What was the average number of candidates who appeared from zone T over all the years together?

810

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815

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825

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805

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820

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Solution

The correct option is D 805
Required average $= \frac{(6.2 + 6.2 + 6.4 + 7.8 + 9.9 + 11.8)}{6} \times 100 = \frac{48.30}{6} \times 100 = \frac{4830}{6} = 805$

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What was the average number of candidates who appeared from zone T over all the years together?

The correct option is D 805Required average =(6.2+6.2+6.4+7.8+9.9+11.8)6×100=48.306×100=48306=805

The correct option is D 805
Required average $= \frac{(6.2 + 6.2 + 6.4 + 7.8 + 9.9 + 11.8)}{6} \times 100 = \frac{48.30}{6} \times 100 = \frac{4830}{6} = 805$