## NCERT solutions for class 9 Maths

** Surface Areas and Volumes**

**Chapter 13 **

**Exercise 13.5**

**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:**

Matchbox is in the form of cuboid.

Length (l) = 4 cm, breadth (b) = 2.5 cm

and height (h) = 1.5 cm

∴ Volume of a matchbox = l x b x h

= 4 x 25 x 1.5 cm^{3}

= 4 x x cm^{3}

= 15 cm^{3}

⇒ Volume of 12 match boxes = 12 x 15 cm^{3}

= 180 cm^{3}

**Ex 13.5 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} = 1000L)**

**Solution:**

Length (l) =6 m, breadth (h) =5 m and

depth (h) = 4.5 m

∴ Capacity =l x b x h = 6 x 5 x 4.5m

^{3}

= 6 x 5 x 4.5 = 135m

^{3}

∵ 1 m

^{3}= 1000 litres

⇒ 135 m

^{3}= 135000 litres

∴ The water in the tank hold = 135000 litres.

**Ex 13.5 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:**

Length (l) = 10 m, breadth (b) = S m

Volume(V) = 380m^{3}

Let the height of the cuboidal vessel = h m

Volume of the cuboidal vessel = l x b x h

⇒ 10 x 8 x h = 80h m^{3}

⇒ 80h = 380

⇒ h = = 4.75m

Thus, the height of the chhodkar vessel = 4.75 m

**Ex 13.5 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:**

Length of cuboidal (i) = 8m

Breadth (b) = 6 m

Depth (h) = 3 m

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

= 8 x 6 x 3 = 144 m^{3}

Hence, the cost of digging a pit = ₹ 144 x 30

= ₹ 4320

**Ex 13.5 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 cuboidal tank (l) = 2.5 m

Depth (h) = 10 m

Let breadth of the tank = b m.

∴ Volume v = 50000 liters = 50 m^{3}. [ ∵ 1000 litres =1 m^{3} ]

⇒50 = 2.5 x b x 10

⇒ 50 = x 10 x b

⇒ 50 = 25b

⇒ b = = 2 m

Thus, the breadth of cuboidal tank = 2 m

**Ex 13.5 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 x 15 m x 6 m. For how many days will the water of this tank last?**

**Solution:**

Length (l) = 20 m

Breadth (b) = 15 m

Height (h) = 6m

∴ Volume of the tank = l x b x h

= 20 x 15 x 6 m^{3} = 1800 m^{3}

= 1800 x 1000 litres = 1800000 litres [ 1m^{3} = 1000 litres]

Daily requirement of water = 150 litres

The number of days to consume all water of tank = volume of tank / volume of water for 4000 person for one day

= 1800000/ 600000 = 3

**Ex 13.5 Question 7.**

**A go down measures 40 m x 25 m x 10 m. Find the maximum number of wooden crates each measuring 15 m x 125 m x 0.5 m that can be stored in the go down.**

**Solution:**

Volume of the go down = 40 x 25 x 10 m^{3}

Volume of a wooden crate = 1.5 x 1.25 x 0.5 m3

∴ Maximum number of wooden crates = 10667.

**Ex 13.5 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 larger cube A = 12 cm

Volume = (side)^{3} = (12)^{3} = 12 x 12 x 12 = 1728cm^{3}

According to question , Volume of small cube = a^{3} = volume of larger cube / 8

= 1728 / 8 = 216 cm^{3}

a^{3} = 216 cm^{3}⇒ a = 6 cm

Thus, the side of new cube = 6 cm

Now,

Ratio of aur face area of two cubes = surface adda of larger cube / surface adda of smaller cube

= 6 A^{2}/ 6a^{2}

= A^{2}/a^{2}

=12^{2} / 6^{2}

= 144 / 36

= 4 : 1

**Ex 13.5 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:**

Length (l) = 2 km = 2000 m

Breadth (b) = 40 m,

depth (h) = 3 m

Volume of water fall into the sea in 1 hr (60 minutes) = 2000 x 40 x 3 = 240000 m^{3}

Volume of water fall into the sea in 1 minute

= 240000/ 60 = 4000m^{3}