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

Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.5 Textbook Exercise Questions and Answers.

RBSE Class 9 Maths Solutions 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:
Here l = 4 cm, b = 2.5 cm and h = 1.5 cm.
∴ Volume of one matchbox = l × b × h
= (4 × 2.5 × 1.5) cm³ = 15 cm³
∴ Volume of a packet containing 12 such boxes
= (12 × 15) cm³ = 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:
Here l = 6 m, b = 5 m and h = 4.5 m.
∴ Volume of the tank = lbh
= (6 × 5 × 4.5) m³ = 135 m³
The tank can hold = 135 × 1000 litres (∵ 1 m³ = 1000 litres)
= 135000 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:
Here, length = 10 m, breadth = 8 m and volume = 380 m³
Height = \(\frac{\text { Volume of cuboid }}{\text { Length } \times \text { Breadth }}=\left(\frac{380}{10 \times 8}\right)\)m = 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:
Here, l = 8 m, 6 = 6 m and h = 3 m.
∴ Volume of the pit = lbh
= (8 × 6 × 3) m³ = 144 m³

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 mand 10 in.
Answer:
Here, length = 2.5 m, depth = 10 m
volume = 50000 litres
= (5000 × \(\frac{1}{1000}\))m³
= 50 m³
∴ Breath = \(\frac{\text { Volume of cuboid }}{\text { Length } \times \text { Depth }}=\left(\frac{50}{2.5 \times 10}\right)\)m = 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:
Here, l = 20 m, b = 15 m and h = 6 m.
Capacity of the tank = lbh
= (20 × 15 × 6) m³ = 1800 m³

Water requirement per person per day = 150 litres Water required for 4000 persons per day = (4000 × 150)
= \(\left(\frac{4000 \times 150}{1000}\right)\)m³ = 600 m³

∴Number of days the water will last = \(\frac{\text { Capacity of tank }}{\text {Total water required per day }}\)
= \(\left(\frac{1800}{600}\right)\) = 3

Thus, the water will last for 3 days.

Question 7.
A godown measures 40 m × 25 m × 15 wooden crates each measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown.
Answer:
Volume of the godown = (40 × 25 × 15) m³ = 15000 m³
Volume of 1 crate = (1.5 × 1.25 × 0.5) m³ = 0.9375 m³
Number of crates that can be stored in the godown
= \(\frac{\text { Volume of the godown }}{\text { Volume of } 1 \text { crate }}\)
= \(\frac{15000}{0.9375}\) = 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.
Answer:
Let V₁ = Volume of the cube of edge 12 cm
= (12 × 12 × 12) cm³
and V₂ = Volume of each of the small cubes cut out of the first one
= \(\frac{1}{8}\) × V₁ = (\(\frac{1}{8}\) × 12 × 12 × 12)cm³
= (6 × 6 × 6) cm³
∴ Side of the new cube = 6 cm
Ratio of their surface areas = \(\frac{6(\text { side })^{2}}{6(\text { side })^{2}}=\frac{6 \times 12 \times 12}{6 \times 6 \times 6}=\frac{4}{1}\) i.e. 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 a minute?
Answer:
Since the water flows at the rate of 2 km per hour, the water from 2 km of river flows into the sea in one hour.
The volume of water flowing into the sea in one hour
= Volume of the cuboid
= l × b × h = (2000 × 40 × 3) m³
∴ The volume of water flowing into the sea in one minute
= \(\left(\frac{2000 \times 40 \times 3}{60}\right)\) m³ = 4000 m³