1. The sum of a series of 5 consecutive odd numbers is 195. The second lowest number of this series is 9 less than the second highest number of another series of 5 consecutive even numbers. What is 40% of the second lowest number of the series of consecutive even numbers?
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By: anil on 05 May 2019 01.37 am
Let the five consecutive odd numbers in increasing order = $$(x-4) , (x-2) , (x) , (x+2) , (x+4)$$ Sum of these numbers = $$(x-4) + (x-2) + (x) + (x+2) + (x+4) = 195$$ => $$5x = 195$$ => $$x = frac{195}{5} = 39$$ Thus, the odd numbers are = 35 , 37 , 39 , 41 , 43 Let another series of even numbers in increasing order = $$(y-4) , (y-2) , (y) , (y+2) , (y+4)$$ Also, $$37 = (y + 2) - 9$$ => $$y = 37 + 9 - 2 = 44$$ Thus, second lowest number of the even series = 44 - 2 = 42 $$ herefore$$ 40% of 42 = $$frac{40}{100} imes 42 = 16.8$$
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