Question

Ravi donates 10 aluminum buckets to an orphanage. A bucket made of aluminum of
height 20 cm and has its upper and lower ends of radius 36 cm and 21 cm respectively.
(i) Find cost of preparing 10 buckets if the cost of aluminum sheets is 42 per
100cm2.
(ii) Find the quantity of water store in a bucket (In liter).​

Answer 1

Given:

Ravi donates 10 aluminium buckets to an orphanage

The dimensions of the frustum-shaped bucket:

Height = 20 cm

Upper-end radius = 36 cm

Lower-end radius = 21 cm

To find:

(i) Find the cost of preparing 10 buckets if the cost of aluminium sheets is 42 per  100 cm².

(ii) Find the quantity of water stored in a bucket (In litre).​

Solution:

(i) Finding the cost of preparing 10 buckets if the cost of aluminium sheets is 42 per  100 cm²:

Let's assume,

h = height of the frustum-shaped bucket = 20 cm

l = slant height of the frustum -shaped bucket

R = radius of the upper-end of the frustum = 36 cm

r = radius of the lower-end of the frustum = 21 cm

We will find the slant height of the frustum,  i.e.,

$l = \sqrt{(R -r)^2 + h^2} = \sqrt{(36-21)^2 + 20^2} = \sqrt{15^2 + 20^2} = \sqrt{625} = 25\: cm$

 

So,

The surface area of the aluminium sheet for making 1 bucket is given by,

= [Lateral surface area of the frustum] + [Area of the circular bottom]

= [ $\pi (R + r) l$ ] + [ $\pi r^2$ ]

=  $[\frac{22}{7} \times (36+21)\times 25] + [\frac{22}{7} \times 21^2]$

=   $[\frac{22}{7} \times 57\times 25] + [\frac{22}{7} \times 21^2]$

=   $[4478.57\:cm^2] + [1386 \:cm^2]$

=   $5864.57 \:cm^2$

∴ The surface area of the aluminium sheet for making 10 buckets is,

= 5864.57 cm² × 10

= 58645.71 cm²

Now,

If 100 cm² = Rs. 42

Then 58645.71 cm² = $\frac{42}{100} \times 58645.71$ = Rs. 24,631.20

Thus, the cost of preparing 10 buckets if the cost of aluminium sheets is 42 per  100 cm² is Rs. 24,631.20.

(ii) Finding the quantity of water stored in a bucket (In litre):

To find the quantity of water in the bucket, we have to find the volume of the frustum-shaped bucket.

∴ The volume of the bucket is,

= $\frac{1}{3} \pi h(r^2 + rR + R^2)$

= $\frac{1}{3} \times \frac{22}{7} \times 20\times (21^2 + (36\times 21) + 36^2)$

= $\frac{1}{3} \times \frac{22}{7} \times 20\times (441 + 756 + 1296)$

= $\frac{1}{3} \times \frac{22}{7} \times 20\times 2493$

= $52234.28 \:cm^3$

We know that,

If 1 cm³ = $\frac{1}{1000}$ litre

Then, 52234.28 cm³ = $\frac{52234.28}{1000}$ = 52.23 litres

Thus, the quantity of water stored in a bucket (in litre) is 52.23 litres.

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