Question

Hameed has built a cubical water tank with lid for his house, with each outer edge 1.5 m long. He gets the outer surface of the tank excluding the base, covered with square tiles of side 25 cm (see in the figure below). Find how much he would spend on the tiles if the cost of the tiles is Rs.360 per dozen.

$\huge \red{ \mid{ \underline{ \overline{ \tt SPAMMERS STAY AWAY ❌ }} \mid}}$

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Answer 1

★ Concept :-

Here the concept of CSA of Cube and Area of square has been used. We see that we are given the length of each side of the cubical tank and also we known that each dimension is equal in length. So firstly we can find the CSA of cubical tank which will give area of four sides and add the area of one more side that is top side. Then we will divide it with area of each tile to get number of tiles.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{\pink{CSA\;of\;Cube\;=\;\bf{4\:a^{2}}}}}$

$\\\;\boxed{\sf{\pink{Area\;of\;Square\;=\;\bf{(Side)^{2}}}}}$

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★ Solution :-

» Length of each edge of tank = a = 1.5 m

» Side of each tile = 25 cm = 0.25 m

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~ For Area to be Tiled ::

✒ Area to be tiled = CSA of Tank + Area of one side

• CSA of Tank ::

We know that,

$\\\;\sf{\rightarrow\;\;CSA\;of\;Cube\;=\;\bf{4\:a^{2}}}$

By applying values, we get

$\\\;\sf{\rightarrow\;\;CSA\;of\;Cubical\;Tank\;=\;\bf{4\:(1.5)^{2}}}$

$\\\;\sf{\rightarrow\;\;CSA\;of\;Cubical\;Tank\;=\;\bf{4\:(2.25)}}$

$\\\;\bf{\rightarrow\;\;CSA\;of\;Cubical\;Tank\;=\;\bf{\orange{9\;\;cm^{2}}}}$

• Area of one side of tank ::

We know that,

$\\\;\sf{\rightarrow\;\;Area\;of\;Square\;=\;\bf{(Side)^{2}}}$

By applying values, we get

$\\\;\sf{\rightarrow\;\;Area\;of\;one\;side\;of\;Tank\;=\;\bf{(1.5)^{2}}}$

$\\\;\bf{\rightarrow\;\;Area\;of\;one\;side\;of\;Tank\;=\;\bf{\blue{2.25\;\;cm^{2}}}}$

Now,

✒ Area to be tiled = 9 cm² + 2.25 cm²

✒ Area to be tiled = 11.25 cm²

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~ For the Area of each tile ::

We know that,

$\\\;\sf{\rightarrow\;\;Area\;of\;Square\;=\;\bf{(Side)^{2}}}$

By applying values,

$\\\;\sf{\rightarrow\;\;Area\;of\;each\;Tile\;=\;\bf{(0.25)^{2}}}$

$\\\;\bf{\rightarrow\;\;Area\;of\;each\;Tile\;=\;\bf{\red{0.0625\;\;cm^{2}}}}$

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~ For number of dozens tiles required ::

Firstly we shall find out the number of tiles required. This is given as,

$\\\;\sf{\Longrightarrow\;\;Number\;of\;Tiles\;=\;\bf{\dfrac{Area\;to\;be\;tiled}{Area\;of\;each\;tile}}}$

By applying values, we get

$\\\;\sf{\Longrightarrow\;\;Number\;of\;Tiles\;=\;\bf{\dfrac{11.25}{0.0625}}}$

$\\\;\bf{\Longrightarrow\;\;Number\;of\;Tiles\;=\;\bf{\green{180\;\;Tiles}}}$

• Now we need to find out the number of dozens of tiles. This is given as,

$\\\;\sf{\Longrightarrow\;\;Number\;of\;dozens\;of\;tiles\;=\;\bf{\dfrac{Number\;of\;Tiles}{12}}}$

By applying values, we get

$\\\;\sf{\Longrightarrow\;\;Number\;of\;dozens\;of\;tiles\;=\;\bf{\dfrac{180}{12}}}$

$\\\;\sf{\Longrightarrow\;\;Number\;of\;dozens\;of\;tiles\;=\;\bf{15\;\;Dozens}}$

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~ For Total cost of tiles ::

This is given as,

✒ Total cost of tiles = Number of dozens × Rate

By applying values, we get

✒ Total cost of tiles = 15 × 360

✒ Total cost of tiles = Rs. 5400

$\\\;\underline{\boxed{\tt{Total\;\:cost\:\;of\:\;tiles\;=\;\bf{\purple{Rs.\:\;5400}}}}}$

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★ More to know :-

$\\\;\sf{\leadsto\;\;TSA\;of\;Cuboid\;=\;6a^{2}}$

$\\\;\sf{\leadsto\;\;Volume\;of\;Cuboid\;=\;a^{3}}$

$\\\;\sf{\leadsto\;\;Perimeter\;of\;Square\;=\;4\:\times\:Side}$

$\\\;\sf{\leadsto\;\;Diagonal\;of\;Cuboid\;=\;\sqrt{3a^{3}}}$

$\\\;\sf{\leadsto\;\;Diagonal\;of\;Square\;=\;a\sqrt{2}}$

Answer 2

$\begin{gathered}\begin{gathered}\bf Given \: - \begin{cases} &\sf{Edge \: of \: water \: tank, \: a = \: 1.5 m} \\ &\sf{Side \: of \: square \: tile, \: x = 25 cm } \\ &\sf{cost \: of \: the \: tiles \: is \: Rs. \: 360 \: per \: dozen.} \end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\begin{gathered}\bf To \: Find \: - \begin{cases} &\sf{Amount \: spend \: on \: tiles } \end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\begin{gathered}\bf Formula \: Used :- \begin{cases} &\sf{Curved \: Surface \: Area_{(Cube)} =4 {a}^{2} } \\ &\sf{Area \: of \: square \: = {x}^{2} } \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf where= \begin{cases} &\sf{a \: = \: edge \: of \: cube} \\ &\sf{x \: = \: side \: of \: square} \end{cases}\end{gathered}\end{gathered}$

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$\large\underline\purple{\bold{Solution :- }}$

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$\large\underline\red{\bold{❥︎Step :- 1 }}$

☆ Edge of cubical water tank, a = 1.5 m = 150 cm.

☆Since, Hameed have to put tiles to the four walls and the upper lid. So area to be covered with tiles is

$\bf \:  ⟼ Area_{(covered \: with \: tiles )} = 5 {a}^{2}$

$\bf \:  ⟼ Area_{(covered \: with \: tiles )} = 5 \times {(150)}^{2} \: {cm}^{2}$

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$\large\underline\red{\bold{❥︎Step :- 2 }}$

☆ Edge of square tile, x = 25 cm

$\bf \:  ⟼ ☆ Area_{(square \: tile)} = {x}^{2} = {(25)}^{2} \: {cm}^{2}$

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$\large\underline\red{\bold{❥︎Step :- 3 }}$

$\sf \:  ⟼Number \: of \: tiles \: required \: = \dfrac{ Area_{(covered \: with \: tiles )} }{Area_{(square \: tile)}}$

$\sf \:  ⟼Number \: of \: tiles \: required \: = \dfrac{5 \times \: ^6 \: \: \cancel{150} \times \: ^6\cancel{150}}{ \cancel{25} \times \cancel{25}}$

$\sf \:  ⟼Number \: of \: tiles \: required \: = 180$

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$\large\underline\red{\bold{❥︎Step :- 4 }}$

$\sf \:  ⟼☆ \: Cost \: of \: tile \: is \: ₹ 360 \: per \: dozen$

$\sf \:  ⟼☆Cost \: of \: 1 \: tile \: = \dfrac{ 360 }{12} = ₹ \: 30$

$\sf \:  ⟼☆Cost \: of \: 180 \: tiles \: =180 \times 30 = ₹ \: 5400$

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$\large{\boxed{\boxed{\bf{Hence, \: Amount \: spend \: = ₹ \: 5400}}}}$

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