A well of 4m diameter and 35m deep is dugout and the excavated soil is transported in a rectangular parallelopiped shaped truck with dimensions $$5m\times2m\times0.5m$$. To avoid over during transportation only 80% of its capacity is filled. If the loose soil occupies 20% more space while filling into the truck, then the number of trips required to transport the soil completely away from the place of digging is
Solution
Volume of well$$=Ο2^2Γ35=440m^3.$$
Volume of truck$$=5Γ2Γ0.5=5m^3.$$
but the truck is only able to fill its 80% of capacity to fill the space.
So, effective amount was$$=0.8Γ5=4m^3.$$
So,20% of wasted amount is
$$=0.2Γ440=88m^3.$$
So, number of trips required are
$$=(440/4)+(88/4)=110+22=132.$$
D is correct choice.
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