Question

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?

Answer 1

Area of field=13:5×2:5=33:75

Upper area of tank=5×4:5=22:5

Area of field on which soil is spread=33:75_22:5

=11:25m   Volume of tank=5×4:5×2:1=47:25m^3

Raised in level=volume of tank/area of field on which

soil spread=47:25/11:25=4:2

So,raised  level=4:2m

Answer 2

Answer:

We have ,

Area of the field = (13.5 × 2.5) m² = 33.75 m²

Volume of the earth dug out = Vol. of the tank

= (5 × 4.5 × 2.1) m³ = 47.25 m³

Area of the tank = (5 × 4.5) m² = 22.5 m²

_____________________

Area on which the earth dug out has been spread

= area of the field - area of the tank = (33.75 - 22.5) m² = 11.25 m²

$\therefore$ Level raised

$\sf = \frac{area \: of \: the \: earth \: dug \: out}{area \: of \: which \: the \: earth \: is \: spread } \\ \\ = \sf \frac{47.25}{11.25} = 4.2m$