Question

hey guys i wanna quick solution with full explanation if dont know no need to answer ​

Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}$

Here the concept of TSA of Cuboid and Area of Rectangle has been used. We are given the dimensions of the room. First we need to find the area to be painted. For area required to paint we will calculate TSA of Cuboid and subtract area of base because is not painted. Then we shall apply it in formula and find the answer.

Let's do it !!

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★ Equations Used :-

$\\\;\boxed{\sf{TSA\;of\;Cuboid\;=\;\bf{2(LB\;+\;BH\;+\;LH)}}}$

$\\\;\boxed{\sf{Area\;of\;Rectangle\;=\;\bf{Length\;\times\;Breadth}}}$

$\\\;\boxed{\sf{Number\;of\;baskets\;needed\;=\;\bf{\dfrac{Area\;to\;be\;painted}{Area\;covered\;by\;1\;Bucket}}}}$

$\\\;\boxed{\sf{Total\;cost\;=\;\bf{Cost\;per\;Bucket\;\times\;No.\;of\;Bucket}}}$

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★ Solution :-

Given,

» Length of Room = L = 20 m

» Breadth of Room = B = 5 m

» Height of Room = H = 10 m

» Area covered by each bucket = 5 m²

» Cost of Each Bucket = Rs. 20

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~ For Area to be Painted ::

$\\\;\;\;\sf{:\Rightarrow\;\;Area\;to\;be\;Painted\;=\;\bf{TSA\;of\;Room\;-\;Area\;of\;Base}}$

We also know that,

$\\\sf{:\rightarrow\;\;TSA\;of\;Cuboid\;=\;\bf{2(LB\;+\;BH\;+\;LH)}}$

$\\\sf{:\rightarrow\;\;Area\;of\;Rectangle_{(Base)}\;=\;\bf{Length\;\times\;Breadth}}$

Now by applying values in main equation, we get,

$\\\;\;\;\sf{:\Rightarrow\;\;Area\;to\;be\;Painted\;=\;\bf{2(LB\;+\;BH\;+\;LH)\;-\;(L\;\times\;B)}}$

$\\\;\;\;\sf{:\Rightarrow\;\;Area\;to\;be\;Painted\;=\;\bf{2[(20\;\times\;5)\;+\;(5\;\times\;10)\;+\;(20\;\times\;10)]\;-\;(20\;\times\;5)}}$

$\\\;\;\;\sf{:\Rightarrow\;\;Area\;to\;be\;Painted\;=\;\bf{2[100\;+\;50\;+\;200]\;-\;(20\;\times\;5)}}$

$\\\;\;\;\sf{:\Rightarrow\;\;Area\;to\;be\;Painted\;=\;\bf{2[350]\;-\;(100)}}$

$\\\;\;\;\sf{:\Rightarrow\;\;Area\;to\;be\;Painted\;=\;\bf{700\;-\;(100)}}$

$\\\;\;\;\bf{:\Rightarrow\;\;Area\;to\;be\;Painted\;=\;\bf{600\;\;m^{2}}}$

$\\\;\underline{\boxed{\tt{Area\;\;to\;\;Painted\;=\;\bf{600\;\;m^{2}}}}}$

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~ For Number of Buckets required ::

$\\\;\;\;\sf{:\Longrightarrow\;\;Number\;of\;baskets\;needed\;=\;\bf{\dfrac{Area\;to\;be\;painted}{Area\;covered\;by\;1\;Bucket}}}$

$\\\;\;\;\sf{:\Longrightarrow\;\;Number\;of\;baskets\;needed\;=\;\bf{\dfrac{600}{5}}}$

$\\\;\;\;\bf{:\Longrightarrow\;\;Number\;of\;baskets\;needed\;=\;\bf{120\;\;Buckets}}$

$\\\;\underline{\boxed{\tt{No.\;of\;\;Buckets\;\;required\;=\;\bf{120}}}}$

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~ For the Total Cost of Painting ::

$\\\;\;\;\sf{:\mapsto\;\;Total\;cost\;=\;\bf{Cost\;per\;Bucket\;\times\;No.\;of\;Bucket}}$

$\\\;\;\;\sf{:\mapsto\;\;Total\;cost\;=\;\bf{20\;\times\;120}}$

$\\\;\;\;\sf{:\mapsto\;\;Total\;cost\;=\;\bf{Rs.\;\;2400}}$

Hence, Option d.) 2400 is correct option.

$\\\;\large{\underline{\underline{\rm{Hence,\;total\;cost\;of\;painting\;is\;\;\boxed{\bf{Rs.\;\;2400}}}}}}$

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★ More to know :-

$\\\;\sf{\leadsto\;\;Volume\;of\;Cuboid\;=\;Length\;\times\;Breadth\;\times\;Height}$

$\\\;\sf{\leadsto\;\;Volume\;of\;Cube\;=\;(Side)^{3}}$

$\\\;\sf{\leadsto\;\;CSA\;of\;Cuboid\;=\;2(L\;+\;B)\;\times\;Height}$

$\\\;\sf{\leadsto\;\;CSA\;of\;Cube\;=\;4\;\times\;(Side)^{2}}$

$\\\;\sf{\leadsto\;\;TSA\;of\;Cube\;=\;6\;\times\;(Side)^{2}}$

Answer 2

$\large\purple{\boxed{\tt{Answer⤵}}}$

Area To be painted = 2[(20×5) + (5×10) + 20×5)]

Area to be painted = 2(350) - 100

Area to be painted = 700 - 100

$\large\pink{\boxed{\tt{Area\: to\: be\: painted=600m²}}}$

Buckets required :-

$\large\sf number \: of \: buckets \: required= \frac{area \: to \: be \: painted}{area \: covered \: by \: a \: bucket}$

$\large\tt buckets \: needed = \frac{600}{5}$

$\large\orange{\boxed{\tt{Buckets\: required=120\: Buckets}}}$

Total cost of painting :-

Total cost = Cost per bucket × Number of buckets

Total cost = 20 × 120

$\large\red{\boxed{\tt{Total\: Cost=Rs.2400}}}$