Question

The circumference of the road roller is 30.8 m and its length is 7 m. Find the
of revolutions made by it to level an area of 11858 m.
55 m by 5 m by 6 m is dug and the earth taken outis​

Answer 1

Answer:

$\red {number \: revolutions \: made}\green {= 55}$

Step-by-step explanation:

$Given \: Circumference \: of \: the \: road \\roller (C) = 30.8 \:m$

$Length (l) = 7\:m$

$Area \: levelled (A) = 11858 \: m^{2}$

$Let \: number \: revolutions \: made = n$

$\implies n = \frac{A}{C\times l}\\= \frac{11858}{30.8\times 7} \\= 55$

Therefore.,

$\red {number \: revolutions \: made}\green {= 55}$

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