Directions (1 - 5): Study the given table carefully and answer the question that follow:
Number of candidates appeared and qualified for a test (in hundreds) in 6 different years from 5 different zones.
Note: Here App. means Appeared and Qual. Means qualified.
| Year | P | Q | R | S | T | |||||
| App. | Qual. | App. | Qual. | App. | Qual. | App. | Qual. | App. | Qual. | |
| 2005 | 3.2 | 2.5 | 3.5 | 1.4 | 3.8 | 2.2 | 4.2 | 2.4 | 6.2 | 2.6 |
| 2006 | 4.6 | 3.4 | 6.9 | 4.2 | 6.9 | 4.4 | 7.4 | 3.3 | 6.2 | 4.8 |
| 2007 | 6.5 | 4.9 | 7.7 | 4.5 | 5.9 | 4.8 | 8.3 | 5.6 | 6.4 | 4.2 |
| 2008 | 7.4 | 5.7 | 5.4 | 3.4 | 7.2 | 3.2 | 9.3 | 6.4 | 7.8 | 6.2 |
| 2009 | 8.8 | 4.8 | 6.6 | 5.2 | 8.6 | 6.8 | 11.4 | 5.2 | 9.9 | 6.9 |
| 2010 | 9.2 | 5.6 | 10.6 | 6.4 | 10.3 | 7.4 | 14.2 | 11.4 | 11.8 | 9.4 |
What was the average number of candidates appeared from zone T over all the years together?
A. 810
B. 815
C. 825
D. 805
E. 820
Answer: Option D
Solution(By Examveda Team)
Average number of candidates appeared from T all over the years$$\eqalign{ & = \frac{\left(6.2+6.2+6.4+7.8+9.9+11.8\right)\times100}{6} \cr & = \frac{48.3\times100}{6} \cr & = \frac{4830}{6} \cr & = 805 \cr} $$
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