# A rectangular tank measuring 5m * 4.5m * 2.1m is dug in the centre of the field measuring 13.5m * 2.5m . The earth dug out is spread evenly over the remaining portion of the field . How much is the level of the field raised ?

Solution:  We have ,

Area of the field $=(13.5\times2.5)m^2=33.75m^2$

Volume of the earth dug out = volume of the tank $=$ $(5\times 4.5 \times 2.1)m^3=47.25m^3$

Area of the tank $=(5\times 4.5)m^2=22.5m^2$

Area on which the earth dug out has been spread  $=$ Area of the field $-$ Area of the tank $=(33.75-22.5)m^2=11.25m^2$

$\therefore$          Level raised = volume of the earth dug out / Area of which the earth is spread    $=\frac{47.25}{11.25}=4.2m$

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