5 cm = m
𝟐 𝟏𝟎𝟎
(i) Speed at which the water is discharged = 5 m/s
∴ height of the cylindrical pipe = 5 m
Volume of water flowing in 1 sec = 𝝅r2h
𝟐𝟐 𝟑.𝟓 𝟑.𝟓
= x x ×5
𝟕 𝟏𝟎𝟎 𝟏𝟎𝟎
𝟏𝟏 𝐱 𝟎.𝟓 𝐱 𝟑.𝟓
= m3
𝟏𝟎𝟎𝟎
𝟔𝟎 𝐱 𝟏𝟏 𝐱 𝟎.𝟓 𝐱 𝟑.𝟓
Volume of water flowing in 1 minute = m3
𝟏𝟎𝟎𝟎

1 m3 = 1000 𝒍
𝟔𝟎 𝐱 𝟏𝟏 𝐱 𝟎.𝟓 𝐱 𝟑.𝟓 𝟔𝟎 𝐱 𝟏𝟏 𝐱 𝟎.𝟓 𝐱 𝟑.𝟓 𝐱 𝟏𝟎𝟎𝟎
∴ m3 =
𝟏𝟎𝟎𝟎 𝟏𝟎𝟎𝟎
= 60 x 11 x 0.5 x 3.5
= 1155 𝒍
(ii) Dimensions of the rectangular tank = 4 m x 3 m x 2.31 m
Volume = 𝒍 x b x h
= (4 x 3 x 2.31) m3
𝟔𝟎 𝐱 𝟏𝟏 𝐱 𝟎.𝟓 𝐱 𝟑.𝟓
m3 water flows in 1 min.
𝟏𝟎𝟎𝟎
𝟔𝟎 𝐱 𝟏𝟏 𝐱 𝟎.𝟓 𝐱 𝟑.𝟓
∴ 1 m3 water will flow in 1÷ min.
𝟏𝟎𝟎𝟎
𝟏 𝐱 𝟏𝟎𝟎𝟎 𝐱 𝟒 𝐱 𝟑 𝐱 𝟐.𝟑𝟏
(4 x 3 x2.31) m3 water will flow in min.
𝟔𝟎 𝐱 𝟏𝟏 𝐱 𝟎.𝟓 𝐱 𝟑.𝟓

= 2 x 4 x 3 = 24 min.
Q13. The radius and height (R and H) of the first cylindrical
Vessel = 15 cm and 40 cm respectively
The radius and height (r and h) of the second cylindrical
Vessel = 20 cm and 45 cm respectively
Total volume of the milk
= Volume of vessel 1 + Volume of Vessel 2
= 𝝅R2H + 𝝅r2h
= 𝝅(R2H + r2h)
𝟐𝟐
= (15 x 15 x 40 + 20 x 20 x 45)
𝟕
𝟐𝟐
= (9000 + 18000)
𝟕
𝟐𝟐
= x 27000 cm3
𝟕

Volume of cylindrical vessel in which the milk is poured


= Volume of the milk.
𝟐𝟐
𝝅r2h = x 27000
𝟕
𝟐𝟐 𝟐𝟐
x r2 x 30 = x 27000
𝟕 𝟕
𝟐𝟕𝟎𝟎𝟎
r2 =
𝟑𝟎
r2 = 900

r = √𝟗𝟎𝟎
∴ r = 30 cm
Q15. Length of the cuboid from which cylinder is cut = 30 cm
Side of the cross-section of a square = 14 cm
Height of the cylinder = 30 cm
𝒅
Radius =
𝟐
𝟏𝟒
=
𝟐

= 7 cm
(i) Volume = 𝝅r2h
𝟐𝟐
= x 7 x 7 x 30
𝟕

= 4620 cm3
(ii) Volume of the wood wasted
= Volume of Cuboid – Volume of cylinder
= (𝒍 x b x h) – 4620
= (30 x 14 x 14) – 4620
= 5880 – 4620
= 1260 cm3
H.W.: Do Q.1, 3, 11 and 14 of Ex. 18.3

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