Question

A cuboidal oil tin is 30cm by 40cm by 50cm find the cost of the tin required for making 20 such tins if the cost of tin sheet is Rs, 20 per Square metre.​

Answer 1

Answer:

Your Answer:-

Step-by-step explanation:

Cost required for making 20 tins with length, breadth and height of 30 cm, 40 cm and 50 cm respectively. Now multiply the total surface area of one tin with 20 tins. °•cost of making 20 cuboid tins of dimension 30cm × 40cm × 50cm is ₹ 376.

Answer 2

$\large\underline{ \underline{ \sf \maltese{ \: Question:- }}}$

• A cuboidal oil tin is 30cm by 40cm by 50cm find the cost of the tin required for making 20 such tins if the cost of tin sheet is Rs, 20 per Square metre.

$\large\underline{ \underline{ \sf \maltese\red{ \: Diagram:- }}}$

$\setlength{\unitlength}{0.74 cm}\begin{picture}\thicklines\put(5.6,5.4){\bf A}\put(11.1,5.4){\bf B}\put(11.2,9){\bf C}\put(5.3,8.6){\bf D}\put(3.3,10.2){\bf E}\put(3.3,7){\bf F}\put(9.25,10.35){\bf H}\put(9.35,7.35){\bf G}\put(3.5,6.1){\sf 30\:cm}\put(7.7,6.3){\sf 40\:cm}\put(11.3,7.45){\sf 50\:cm}\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\put(4,7.3){\line(1,0){5}}\put(4,10.3){\line(1,0){5}}\put(9,10.3){\line(0,-1){3}}\put(4,7.3){\line(0,1){3}}\put(6,6){\line(-3,2){2}}\put(6,9){\line(-3,2){2}}\put(11,9){\line(-3,2){2}}\put(11,6){\line(-3,2){2}}\end{picture}$

$\large\underline{ \underline{ \sf \maltese{ \: Solution:- }}}$

The cost of tins depend upon their Total Surface Area . It is given that a tin is in the shape of a Cuboid such that :-

• Length =$\red{\boxed{ \sf \blue{ 30cm }}}$
• Breadth =$\red{\boxed{ \sf \blue{ 40cm }}}$
• Height = $\red{\boxed{ \sf \blue{ 50cm }}}$

$\begin{gathered}\begin{gathered}\begin{gathered}: \implies \underline\blue{ \boxed{\displaystyle \sf \bold\orange{\: Surface \: Area \: of \: 1 \: Tin \: = 2(lb \: + \: bh \: + \: lh \: ) }} }\\ \end{gathered}\end{gathered}\end{gathered}$

$\quad {:} \longrightarrow \sf{\sf{2(30 \times 40 \: + \: 40 \times 50 \: + \: \: 30 \times 50) {cm}^{2} }}$

$\quad {:} \longrightarrow \sf{\sf{2(1200\: + \: 2000 \: + \: 1500) {cm}^{2} }}$

$\quad {:} \longrightarrow \sf{\sf{( 2 \: \times \: 4700 ) {cm}^{2} }}$

$\quad {:} \longrightarrow \sf{\sf{9400{cm}^{2} }}$

$\qquad\quad {:} \longrightarrow \underline \red{\boxed{\sf{Surface \: Area \: of \: 1 \: Tin \: = \: 9400 {cm}^{2} }}}$

$\begin{gathered}\begin{gathered}\begin{gathered}: \implies \underline\blue{ \boxed{\displaystyle \sf \bold\orange{\: Surface \: Area \: of \: 20 \: such \: Tins \: = \: 20 \: \times \: 9400 {cm}^{2} }} }\\ \end{gathered}\end{gathered}\end{gathered}$

$\quad {:} \longrightarrow \sf{\sf{188000 {cm}^{2} }}$

10000cm² = 1 m²

$\quad {:} \longrightarrow \sf{\sf{ \frac{188000}{10000} }}$

$\quad {:} \longrightarrow \sf{\sf{ \frac{188\cancel{000}}{10\cancel{000}}{m}^{2} }}$

$\qquad\quad {:} \longrightarrow \underline \red{\boxed{\sf{Surface \: Area \: of \: 20\: Tins \: = \: 18.8 {m}^{2} }}}$

Now, cost of 1 square metre of tin sheet = Rs, 20

Cost of 18.8m² of tin sheet = 20 × 18.8 = Rs, 376

$\begin{gathered}\begin{gathered}\qquad \therefore\: \sf{ </strong><strong>Cost</strong><strong> \:</strong><strong>of </strong><strong>\</strong><strong>:</strong><strong> </strong><strong>making</strong><strong> </strong><strong>\</strong><strong>:</strong><strong> </strong><strong>2</strong><strong>0</strong><strong> </strong><strong>\</strong><strong>:</strong><strong> </strong><strong>Tins </strong><strong>\</strong><strong>:</strong><strong> </strong><strong> = \underline {\underline{ </strong><strong>Rs,</strong><strong>3</strong><strong>7</strong><strong>6</strong><strong>}}}\\\end{gathered}\end{gathered}$