PSEB 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.5 Textbook Exercise Questions and Answers.

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

Question 1.
A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What will be the volume of a packet containing 12 such boxes ?
Answer:
For the cuboidal matchbox, length l = 4 cm; breadth b = 2.5 cm and height h = 1.5 cm.
Volume of a cuboidal matchbox
= l × b × h
= 4 × 2.5 × 1.5 cm³
= 15 cm³
Then, volume of 12 matchboxes = 12 × 15 cm³ = 180 cm³
Thus, the volume of a packet containing 12 matchboxes is 180 cm³.

Question 2.
A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (1 m³ = 1000 l)
Answer:
For the cuboidal water tank, length l = 6m; breadth b = 5 m and height h = 4.5 m.
Capacity of the cuboidal tank = l × b × h
= 6 × 5 × 4.5 m³
= 135 m³
1 m³ = 1000 litres
∴ 135 m³ = 135000 litres
Thus, the given cuboidal water tank can hold 1,35,000 litres of water.

Question 3.
A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid?
Answer:
For the cuboidal vessel, length l = 10 m;
breadth b = 8 m and capacity = 380 m³.
Capacity of a cuboidal vessel = l × b × h
∴ 380 m³ = 10 m × 8 m × h m
∴ h = \(\frac{380}{10 \times 8}\) m
∴ h = 4.75 m
The height of the cuboidal vessel must be made 4.75 m.

Question 4.
Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of ₹ 30 per m³.
Answer:
For the cuboidal pit, length l = 8m; breadth b = 6 m and height (depth) h = 3 m.
Volume of the earth to be dugout to make the cuboidal pit = Volume of a cuboid
= l × b × h
= 8 × 6 × 3 m³
= 144 m³
Cost of digging out 1 m³ of earth = ₹ 30
∴ Cost of digging out 144 m³ of earth
= ₹ (30 × 144)
= ₹ 4320
Thus, the cost of digging the cuboidal pit is ₹ 4320.

Question 5.
The capacity of a cuboidal tank is 50,000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.
Answer:
For the cuboidal tank, length l = 2.5 m;
height (depth) h = 10 m and
capacity = 50,000 litres.
1000 litres = 1 m³
∴ 50,000 litres = \(\frac{50,000}{1000}\) m³ = 50 m³
Capacity of cuboidal tank = l × b × h
∴ 50 m³ = 2.5 m × b m × 10 m
∴ b = \(\frac{50}{2.5 \times 10}\) m
∴ b = 2 m
Thus, the breadth of the cuboidal tank is 2 m.

Question 6.
A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20 m × 15 m × 6 m. For how many days will the water of this tank last?
Answer:
Total requirement of water per day
= No. of people × daily requirement per person
= 4000 × 150 litres
= 6,00,000 litres
= \(\frac{6,00,000}{1000}\) m³
= 600 m³
For the cuboidal tank, length l = 20 m;
breadth b = 15 m and height h = 6 m
Capacity of the cuboidal tank = l × b × h
= 20 × 15 × 6 m³
= 1800 m³
600 m³ of water can last for 1 day in the village.
∴ 1800 m³ of water can last for \(\frac{1800}{600}\) = 3 days in the village.

Question 7.
A godown measures 40 m × 25 m × 15 m. Find the maximum number of wooden crates each measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown.
Answer:
For the cuboidal godown, length l = 40 m;
breadth b = 25 m and height h = 15 m.
Capacity of cuboidal godown = l × b × h
= 40 × 25 × 15 m³
For the wooden cuboidal crate, length l = 1.5 m; breadth b = 1.25 m and height h = 0,5 m.
Volume of 1 cuboidal crate
= l × b × h
= 1.5 × 1.25 × 0.5 m³
∴The no. of crates that can be stored in the godown = \(\)
= \(\left(\frac{40}{1.25}\right) \times\left(\frac{25}{0.5}\right) \times\left(\frac{15}{1.5}\right)\)
= 32 × 50 × 10
= 16,000

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.
Answer:
For the original cube, edge a =12 cm.
Volume of original cube = a³ = 123 cm³
= 1728 cm³
8 cubes of equal volume are made from the original cube.
∴ Volume of each new cube = \(\frac{1728}{8}\) cm³
= 216 cm³
Let the edge of new cube be A cm.
Volume of new cube = A³
∴ 216 cm³ = A³
∴ A = \(\sqrt[3]{216}\) cm = 6 cm
Thus, the side of each new cube is 6 cm.
Total surface area of original cube
= 6a²
= 6 (12)² cm²
Total surface area of a new cube = 6A²
= 6 (6)² cm²
\(\frac{\text { Total surface area of original cube }}{\text { Total surface area of a new cube }}\) = \(\frac{6(12)^{2} \mathrm{~cm}^{2}}{6(6)^{2} \mathrm{~cm}^{2}}\)
= \(\left(\frac{12}{6}\right)^{2}\)
= 4
= 4:1
Thus, the required ratio of the total surface area of the original cube and the total surface area of a new cube is 4:1.
Note: If the ratio of TSA of the original ‘ cube and TSA of all the new cubes is required, then it will be 1 : 2.

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?
Answer:
2 km = 2000 m and 1 hour = 60 minutes
Rate of flow of water in the river
= 2 km/hour
= \(\frac{2000}{60}\) m/min
Thus, during 1 minute, water of length will flow in the sea.
Then, the water falling in sea per minute takes cuboidal shape with length l = \(\frac{2000}{60}\) m,
breadth b = 40 m and height (depth) h = 3 m.
Volume of water falling in sea per minute
= l × b × h
= \(\frac{2000}{60}\) × 40 × 3 m³
= 4000 m³
Thus, 4000 m³ of water will fall into the sea in a minute.