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Bunuel
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Given weekend attendence is 50% of average attendence for preciding 5 days
If 5x was the attendence for 5 days then average attendence is x. Hence for weekend the attendence is 0.5x
total attendence 5.5x

Hence percent of weekend address as a whole is 0.5x/5.5x = 1/11 = 9 percent (Option A)
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Language trap question, better to solve by creating table

let Avg attendance for 5 day=X
total attendance for 5 day =5x
weekend attendance=x*.50=.50x
total attendance= 5x+.50x=5.50x
% of weekend attendance= .50x/5.50x*100= 9%
option A
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Great breakdown of the classic average vs. total trap. The key trick here is remembering that the 5-day total volume is 5 × Average, so the overall 7-day sum becomes 5.5A.

Misinterpreting averages as total counts is actually a very common error when setting up the underlying math for an attendance calculator, as averaging raw percentages always distorts the true weighted ratio. Spotting that 0.5 / 5.5 = 1/11 ≈ 9.09% makes Option A a quick pick without needing long division.
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