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Question -

A field is in the form of a rectangle of length 18 m and width 15 m. A pit, 7.5 m long, 6 m broad and 0.8 m deep, is dug in a comer of the field and the earth taken out is spread- over the remaining area of the field. Find out the extent to which the level of the field has been raised.



Answer -

Length of a field (L) = 18m
and width (B) = 15 m
Length of pit (l) = 7.5 m
Breadth (b) = 6 m
and depth (h) = 0.8 m
∴ Volume of earth dugout = lbh
= 7.5 x 6 x 0.8 m3
= 45 x 0.8 == 36 m3
Total area of the field = L x B
= 18 x 15 = 270 m2
and area of pit = lb = 7.5 x 6 = 45 m2
∴ Remaining area of the field excluding pit
= 270 – 45 = 225 m2
Let by spreading the earth on the remaining part of the field, the height = h
= 225 x h = 36
⇒ h =  = 0.16 m = 16 cm
∴ Level of field raised = 16 cm

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