**1. What is the percentage of candidates qualified from State N for all the years together, over the candidates appeared from State N during all the years together?**

**Answer**: Option D

**Explanation**:

Required Percentage

=\(\frac{(840 + 1050 + 920 + 980 + 1020)}{(7500 + 9200 + 8450 + 9200 + 8800) * 100}\)

= \(\frac{4810}{43150} * 100\) = 11.15%

**2. What is the average of candidates who appeared from State Q during the given years?**

**Answer**: Option D

**Explanation**:

\(\frac{(8100 + 9500 + 8700 + 9700 + 8950)}{15}\)

= \(\frac{44950}{5}\)= 8990

**3. Total number of candidates qualified from all the states together in 1997 is approximately what percentage of the total number of candidates qualified from all the states together in 1998?**

**Answer**: Option C

**Explanation**:

\(\frac{(720 + 840 + 780 + 950 + 870)}{(980 + 1050 + 1020 + 1240 + 940) * 100}\)%

= \(\frac{4160}{5230) * 100}\)* 100 % = 79.54% = 80%

**4. The percentage of the total number of qualified candidates to the total number of appeared candidates among all the five states in 1999 is:**

**Answer**: Option B

**Explanation**:

Required Percentage

= \(\frac{(850 + 920 + 890 + 980 + 1350)}{(7400 + 8450 + 7800 + 8700 + 9800) * 100}\)%

= \(\frac{4990}{42150}\)% = 11.84%

**5. Combining the states P and Q together in 1998, what is the percentage of the candidates qualified to that of the candidates appeared?**

**Answer**: Option C

**Explanation**:

Required percentage

= \(\frac{((1020 + 1240)}{(8800 + 9500) * 100}\)%

= (\(\frac{2260}{18300 * 100}\)* 100)% = 12.35%

**1. How many girls more participated in year 2004 as compared to 2003 for all schools taken together?**

**Answer**: Option A

**Explanation**:

Number of girls participating in 2004

= 100 + 140 + 120 + 120 = 480

number of girls participants in 2003

= 90 + 120 + 70 + 90 = 370

required number = 480 – 370 = 110

**2. The ratio of boys: girls for school B over the years 2001-2004 is**

**Answer**: Option C

**Explanation**:

Number of boys participants for school B = 280 + 300 + 420 + 480 = 1480

number of girls participants for school B = 60 + 80 + 120 + 140 = 400

required ratio = 1480 : 400 = 37:10

**3. The percentage increase in participation of boys from school B in 2004 over those in 2001 is nearly**

**Answer**: Option B

**Explanation**:

Number of boys participants in 2004 from school B = 480

Number of boys participants in 2001 from school B = 280

Percentage increase = \(\frac{(480 – 280)}{280}\) * 100% = 71%

**4. The total number of participants in year 2003 is**

**Answer**: Option C

**Explanation**:

Number of participation in 2003

= (340 + 90) + (420 + 120) + (230 + 70) + (360 + 90)

= 1720

= (\(\frac{495.4}{713.36}\) * 100) % Ëœ 69.45%

**5. In year 2004 the number of participating girls is what percent of number of participating boys?**

**Answer**: Option D

**Explanation**:

Number of girls participants in 2004 = 100 + 140 + 120 + 120 = 480

number of boys participants in 2004 = 370 + 480 + 360 + 500 = 1710

required ratio = 480 : 1710 = 16 : 57

**1. For the candidates from which of the following locations was there continuous increase both in appeared and passed?**

**Answer**: Option D

**Explanation**:

It is clear from visual inspection that there is no continuous increase both in appeared and passed in any location.

**2. In which of the following years was the percentage qualified to appeared candidates from Semi-urban area the least?**

**Answer**: Option D

**Explanation**:

Percentage of qualified to appeared candidates from Semi-urban area in 2005 = \(\frac{2513}{7894}\)* 100 = 31.83% Percentage of qualified to appeared candidates from Semi-urban area in 2006 = \(\frac{2993}{8562}\)* 100 = 34.25% Percentage of qualified to appeared candidates from Semi-urban area in 2007 = 2468/8139 * 100 = 30.32% Percentage of qualified to appeared candidates from Semi-urban area in 2008 = 3528/9432 * 100 = 37.40% Percentage of qualified to appeared candidates from Semi-urban area in 2009 = 4015/9784 * 100 = 41.03% Percentage of qualified to appeared candidates from Semi-urban area in 2010 = 4263/9969 * 100 = 42.76%

**3. Approximately, what was the percentage drop in the number of Semi-urban candidates appeared from 2006 to 2007?**

**Answer**: Option A

**Explanation**:

Required percentage

= \(\frac{(8562 – 8139)}{8562}\) * 100 = 4.94% = 5%

**4. In 2008 the percentage of candidates qualified to appeared was approximately 35 from which location?**

**Answer**: Option A

**Explanation**:

In 2008 the percentage of candidates qualified to appeared in Rural area = \(\frac{1798}{50322}\) * 100 = 35.73%

In 2008 the percentage of candidates qualified to appeared in Semi-urban area = \(\frac{3528}{9432}\) * 100 = 37.40%

In 2008 the percentage of candidates qualified to appeared in State capital area = \(\frac{3628}{8529}\) * 100 = 42.53%

In 2008 the percentage of candidates qualified to appeared in Metropolises area = \(\frac{5138}{11247}\) * 100 = 45.86%

Therefore, the most appropriate answer is Rural area.

**5. The total number of candidates qualified from Rural in 2008 and Semi-urban in 2005 was exactly equal to the total number of candidates qualified from State-capitals in which of the following years?**

**Answer**: Option C

**Explanation**:

The number of candidates qualified from Rural area in 2008 = 1798

The number of candidates qualified from Semi-urban area in 2005 = 2513

Total sum = 1798 + 2513 = 4311

The number of candidates qualified from State-capitals in 2009 = 4311