**QUESTIONS-:**

*1. A matchbox measures 5cm × 1cm × 3.5cm. Determine the volume of a packet containing 14 such boxes.*

*Soln.*

Given the dimension of the matchbox = 5cm × 1cm × 3.5cm

Let us assume, l = 5cm, b = 1cm, h = 3.5cm

As we know that, Volume of one matchbox = (l × b × h)

= \(\left ( 5\times 1\times 3.5 \right )cm^{3}=17.5cm^{3}\)

\(∴ The\:volume\:of\:a\:packet\:containing\:14\:such\:boxes=\left ( 17.5\times 14 \right )cm^{3}=245cm^{3}\)

**2 . **

*A cuboidal water tank is 10 m long, 2 m wide and 7.5 m deep. How many litres of water can it hold? (\(1m^{3}\) = 1000 l)**Soln.*

Dimensions of water tank = 10m × 2m × 7.5m

Let us assume, l = 10m, b = 2m, h = 7.5m

Therefore Volume of the tank = \(\left ( l\times b\times h \right )m^{3}\)

= \(\left ( 10\times 2\times 7.5 \right )m^{3}=150m^{3}\)

Hence, the tank can hold = 150 x 1000 litres = 150000 litres of water.

*3. A cuboidal vessel is 25 m long and 12 m wide. Determine the height that must be made to hold 400 cubic metres of a liquid.*

*Soln.*

Given, Length = 25 m , Breadth = 12 m and Volume = 400\(m^{3}\)

As we know, Volume of cuboid = Length x Breadth x Height

Therefore, Height = Volume of cuboid/(Length × Breadth)

= \(\frac{400}{25\times 12}m=1.33m\)

4*. Find the value of digging a cuboidal pit 15 m long, 5 m broad and 3 m deep at the rate of Rs.50 per \(m^{3}\).*

*Soln.*

Here, length = 15m, breadth = 5m and height = 3m

As we know that, Volume of the pit = \(\left ( l\times b\times h \right )m^{3}\)

= \(\left (15\times 5\times 3 \right )m^{3}=225m^{3}\)

The rate of digging is = Rs.50 per \(m^{3}\)

∴ The total value of digging the pit = Rs.(225 x 50)

= Rs.11250

*5. The capacity of a cuboidal tank is 2,10000 litres of water. Calculate the breadth, given that the length is 3.5m and depth is 20m.*

*Soln.*

Given, length = 3.5m, depth = 15m and volume = 30000 litres

As we know that, \(1m^{3}=1000\:litres\)

\(∴ 210000\:litres=\frac{210000}{1000}m^{3}\)= 210\(1m^{3}\)

Breadth = \(\frac{volume\:of\:cuboid}{length\times depth}\)

= \(\frac{210}{\left ( 3.5\times 20 \right )}m\)

= 3m

*6. A village, with a population of 6000, requires 200 litres of water per head per day. It has a tank measuring 30m × 25m × 8m. Justify the number of days that will take to empty the water tank.*

*Soln.*

** **Given, the dimension of the tank = 30m × 25m × 8m

So, l = 30m, b = 25m and h = 8m

As we know that, the total capacity of the tank = \(\left ( 30\times25\times 8 \right )m^{3}=6000m^{3}\)

Water required for a single person per day = 200 litres

The requirement of water for 6000 person in a single day = (6000 x 200) litres

= \(\frac{\left ( 6000\times 200 \right )}{1000}=1200m^{3}\)

Hence, the number of days the water will last = (the capacity of the tank /water required per day) = \(\left ( \frac{6000}{1200} \right )=5\)

∴ The water lasts for 5 days.

*7. A warehouse measures 50m × 35m × 25m. Calculate the maximum number of wooden boxes each measuring 2.5m × 1.5m × 1m that can be stored in the warehouse.*

*Soln.*

** **Given the dimensions of the warehouse = 50m x 35m x 25m

As we know that, the volume of the warehouse will be = \(\left ( lbh \right ) m^{3}\)

= \(\left (50\times 35\times 25 \right ) m^{3}=43750m^{3}\)

Now, the dimension of box = 2.5m x 1.5m x 1m

Similarly, volume of 1 box = \(\left (2.5\times 1.5\times 1 \right ) m^{3}=3.75m^{3}\)

Hence, Number of box that can be stored = volume of warehouse / volume of 1 box = \(\frac{43750}{3.75}=11666.666=11666\)

*8. A solid cuboid having side 20 cm is cut into 16 cubes of equal volume. Calculate the side of the new cuboid and also calculate the ratio between their surface areas.*

*Soln.*

Here the edge of the cube = 20cm

So, Volume of the cuboid = \(\left ( edge \right )^{3}cm^{3}\)

= \(\left ( 20\times20\times 20 \right )cm^{3}=8000cm^{3}\)

Now, The number of smaller cube = 16

So, the volume of 1 small cube = \(\frac{8000}{16}cm^{3}=500cm^{3}\)

Let us assume the side of small cube as ‘p’

\(p^{3}=500\:\:\Rightarrow p=7.937\:\left ( approx \right )\)

Hence, the surface area the cube = \(7.937\left ( side \right )^{2}\)

Therefore, the ratio of their surface area

= (7.937 x 20 x 20)/(7.937 x 7.937 x 7.937)

= \(\frac{40}{1.585}\) = 40 :1.585

*9. A river 5 m deep and 60 m wide is flowing at a rate of 6 km per hour. Estimate the amount of water that will fall into the sea in a minute.*

*Soln.*

Given, Depth (h) = 5m

Width (b) = 60m

So, the rate of flow of water (l) = 6km per hour= \(\left ( \frac{6000}{60} \right )m\:per\:minute=100m\:per\:minute\)

Therefore, the volume of water flowing into the sea in a minute = \(lbh\:m^{3}\)

\(\left ( 100\times 60\times 5 \right )m^{3}=30000m^{3}\)