1. Merlin walks certain distance at (9/10)th of the usual speed and takes 15 minutes more than the usual time. Find the usual time taken.
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By: anil on 05 May 2019 01.14 am
Let usual speed = $$10$$ m/min and usual time taken = $$t$$ min => New speed = $$9$$ m/min and new time = $$(t+15)$$ min Also, speed is inversely proportional to time. => $$frac{10}{9}=frac{t+15}{t}$$ => $$10t=9t+135$$ => $$10t-9t=t=135$$ $$ herefore$$ Usual time taken = 135 minutes => Ans - (B)
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