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(D) Let the number of bricks be $n$.Volume of the pile $= 20 \, m^3 = 20 	imes 100 	imes 100 	imes 100 \, cm^3 = 20,000,000 \, cm^3$.Volume of one

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$8000$

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(D) Let the number of bricks be $n$.Volume of the pile $= 20 \, m^3 = 20 	imes 100 	imes 100 	imes 100 \, cm^3 = 20,000,000 \, cm^3$.Volume of one brick $= 	ext{length} 	imes 	ext{breadth} 	imes 	ext{height} = 25 \, cm 	imes 12.5 \, cm 	imes 8 \, cm$.$25 	imes 12.5 = 312.5$.$312.5 	imes 8 = 2500 \, cm^3$.Since there is no gap between the bricks, the total volume of the pile is equal to the total volume of $n$ bricks:$n 	imes (	ext{Volume of one brick}) = 	ext{Volume of the pile}$.$n 	imes 2500 = 20,000,000$.$n = \frac{20,000,000}{2500} = \frac{200,000}{25} = 8000$.Therefore, the number of bricks is $8000$.

Some bricks are arranged in an area measuring $20 \, m^3$. If the length, breadth, and height of each brick are $25 \, cm$, $12.5 \, cm$, and $8 \, cm$ respectively, then the number of bricks in that pile is (assume there is no gap between two bricks):

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