A field is $1700 c m$ long and $11$ m broad. A rectangular pit of dimensions $6 m \times 3.5 m \times 2 m$ is dug anywhere in this field and dug out soil is spread evenly over the remaining part of the field. The approximate rise in the level of the rest of the field is

11.23 c m

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12.65 c m

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9.57 c m

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10.1 c m

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Solution

The correct option is A $11.23 c m$
Length of the field $= 1700 c m$ $= 17 m$

Breadth of the field $= 11 m$

Area of the field $= 17 \times 11$

$= 187 m^{2}$

Volume of rectangular pit $= 6 m \times 3.5 m \times 2 m$

$= 42 m^{3}$

Area of rectangular pit $= 6 m \times 3.5 m$

$= 21 m^{2}$

Remaining area $= 187 - 21$

$= 166 m^{2}$

Rise in the field $= \frac{v o l u m e o f t h e s o i l}{r e m a i n i n g a r e a}$

$= \frac{42}{166}$

$= .253 m$

$= 25.3 c m$

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A field is 1700cm long and 11 m broad. A rectangular pit of dimensions 6m×3.5m×2m is dug anywhere in this field and dug out soil is spread evenly over the remaining part of the field. The approximate rise in the level of the rest of the field is

A field is $1700 c m$ long and $11$ m broad. A rectangular pit of dimensions $6 m \times 3.5 m \times 2 m$ is dug anywhere in this field and dug out soil is spread evenly over the remaining part of the field. The approximate rise in the level of the rest of the field is