Tabulation DI (Data Interpretation) Online Test - 1

Ans: D

Number of boys in School D in 2007 = (6.4 x 100) = 640.

Number of boys in School D in 2008 = (5.3 x 100) = 530.

Decrease % = $$({110} / {640} x 100)$$% = $${275} / {16}$$% = 17.1% = 17%.

Ans: D

Number of girls in School B in 2009 = (5.9 x 100) = 590.

Total number of students in School E in 2006 = (5.5 + 3.6) × 100 = (9.1 × 100) = 910.

Required % = $$({590} / {910} × 100)$$% = $${5900} / {91}$$%. = 64.8% = 65%(nearly).

Ans: D

Average number of girls in School A over the years

=$${(3.6 + 4.2 + 6.9 + 9.6 + 12.9 + 14.4) × 100} /{6}$$

= $$({51.6 × 100} / {6})$$ = $${5160} / {6}$$ = 860.

Ans: B

(Number of boys in C in 2009) : (Number of girls in A in 2009)

= (8.7 x 100) : (12.9 x 100)

= $${870} / {1290}$$ = $${29} / {43}$$ = 29 : 43.

Ans: B

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, 2nd highest in 2009 and third highest in 2005.