The dimensions of a field 15 m by 12 m. A pit 7.5 M long 6 metre wide and 1.5 m Deep is dug at one corner off the field, the earth removed is evenly spread over the remaining area of the field calculate the rise in the level of the field.

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Solution

Given a pit 7.5 M long 6 metre wide and 1.5 m Deep.
So, volume of earth in pit $= 7.5 \times 6 \times 1.5$
$= 67.5 m^{3}$
And also area of the pit $= 7.5 \times 6 = 45 m^{2}$
The dimensions of a field 15 m by 12 m.
So, area of the field = $15 \times 12 = 180 m^{2}$
Now, Area of remaining field A = $180 - 45 = 135 m^{2}$
Let height of field increased be x
Therefore, Volume of pit = $A \times x$
$\Rightarrow 67.5 = 135 \times x$
$x = \frac{67.5}{135}$
$∴ x = 0.5 m$

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