NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.5

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.5 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.5.

• Surface Areas and Volumes Class 9 Ex 13.1
• Surface Areas and Volumes Class 9 Ex 13.2
• Surface Areas and Volumes Class 9 Ex 13.3
• Surface Areas and Volumes Class 9 Ex 13.4
• Surface Areas and Volumes Class 9 Ex 13.5
• Surface Areas and Volumes Class 9 Ex 13.6
• Surface Areas and Volumes Class 9 Ex 13.7
• Surface Areas and Volumes Class 9 Ex 13.8
• Surface Areas and Volumes Class 9 Ex 13.9

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 13
Chapter Name	Surface Areas and Volumes
Exercise	Ex 13.5
Number of Questions Solved	9
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.5

Question 1.
A matchbox measures 4 cm x 2.5 cm x 1.5 cm. What will be the volume of a packet containing 12 such boxes?
Solution.
Given, dimensions of match box = 4 cm x 2.5 cm x 1.5 cm
So, length (l) = 4 cm, breadth (b) = 2.5 cm and height (h) = 1.5 cm
∴ Volume of a match box = Volume of a cuboid (V)
= 4 cm x 2.5 cm x 1.5 cm
= 15 cm³     [∵ volume of cuboid (V) = Ibh]
Volume of a packet containing 12 such boxes
= 12 x Volume of one match box = 12 x 15 = 180 cm³

Question 2.
A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many liters of water can it hold?
(1m ³ = 1000 l)
Solution.
Given, length (l) = 6 m, breadth (b) = 5m and depth or height (b) = 4.5m Volume of a cuboidal water tank = 6 m x 5 m x 4.5 m [∵  volume of cuboid (V) = Ibh]
= 30 x 4.5m³ = 135m³
= 135 x 1000 L = 135000 L [∵ 1 m³ = 1000 L]
Hence, tank can hold 135000 L of water.

Question 3.
A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 m³ of a liquid?
Solution.
Let h be the height of vessel.
Given, length (l) = 10 m and breadth (b) = 8 m
Volume of a cuboidal vessel = Liquid to be hold
∴  I x b x h = 380    [given]
⇒ 10 x 8 x h = 380
= h = = 4.75m
Hence, the cuboidal vessel must be made 4.75 m high.

Question 4.
Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of Rs. 30 per m³
Solution.
Volume of a cuboidal pit = l x b x h = (8 x 6 x 3) = 144 m³
[∵ l = 8 m, b = 6 m and h- 3 m, given]
∵ Cost of digging 1 m³ cuboidal pit = Rs. 30
∴ Cost of digging 144 m³ cuboidal pit = 30 x 144 = Rs. 4320

Question 5.
The capacity of a cuboidal tank is 50000 liters of water. Find the breadth of the tank, if its length and depth are 2.5 m and 10 m, respectively.
Solution.
Given, l = 2.5 m and h = 10 m
Let breadth of tank be b.

Question 6.
A village, having a population of 4000, requires 150 liters of water per head per day. It has a tank measuring 20 m x 15 m x 6 m. For how many days, will the water of this tank last?
Solution.
Given, length (l) = 20 m, breadth (b) = 15 m and height (h) = 6 m
∴ Capacity of the tank = Volume of the tank
= Ibh = (20 x 15 x 6) = 1800 m³
∵ Water required for one person per day = 150 L
∴  Water required for 4000 persons per day

Question 7.
A godown measures 40 m x 25 m x 15 m. Find the maximum number of wooden crates each measuring 1.5 m x 1.25 m x 0.5 m that can be stored in the godown.
Solution.
Given, dimensions for godown are as length (l) = 40 m, breadth (b) = 25 m and height (b) = 15 m
Volume of the godown = l x b x h = 40m x 25m x 15m
Dimensions for each wooden crates are as

Question 8.
A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.
Solution.
Volume of a solid cube = 12 cm x 12 cm x 12 cm = 1728 cm³
The solid cube is cut into eight cubes of equal volume.
Hence, the volume of the new cube =  cm³ =216cm²
(side)³ = 216 cm³
side = 6 cm
Hence, side of the new cube = 6 cm
Surface area of solid cube = 6(side)² = 6(12)² cm²
S₁= 6 x 144 = 864 cm²
Surface area of new cube = 6 (side)² = 6 (6)² cm²
S₂ = 216 cm²
∴ Required ratio = S₂ : S₂
= 864 : 216 = 4:1

Question 9.
A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?
Solution.
Here, breadth of the river = 40 m
Depth of the river = 3 m
and length per hour of the river = 2 km = 2000 m
or length per minute of the river =  m
∴ Water flowing per minute =  x 40 x 3 = 4000 m³

 

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