1. The outer circumference of a circular race track is 528 metre. The track is every where 14 metre wide. Cost of levelling the track at the rate of ₹10 per sq. metre is:
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By: anil on 05 May 2019 02.19 am
Let outer radius = $$R$$ m and inner radius = $$r=(R-14)$$ m Outer circumference = $$2pi R=528$$
=> $$2 imesfrac{22}{7} imes R=528$$ => $$R=528 imesfrac{7}{44}=84$$ m Thus, inner radius = $$84-14=70$$ m => Area of track = $$pi(R^2-r^2)$$ = $$frac{22}{7}(R+r)(R-r)$$ = $$frac{22}{7}(84+70)(84-70)$$ = $$frac{22}{7} imes154 imes14=6776$$ $$m^2$$ $$ herefore$$ Total cost of levelling = $$6776 imes10=Rs.$$ $$67,760$$ => Ans - (D)
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=> $$2 imesfrac{22}{7} imes R=528$$ => $$R=528 imesfrac{7}{44}=84$$ m Thus, inner radius = $$84-14=70$$ m => Area of track = $$pi(R^2-r^2)$$ = $$frac{22}{7}(R+r)(R-r)$$ = $$frac{22}{7}(84+70)(84-70)$$ = $$frac{22}{7} imes154 imes14=6776$$ $$m^2$$ $$ herefore$$ Total cost of levelling = $$6776 imes10=Rs.$$ $$67,760$$ => Ans - (D)