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

## Haryana Board 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Exercise 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?

Solution:

Dimensions of the matchbox are 4 cm × 2.5 cm × 1.5 cm. Since a matchbox is in the shape of a cuboid.

∴ Volume of the 1 matchbox = 4 × 2.5 × 1.5 = 15 cm^{3}

∴ Volume of a packet containing 12 such boxes = 15 × 12 = 180 cm^{3}

Hence, volume of a packet containing 12 such boxes = 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 ?)

Solution:

Length of cuboidal water tank (l) = 6 m

Breadth of cuboidal water tank (b) = 5 m

and depth of cuboidal water tank (h) = 4.5 m

∴ Volume of the cuboidal water tank = l × b × h

= 6 × 5 × 4.5

= 135 m^{3}

= 135 × 1000 litres

= 135000 litres

Hence,water tank can hold volume of water = 135000 litres.

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?

Solution:

We have,

Length of cuboidal vessel (l) = 10 m

Breadth of cuboidal vessel (b)= 8 m

Let the height of cuboidal vessel be h m and volume of cuboidal vessel = 380 m^{3}

⇒ l × b × h = 380

⇒ 10 × 8 × h = 380

⇒ 80 × h = 380

⇒ h = \(\frac{380}{80}\)

⇒ h = 4.75 m

Hence,height of the cuboidal vessel = 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 Rs. 30 per m^{3}.

Solution:

We have, Length of cuboidal pit (l) = 8 m

Breadth of cuboidal pit (b) = 6 m

and depth of cuboidal pit (h) = 3 m

∴ Volume of the cuboidal pit = l × b × h

= 8 × 6 × 3 = 144 m^{3}

Rate of the digging the pit = Rs. 30 per m^{3}

∴ Cost of digging the pit = Rs. 144 × 30

= Rs. 4320

Hence,cost of digging the pit = Rs. 4320.

Question 5.

The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.

Solution:

We have,

Length of cuboidal tank (l) = 2.5 m

Depth of cuboidal tank (h) = 10 m

Let the breadth of cuboidal tank be b m.

and capacity of cuboidal tank = 50000 litres

⇒ l × b × h = \(\frac{50000}{1000}\)

⇒ 2.5 × 6 × 10 = 50 m^{3}

⇒ b = \(\frac{50}{2.5 \times 10}\)

⇒ b = 2 m

Hence,breadth of the cuboidal tank = 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?

Solution:

Dimensions of the water tank are 20 m × 15 m × 6 m.

Volume of the water tank = 20 × 15 × 6

= 1800 m^{3}

Requirement of water per head per day = 150 litres = \(\frac{150}{1000}\) m^{3}

Requirement of water for population of 4000 per day

= \(\frac{150 \times 4000}{1000}\)

∵ 600 m^{3} water is sufficient for 4000 persons = 1 day

∴ 1800 m^{3} water is sufficient for 400 persons = \(\frac{1 \times 1800}{600}\) = 3 days

Hence, water of tank is sufficient for days = 3.

Question 7.

A godown measures 40 m × 25 m × 10 m. Find the maximum number of wooden crates each measuring 1.0 m × 1.25 m × 0.5 m that can be stored in the godown.

Solution:

Dimensions of the godown are 40 m × 25 m × 10 m.

∴ Volume of the godown = 40 × 25 × 10

= 10000 m^{3}

Dimension of each wooden crate is 1.0 m × 1.25 m × 0.5

∴ Volume of each wooden crate = 1.0 × 1.25 × 0.5

= 0.625 m^{3}

= \(\frac{\text { Volume of the godown }}{\text { Volume of } 1 \text { wooden crate }}\)

\(\frac{10000}{0.625}\) = 16000

Hence,number of wooden crates = 16000.

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:

We have,

Edge of big cube (a) = 12 cm

∴ Volume of big cube = a^{3} = (12)^{3} = 1728 cm^{3}

∵ It cut into eight cubes of equal volume.

∴ Volume of new (small) cube = \(\frac{1728}{8}\)

= 216 cm^{3}

Let the edge of new cube be x cm.

∴ Volume of new cube = x^{3}

⇒ 216 = x^{3}

⇒ \(\sqrt[3]{216}\) = \(\sqrt[3]{x^3}\)

⇒ \(\sqrt[3]{6 \times 6 \times 6}\) = x

⇒ x = 6 cm

Surface area of new cube = 6x^{2} = 6 × (6)^{2}

= 216 cm^{2}

Surface area of big cube = 6a^{2} = 6 × (12)^{2}

= 864 cm^{2}

\(\frac{\text { Surface area of big cube }}{\text { Surface area of a new cube }}=\frac{864}{216}=\frac{4}{1}\)

Hence side of the cube = 6 cm.

Surface area of big cube: Surface area of new cube = 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 one minute?

Solution:

We have,

Depth of river (h) = 3 m

Width of river (b)= 40 m

Speed of water = 2 km/hr = \(\frac{2000}{60}\) m/min.

∴ The length of water flow in 1 min.

(l) = \(\frac{2000}{60}=\frac{100}{3} \mathrm{~m}\)

Thus,required volume of water = l × b × h

= \(\frac{100}{3}\) × 40 × 3

= 4000 m^{3}

Hence,required volume of water = 4000 m^{3}.