A rectangular field is 120 m*60m.a tank 25 m*8 m*4 m is dug in the field.The earth taken out is spread evenly over the remaining area of the field.How much is the level of field raised if on digging the volume of earth is increased by 1/20 of its volume

Answer :

As given dimension of Tank  = 25 m $\times$ 8 m $\times$ 4 m 
So,
Volume of tank  =  Volume of earth increased

And

Volume of tank  =  25 $\times$  8 $\times$  4   =   800 m³

Let Height increased =  h

So,

Volume of rectangular field  = 120 $\times$ 60  $\times$  h   =  7200h

As given : The volume of earth is increased by $\frac{1}{20}$ of its volume  , So
$$\begin{gathered} \frac{1}{20}\times 7200h = 800 \\ \Rightarrow 360h = 800 \\ \Rightarrow h =\frac{ 800}{360} \\ \mathbf{\Rightarrow }\mathbf{ }\mathit{h}\mathbf{ }\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{22}\mathbf{ }\mathit{m}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathit{A}\mathit{n}\mathit{s}\mathbf{ }\mathbf{)} \end{gathered}$$