Data Interpretation MCQ Questions and Answers - Table Chart

Average number of candidates appeared from T all over the years
$$\eqalign{ & = \frac{\left(6.2+6.2+6.4+7.8+9.9+11.8\right)\times100}{6} \cr & = \frac{48.3\times100}{6} \cr & = \frac{4830}{6} \cr & = 805 \cr} $$

In zone S, the difference between the appeared candidates and the qualified candidates in various years is given below:
2005 → (4.2 - 2.4) = 1.8
2006 → (7.4 - 3.3) = 4.1
2007 → (8.3 - 5.6) = 2.7
2008 → (9.3 - 6.4) = 2.9
2009 → (11.4 - 5.2) = 6.2
2010 → (14.2 - 11.4) = 2.8
If was lowest in 2005 and second lowest in 2007

Total number of candidates who qualified the test in various zones are:
Total number of qualified students in P = 2.5 + 3.4 + 4.9 + 5.7 + 4.8 + 5.6 ⇒ 26.9

Total number of qualified students in Q = 1.4 + 4.2 + 4.5 + 3.4 + 5.2 + 6.2 ⇒ 25.1

Total number of qualified students in R = 2.2 + 4.2 +4.8 + 3.2 + 6.8 + 7.4 ⇒ 28.8

Total number of qualified students in S = 2.4 + 3.3 + 5.6 + 6.4 + 5.2 + 11.4 ⇒ 34.3

Total number of qualified students in T = 2.6 +4.8 +4.2 + 6.2 + 9.4 + 6.9 ⇒ 34.1

 34.1 is the second highest

∴ Zone T is the second highest in total number of qualified students. 

$$\eqalign{ & \text{Required percentage} \cr & = \left(\frac{7.4}{5.4}\times100\right)\% \cr & = \frac{3700}{27}\% \cr & = 137\% \cr} $$

$$\eqalign{ & \text{Required ratio} \cr & = \frac{3.2}{5.6} \cr & = \frac{32}{56} \cr & = \frac{4}{7} \cr & = 4:7 \cr} $$