**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 the part
of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for
Chapter 13 Surface
Areas and Volumes Ex 13.5.

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

**Ex
13.5 Class 9 Maths 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:
**Since, a matchbox is in the shape of a cuboid.

Here, length (l) = 4 cm, breadth (b) = 2.5 cm and height (h) = 1.5 cm

∴ Volume of a matchbox = l × b × h

= 4 × 2.5 × 1.5 cm

^{3}

= 15 cm

^{3}

⇒ Volume of 12 such boxes = 12 × 15 cm

^{3}= 180 cm

^{3}

**Ex
13.5 Class 9 Maths 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)

**Solution:
**Here, Length (l) = 6 m, breadth (h) = 5 m and depth
(h) = 4.5 m

∴ Capacity = l × b × h = 6 × 5 × 4.5 m

^{3 }

= 30 × 4.5 = 135 m

^{3}

∵ 1 m

^{3}= 1000 litres

⇒ 135 m^{3} = 135000 litres

∴ The required amount of water in the tank = 135000 litres.

**Ex
13.5 Class 9 Maths 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:
**Here, Length (l) = 10 m, breadth (b) = 8 m and Volume
(V) = 380 m

^{3}

Let height of the cuboidal vessel be ‘h’.

Volume of the cuboidal vessel = l × b × h

⇒ 10 × 8 × h m

^{3}= 80h m

^{3}

⇒ 80h = 380

⇒ h = 380/80 = 4.75 m

Thus, the required height of the vessel = 4.75 m

**Ex 13.5 Class 9 Maths 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.

**Solution:
**Here, Length (l) = 8 m

Breadth (b) = 6 m

Depth (h) = 3 m

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

= 8 × 6 × 3 m

^{3}= 144 m

^{3}

Hence, the cost of digging a pit = ₹ (144 × 30)

= ₹ 4320

**Ex
13.5 Class 9 Maths 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 2.5 m and 10
m, respectively.

**Solution:
**Length of the tank (l) = 2.5 m

Depth of the tank (h) = 10 m

Let breadth of the tank be b m.

∴ Volume (capacity) of the tank = l × b × h

= 2.5 × b × 10 m

^{3}

= 25b m

^{3}

But
the capacity of the tank = 50000 litres

= 50 m^{3} [ ∵ 1000 litres = 1 m^{3}]

∴ 25b = 50 ⇒ b = 50/25 = 2

Thus, the breadth of the tank = 2 m

**Ex
13.5 Class 9 Maths 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:
**Length of the tank (l) = 20 m

Breadth of the tank (b) = 15 m

Height of the tank (h) = 6 m

∴ Volume of the tank = l × b × h

= 20 × 15 × 6 m

^{3}= 1800 m

^{3 }

Since, 1 m

^{3}= 1000 litres

∴ Capacity of the tank = 1800 × 1000 litres = 1800000 litres

Since, 150 litres of water is required per head per day.

∴ Amount of water required by 4000 people per day = 150 × 4000 litres

Let the required number of days be x.

∴ 4000 × 150 × x = 1800000

∴ 600000 × x = 1800000

⇒ x = 1800000/600000 =
3

Thus, the required number of days is 3.

**Ex 13.5 Class 9 Maths Question 7.
**A go down measures 40 m × 25 m × 10 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 go down.

**Solution:
**Volume of the go down = 40 × 25 × 10 m

^{3 }

Volume of a wooden crate = 1.5 × 1.25 × 0.5 m

^{3 }

**Ex 13.5 Class 9 Maths 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:
**Side of the given cube = 12 cm

Volume of the given cube = (side)

^{3}= (12)

^{3}cm

^{3 }= 1728 cm

^{3}

Let the side of the new cube be n.

∴ Volume of new cube = n^{3}

⇒ Volume of 8 new cubes = 8n^{3}

According to question,

8n^{3} = 1728

⇒ n^{3} = 1728/8 = 216

⇒ n^{3} = 6^{3}

⇒ n = 6

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

Now, surface area of the given cube = 6 × (side)^{2} = 6 × 12^{2} cm^{2} =
6 × 12 × 12 cm^{2}

Surface area of new cube = 6 × 6^{2} cm^{2}

= 6 × 6 × 6 cm^{2}

Now,

**Ex 13.5 Class 9 Maths 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:****
**The water flowing in a river can be
considered in the form of a cuboid.

Breadth (b) = 40 m,

Depth (h) = 3 m

∴ Volume of water (volume of the cuboid so formed) = l × b × h

= 2000 × 40 × 3 m

^{3 }

Now, volume of water flowing in 1 hr (60 minutes) = 2000 × 40 × 3 m

^{3}

Volume of water that will fall in 1 minute

= [2000 × 40 × 3] ÷ 60 m

^{3}

= 4000 m

^{3}

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