The following pie charts shows the percentage distribution of total students selected in SSC examination and percentage distribution of total boys selected students from six different zones. Study the following pie chart carefully and answer the given questions.

Find the sum of number of boys selected from zone B, zone C and zone E together and number of girls selected from zone A & zone D together?

The correct answer is 5680.

From the given pie charts, we can gather the following primary data points:

1. Total number of students selected = 12000

2. Total number of boys selected = 7000

Step 1: Calculate the total number of boys selected from Zone B, Zone C, and Zone E together.

From the second pie chart (Number of total boys selected = 7000):

- Percentage of boys selected from Zone B = 10%

- Percentage of boys selected from Zone C = 22%

- Percentage of boys selected from Zone E = 15%

Total percentage of boys from Zone B, Zone C, and Zone E together = 10% + 22% + 15% = 47%

Number of boys selected from Zone B, Zone C, and Zone E together:

$\text{Boys}\left(\text{B+C+E}\right)=47\%\text{ of }7000=\frac{47}{100}\times 7000=3290$

Step 2: Calculate the total number of girls selected from Zone A and Zone D together.

To find the number of girls in any zone, we subtract the number of boys from the total number of students selected in that zone.

For Zone A:

- Total students from Zone A = 25% of 12000 =

$\frac{25}{100}\times 12000=3000$

- Boys from Zone A = 30% of 7000 =

$\frac{30}{100}\times 7000=2100$

- Girls from Zone A = Total students in Zone A - Boys in Zone A = 3000 - 2100 = 900

For Zone D:

- Total students from Zone D = 20% of 12000 =

$\frac{20}{100}\times 12000=2400$

- Boys from Zone D = 13% of 7000 =

$\frac{13}{100}\times 7000=910$

- Girls from Zone D = Total students in Zone D - Boys in Zone D = 2400 - 910 = 1490

Total girls selected from Zone A & Zone D together = 900 + 1490 = 2390

Step 3: Find the required sum.

Sum = (Boys from Zone B, C & E) + (Girls from Zone A & D)

$\text{Sum}=3290+2390=5680$

Hence, the required sum is 5680.