A man donates 10 aluminum buckets to an orphanage. A bucket made of aluminum is of height 20 cm and has its upper and lowest ends of radius 36 cm and 21 cm respectively. Find the cost of preparing 10 buckets if the cost of an aluminum sheet is Rs. 42 per $100c{m^2}$.
Answer
644.4k+ views
Hint: The bucket is in the form of a frustum with the base. So, the area of the bucket will be the sum of the area of the frustum and the base. For the area of the frustum, we have to calculate the slant height. After that multiply by 10 to get the surface area of 10 buckets. Then multiply the area by $\dfrac{{42}}{{100}}$ to get the cost of preparing the buckets.
Complete step-by-step solution:
Given: - Height of the bucket is $h = 20$ cm.
The upper radius of the bucket is $R = 36$ cm.
The lower radius of the bucket is $r = 21$ cm.
The cost of an aluminum sheet per 100 sq cm is Rs 42.
Now, the slant height of the frustum is given by
$ \Rightarrow l = \sqrt {{h^2} + {{\left( {R - r} \right)}^2}} $
Substitute the values,
$ \Rightarrow l = \sqrt {{{20}^2} + {{\left( {36 - 21} \right)}^2}} $
Subtract the values in the bracket,
$ \Rightarrow l = \sqrt {{{20}^2} + {{15}^2}} $
Square the terms in the square root,
$ \Rightarrow l = \sqrt {400 + 225} $
Add the terms in the square root,
$ \Rightarrow l = \sqrt {625} $
Simplify the terms,
$ \Rightarrow l = 25$cm
Now, the area of the bucket is given by,
$ \Rightarrow A = \pi l\left( {R + r} \right) + \pi {r^2}$
Substitute the values,
$ \Rightarrow A = \dfrac{{22}}{7} \times 25 \times \left( {36 + 21} \right) + \dfrac{{22}}{7} \times {\left( {21} \right)^2}$
Add the term and square the term in the bracket,
$ \Rightarrow A = \dfrac{{22}}{7} \times 25 \times 57 + \dfrac{{22}}{7} \times 441$
Simplify the terms,
$ \Rightarrow A = \dfrac{{31350}}{7} + \dfrac{{9702}}{7}$
As the denominator is the same, add the terms in the numerator,
$ \Rightarrow A = \dfrac{{41052}}{7}c{m^2}$
Multiply the area by 10 to get the area of 10 buckets,
$ \Rightarrow A = \dfrac{{41052}}{7} \times 10c{m^2}$
Multiply the terms,
$ \Rightarrow A = \dfrac{{410520}}{7}c{m^2}$
As we know that the cost of an aluminum sheet is Rs 42 per 100 sq cm. Then,
$ \Rightarrow $ Cost of preparing 10 buckets = Area of the 10 buckets $ \times $ Cost of aluminum sheet per sq cm
Substitute the values,
$ \Rightarrow $ Cost of preparing 10 buckets $ = \dfrac{{410520}}{7} \times \dfrac{{42}}{{100}}$
Simplify the terms,
$ \Rightarrow $ Cost of preparing 10 buckets $ = 24631.20$
Hence, the cost of preparing 10 buckets is Rs 24631.20.
Note: One should remember that the curved surface area is the area of the solid shape excluding the top and bottom cross-section areas. The common mistake made in these problems is forgetting to put the units.
Complete step-by-step solution:
Given: - Height of the bucket is $h = 20$ cm.
The upper radius of the bucket is $R = 36$ cm.
The lower radius of the bucket is $r = 21$ cm.
The cost of an aluminum sheet per 100 sq cm is Rs 42.
Now, the slant height of the frustum is given by
$ \Rightarrow l = \sqrt {{h^2} + {{\left( {R - r} \right)}^2}} $
Substitute the values,
$ \Rightarrow l = \sqrt {{{20}^2} + {{\left( {36 - 21} \right)}^2}} $
Subtract the values in the bracket,
$ \Rightarrow l = \sqrt {{{20}^2} + {{15}^2}} $
Square the terms in the square root,
$ \Rightarrow l = \sqrt {400 + 225} $
Add the terms in the square root,
$ \Rightarrow l = \sqrt {625} $
Simplify the terms,
$ \Rightarrow l = 25$cm
Now, the area of the bucket is given by,
$ \Rightarrow A = \pi l\left( {R + r} \right) + \pi {r^2}$
Substitute the values,
$ \Rightarrow A = \dfrac{{22}}{7} \times 25 \times \left( {36 + 21} \right) + \dfrac{{22}}{7} \times {\left( {21} \right)^2}$
Add the term and square the term in the bracket,
$ \Rightarrow A = \dfrac{{22}}{7} \times 25 \times 57 + \dfrac{{22}}{7} \times 441$
Simplify the terms,
$ \Rightarrow A = \dfrac{{31350}}{7} + \dfrac{{9702}}{7}$
As the denominator is the same, add the terms in the numerator,
$ \Rightarrow A = \dfrac{{41052}}{7}c{m^2}$
Multiply the area by 10 to get the area of 10 buckets,
$ \Rightarrow A = \dfrac{{41052}}{7} \times 10c{m^2}$
Multiply the terms,
$ \Rightarrow A = \dfrac{{410520}}{7}c{m^2}$
As we know that the cost of an aluminum sheet is Rs 42 per 100 sq cm. Then,
$ \Rightarrow $ Cost of preparing 10 buckets = Area of the 10 buckets $ \times $ Cost of aluminum sheet per sq cm
Substitute the values,
$ \Rightarrow $ Cost of preparing 10 buckets $ = \dfrac{{410520}}{7} \times \dfrac{{42}}{{100}}$
Simplify the terms,
$ \Rightarrow $ Cost of preparing 10 buckets $ = 24631.20$
Hence, the cost of preparing 10 buckets is Rs 24631.20.
Note: One should remember that the curved surface area is the area of the solid shape excluding the top and bottom cross-section areas. The common mistake made in these problems is forgetting to put the units.
Recently Updated Pages
Master Class 10 Computer Science: Engaging Questions & Answers for Success

Differentiate between voluntary action and reflex class 10 biology CBSE

The uses of bleaching powder are A It is used bleaching class 10 chemistry CBSE

Fill in the blanks with abstract nouns of the words class 10 english CBSE

How many threedigit numbers are there class 10 maths CBSE

What is a reflex arc class 10 biology CBSE

Trending doubts
Explain the Treaty of Vienna of 1815 class 10 social science CBSE

10 examples of evaporation in daily life with explanations

Which country is known as "The land of Fire and Ice"?

What is the full form of POSCO class 10 social science CBSE

Make a sketch of the human nerve cell What function class 10 biology CBSE

In which year voting age was reduced from 21 to 18?

