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Exercise 13.5

1. A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What will be the volume of a packet containing 12 such boxes?

Sol. Volume of the packet = 12 × volume of one match box
= 12 × 4 × 2.5 × 1.5 cm³ = 180 cm³.

2. A cuboidal, water tank is 6 m long, 5 in wide and 4.5 m deep. How many litres of water can it hold? (1 m³ = 1000 l)

Sol. Volume of water tank = 6 × 5 × 4.5 = 135 m³.

∴ Water it can hold = 135 × 1000 l = 135000 l.

3. A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic meters of a liquid?

Sol. Let height of the vessel be h metres.

Volume of the vessel = 380 m³

⇒ 10 × 8 × h = 380 ⇒ h =  4.75 m.

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 m³.

Sol. Volume of the cuboidal pit = 8 × 6 × 3 m³ = 144 m³.

Cost of digging the pit = Rs.30 × 144 = Rs.4,320.

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 in and 10 m.

Sol.  Let the breadth of the tank be b metres.
Capacity of the tank = 50000 l.

Volume of the tank = ${{50000} \over {1000}}{m^3} = 50{m^3}$

∴  2.5 × b × 10 = 50 ⇒ b = 2 m.

6. A village, having a population of 4000, requires 150 liters 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?

Sol. Volume of the tank = 20 × 15 × 6  m³  = 1800 m³.

Volume of water required per day = ${{4000 \times 150} \over {1000}}{m^3}$
= 600 m³.

∴ Number of days water will last = ${{1800} \over {600}} = 3days$

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 in that can be stored in the godown.

Sol. Number of wooden crates = ${{40 \times 25 \times 10} \over {1.0 \times 1.25 \times 0.5}} \approx 16000$

Hence, 16000 wooden crates can be stored in the godown.

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.

Sol. Volume of the original cube = (12)³ cm³ = 1728 cm³.

Volume of one small cube = ${{1728} \over 8}$ cm² = 216 cm².

∴ Side of small cube = $\root 3 \of {216}$ cm = 6 cm.

Surface area of the original cube / Surface area of a small cube $= {{6{{(12)}^2}} \over {6{{(6)}^2}}} = {4 \over 1}$
Hence, the required ratio is 4 : 1.

9. A river 3 m deep and 40 m wide is flowing at the rate of 2 km per how: How much water will fall into the sea in a minute?

$= {{2000} \over {60}}m = {{100} \over 3}m$
= $= 3 \times 40 \times {{100} \over 3}{m^3} = 4000{m^3}$