Data Interpretation MCQ Questions and Answers - Table Chart

Number of girls in School B in 2009
= 5.9 × 100
= 590
Total Number of students in School E in 2006
= (5.5 + 3.6) × 100
= 9.1 × 100
= 910
$$\eqalign{ & \text{Required }\% \cr & = \left(\frac{590}{910}\times100\right)\% \cr & = \frac{5900}{91}\% \cr & = 64.8\% \cr & \approx 65\% \cr} $$

Total number of students in School E in various years:
2005 → (6.5 + 6.6) × 100
            = 1310
2006 → (5.5 + 3.6) × 100
            = 910
2007 → (2.7 + 2.4) × 100
            = 510
2008 → (5.4 + 5.7) × 100
            = 1110
2009 → (6.8 + 6.5) × 100
            = 1330
2010 → (10.8 + 12.7) × 100
            = 2350
It is highest in 2010, 2^nd highest in 2009 and 3^rd highest in 2005.

Average number of girls in School A over the years
$$\eqalign{ & = \frac{\left(3.6+4.2+6.9+9.6+12.9+14.4\right)\times100}{6} \cr & = \frac{51.6\times100}{6} \cr & = \frac{5160}{6} \cr & = 860 \cr} $$

Number of boys in School D in 2007
= 6.4 × 100
= 640
Number of boys in School D in 2008
= 5.3 ×100
= 530
$$\eqalign{ & \text{Decrease }\% \cr & = \left(\frac{110}{640}\times100\right)\% \cr & = \frac{275}{16}\% \cr & = 17.1\% \cr & \approx 17\% \cr} $$

(Number of boys in C in 2009) : (Number of girls in A in 2009)
$$\eqalign{ & = \left(8.7\times100\right) : \left(12.9\times100\right) \cr & = \frac{870}{1290} \cr & = \frac{29}{43} \cr & = 29 : 43 \cr} $$