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 ?

Answers (1)

Solution:  We have , 

               Area of the field =(13.5	imes2.5)m^2=33.75m^2

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

               Area of the tank =(5	imes 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

	herefore          Level raised = volume of the earth dug out / Area of which the earth is spread    =frac47.2511.25=4.2m

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