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 cm^{3}

= 15 cm^{3}

Then, volume of 12 matchboxes = 12 × 15 cm^{3} = 180 cm^{3}

Thus, the volume of a packet containing 12 matchboxes is 180 cm^{3}.

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 m^{3} = 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 m^{3}

= 135 m^{3}

1 m^{3} = 1000 litres

∴ 135 m^{3} = 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 m^{3}.

Capacity of a cuboidal vessel = l × b × h

∴ 380 m^{3} = 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 m^{3}.

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 m^{3}

= 144 m^{3}

Cost of digging out 1 m^{3} of earth = ₹ 30

∴ Cost of digging out 144 m^{3} 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 m^{3}

∴ 50,000 litres = \(\frac{50,000}{1000}\) m^{3} = 50 m^{3}

Capacity of cuboidal tank = l × b × h

∴ 50 m^{3} = 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

= \(\frac{6,00,000}{1000}\) m^{3}

= 600 m^{3}

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 m^{3}

= 1800 m^{3}

600 m^{3} of water can last for 1 day in the village.

∴ 1800 m^{3} 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 m^{3}

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 m^{3}

∴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 = a^{3} = 123 cm^{3}

= 1728 cm^{3}

8 cubes of equal volume are made from the original cube.

∴ Volume of each new cube = \(\frac{1728}{8}\) cm^{3}

= 216 cm^{3}

Let the edge of new cube be A cm.

Volume of new cube = A^{3}

∴ 216 cm^{3} = A^{3}

∴ A = \(\sqrt[3]{216}\) cm = 6 cm

Thus, the side of each new cube is 6 cm.

Total surface area of original cube

= 6a^{2}

= 6 (12)^{2} cm^{2}

Total surface area of a new cube = 6A^{2}

= 6 (6)^{2} cm^{2}

\(\frac{\text { Total surface area of original cube }}{\text { Total surface area of a new cube }}\) = \(\frac{6(12)^{2} \mathrm{~cm}^{2}}{6(6)^{2} \mathrm{~cm}^{2}}\)

= \(\left(\frac{12}{6}\right)^{2}\)

= 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 m^{3}

= 4000 m^{3}

Thus, 4000 m^{3} of water will fall into the sea in a minute.