A wall 24 m long- 5 m high and 0-25 m thick to be constructed using bricks measuring 25cm - 12-5cm- 7-5 cm- Find the number of bricks required- if 5

The correct option is A 12,160Volume of cuboid =24 × 100 × 5 × 100 × 0.25 × 100 =30000000 5 percent #160;of cement and sand = 30000000 ×5 ...

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Standard VII

Physics

Volume of Regular Solids

A wall 24 m l...

Question

A wall 24 m long, 5 m high and 0.25 m thick to be constructed using bricks measuring $25 c m \times 12.5 c m \times 7.5$ cm. Find the number of bricks required, if 5% of the wall is occupied by cement and sand mixture.

12,160

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12,100

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12,150

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12,140

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Solution

The correct option is A 12,160
Volume of cuboid $= 24 \times 100 \times 5 \times 100 \times 0.25 \times 100$
$= 30000000$
$5$ percent of cement and sand $= 30000000 \times \frac{5}{100}$
$= 1500000$
Total wall is $30000000 - 1500000 = 28500000$
Volume of brick $= 25 \times 12.5 \times 7.5 = 2343.75$
Total number of bricks $= \frac{28500000}{2343.75} = 12, 160$

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A wall 24 m long, 5 m high and 0.25 m thick to be constructed using bricks measuring 25cm×12.5cm×7.5 cm. Find the number of bricks required, if 5% of the wall is occupied by cement and sand mixture.

The correct option is A 12,160Volume of cuboid =24×100×5×100×0.25×100=300000005 percent of cement and sand =30000000×5100=1500000Total wall is 30000000−1500000=28500000Volume of brick =25×12.5×7.5=2343.75Total number of bricks =285000002343.75=12,160

A wall 24 m long, 5 m high and 0.25 m thick to be constructed using bricks measuring $25 c m \times 12.5 c m \times 7.5$ cm. Find the number of bricks required, if 5% of the wall is occupied by cement and sand mixture.