Question

1.How many bricks each 10 cm long, 7.5 cm wide and 2.5 cm thick will be required to build a wall 30 m long, 2 m high and 0.25 m thick? If the bricks sell at ₹1500 per thousand, how much will it cost to build the wall?

2. How many cubic metres of earth must be dug to construct a 25-m-deep well of diameter 5.6 m? If the earth taken out is spread over a rectangular plot of size 22 m×16 m, find the height of the platform so formed.​

Answer 1

✴ Given :-

• ➬ Dimensions of brick = 10 cm × 7.5 cm × 2.5 cm.
• ➬ Dimensions of wall = 30 m × 2 m × 0.25 m
• ➬ Cost of 1000 bricks = ₹ 1500

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✴ To Find :-

• ➬ Total cost of bricks = ?

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✴ Solution :-

❒ How to Solve :

We will find the Volume of the wall and after that volume of brick. On dividing volume of wall / Volume of brick we will get no. of bricks and than we can find the cost. Let's solve :

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❒ Formula Used :

$\large{\blue{\bigstar}}{\underline{\boxed{\pink{\sf{Volume{\small_{(Cuboid) }} = Length \times Width \times Height}}}}}$

❒ Calculating the volumes :

➢Volume of wall( in cm³) :

$\: \: \: \: \: {\hookrightarrow{\sf{ Volume{\small_{(Wall) }} = L \times W \times H}}}$

$\: \: \: \: \: {\hookrightarrow{\sf{ Volume{\small_{(Wall) }} = 3000\times 200 \times 25}}}$

$\: \: \: \: \: {\hookrightarrow{\sf{ Volume{\small_{(Wall) }} = 600000 \times 25}}}$

$\qquad\large{\red{:\longmapsto{\underline{\overline{\boxed{\green{\sf{15000000\: cm³}}}}}}}}$

➢Volume of Brick :

$\: \: \: \: \: {\hookrightarrow{\sf{ Volume{\small_{(brick) }} = L \times W \times H}}}$

$\: \: \: \: \: {\hookrightarrow{\sf{ Volume{\small_{(brick) }} = 10 \times 7.5\times 2.5}}}$

$\: \: \: \: \: {\hookrightarrow{\sf{ Volume{\small_{(brick) }} = 75\times 2.5}}}$

$\qquad\large{\red{:\longmapsto{\underline{\overline{\boxed{\green{\sf{187.5\: cm³}}}}}}}}$

$\qquad{━━━━━━━━━━━━━━━━━━━━━━━━━━}$

❒ No. of bricks required :

$\qquad{\twoheadrightarrow{\sf{No. \: of \: bricks = \dfrac{Vol. \: of \: wall}{Vol. \: of \: bricks}}}}$

$\qquad{\twoheadrightarrow{\sf{No. \: of \: bricks = \dfrac{15000000}{187.5}}}}$

$\qquad{\twoheadrightarrow{\sf{No. \: of \: bricks = \cancel\dfrac{15000000}{187.5}}}}$

$\qquad\large{\red{:\longmapsto{\underline{\overline{\boxed{\green{\sf{80000\: bricks}}}}}}}}$

$\qquad{━━━━━━━━━━━━━━━━━━━━━━━━━━}$

❒ Cost of Bricks :

$\qquad{\twoheadrightarrow{\sf{ \dfrac{No. \: of \: bricks}{1000} \times 1500}}}$

$\qquad{\twoheadrightarrow{\sf{ \dfrac{80000}{1000} \times 1500}}}$

$\qquad{\twoheadrightarrow{\sf{ \dfrac{80\cancel{000}}{\cancel{1000}} \times 1500}}}$

$\qquad{\twoheadrightarrow{\sf{ 80\times 1500}}}$

$\qquad\large{\red{:\longmapsto{\underline{\overline{\boxed{\green{\sf{₹120000}}}}}}}}$

${\color{darkblue}{\underline{▬▬▬▬▬}}{\pink{\underline{▬▬▬▬▬}}{\orange{\underline{▬▬▬▬▬}}{\purple{\underline{▬▬▬▬▬▬}}}}}}$

Answer 2

★ Concept :-

Here concept of volume is used. In first question firstly we will find volume of each brick and volume of wall, then we will divide volume of wall by volume of brick to get number of bricks used, then we will easily find cost of bricks. In second question firstly we will find volume of earth taken out, then we will compare it with volume of earth spread on rectangular plot to get height of platform.

Let's do it !!

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★ Formulae used :-

$\large\:{\boxed{\sf{\pink{Volume_{(cuboid)} = \bf{LBH}}}}}$

$\large\:{\boxed{\sf{\pink{Volume_{(cylinder)} = \bf{\pi r^2h}}}}}$

$\rule{200}2$

★ Solution :-

⠀⠀⠀⠀⠀⠀☯ Question ①

Given,

~ Dimensions of brick ::

»» Length of brick (L) = 10 cm

»» Breadth of brick (B) = 7.5 cm

»» Height of brick (H) = 2.5 cm

~ Dimensions of wall ::

»» Length of wall (L) = 30 m = 3000 cm

»» Breadth of wall (B) = 0.25 m = 25 cm

»» Height of wall (H) = 2 m = 200 cm

$\rule{200}2$

~ Finding volume of brick ::

$\\ \longrightarrow\quad\sf Volume_{(brick)} = LBH$

$\\ \longrightarrow\quad\sf Volume_{(brick)} = (10)(7.5)(2.5)$

$\\ \longrightarrow\quad\sf Volume_{(brick)} = 10\:\times\:7.5\:\times\:2.5$

$\\ \longrightarrow\quad\bf Volume_{(brick)} = \orange{187.5\:cm^3}$

~ Finding volume of wall ::

$\longrightarrow\quad\sf Volume_{(wall)} = LBH$

$\\ \longrightarrow\quad\sf Volume_{(wall)} = (3000)(25)(200)$

$\\ \longrightarrow\quad\sf Volume_{(wall)} = 3000\:\times\:25\:\times\:200$

$\\ \longrightarrow\quad\bf Volume_{(wall)} = \green{15000000\:cm^3}$

~ Finding number of bricks ::

$\\ \implies\quad\sf Number\:of\:bricks = \dfrac{Volume_{(wall)}}{Volume_{(brick)}}$

$\\ \implies\quad\sf Number\:of\:bricks = {\cancel{\dfrac{15000000}{187.5}}}$

$\\ \implies\quad\bf Number\:of\:bricks = \blue{80000}$

$\\ \underline{\boxed{\tt{Hence,\:number\:of\:bricks = {\bf{\purple{80000}}}}}}$

~ Finding cost of bricks ::

$\\ \implies\quad\sf Cost\:of\:bricks = 80\cancel{000}\:\times\:\dfrac{1500}{1\cancel{000}}$

$\\ \implies\quad\sf Cost\:of\:bricks = 80\:\times\:1500$

$\\ \implies\quad\bf Cost\:of\:bricks = \pink{120000}$

$\\ \underline{\boxed{\tt{Hence,\:Cost\:of\:bricks = {\bf{\purple{Rs.\:120000}}}}}}$

$\rule{200}2$

⠀⠀⠀⠀⠀⠀⠀☯ Question ②

Given,

~ Dimensions of well ::

»» Depth (height) of well (h) = 25 m

»» Diameter of well (d) = 5.6 m

»» Radius of well (r) = 2.8 m

~ Dimensions of plot ::

»» Length of plot (L) = 22 m

»» Breadth of plot (B) = 16 m

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~ Finding volume of earth taken out ::

$\\ \dashrightarrow\quad\sf Volume_{(earth\:taken\:out)} = \pi r^2h$

$\\ \dashrightarrow\quad\sf Volume_{(earth\:taken\:out)} = \dfrac{22}{7}\:\times\:(2.8)^2\:\times\:25$

$\\ \dashrightarrow\quad\sf Volume_{(earth\:taken\:out)} = \dfrac{22}{\cancel{7}}\:\times\:\cancel{7.84}\:\times\:25$

$\\ \dashrightarrow\quad\sf Volume_{(earth\:taken\:out)} = 22\:\times\:1.12\:\times\:25$

$\\ \dashrightarrow\quad\bf Volume_{(earth\:taken\:out)} = 616\:cm^3$

Now, we know that earth taken out is spread on rectangular plot to make a platform. Therefore,

$\\ \longmapsto\quad\sf Volume_{(platform)} = Volume_{(earth\:taken\:out)}$

$\\ \longmapsto\quad\sf LBH = 616$

$\\ \longmapsto\quad\sf (22)(16)H = 616$

$\\ \longmapsto\quad\sf 22\:\times\:16\:\times\:H = 616$

$\\ \longmapsto\quad\sf 352\:\times\:H = 616$

$\\ \longmapsto\quad\sf H = \dfrac{616}{352}$

$\\ \longmapsto\quad\bf H = \red{1.75}$

$\\ \underline{\boxed{\tt{Hence,\:height\:of\:platform = {\bf{\purple{1.75\:m}}}}}}$

$\rule{200}2$

★ More to know :-

↠ TSA of cube = 6a²

↠ CSA of cube = 4a²

↠ Volume of cube = a³

↠ TSA of cuboid = 2(lb + bh + hl)

↠ CSA of cuboid = 2(l + b)h

↠ Volume of cuboid = l × b × h

↠ TSA of cylinder = 2πr(r + h)

↠ CSA of cylinder = 2πrh

↠ Volume of cylinder = πr²h

↠ TSA of cone = πr(l + r)

↠ CSA of cone = πrl

↠ Volume of cone = 1/3πr²h

↠ SA of sphere = 4πr²

↠ Volume of sphere = 4/3πr³

↠ TSA of hemisphere = 3πr²

↠ CSA of hemisphere = 2πr²

↠ Volume of hemisphere = 2/3πr³