Answer:
Option D
Explanation:
Number of students in:
A → $\left(\frac{12}{100}\times6000\right)$ = 720
B → $\left(\frac{9}{100}\times6000\right)$ = 540
C → $\left(\frac{26}{100}\times6000\right)$ = 1560
D → $\left(\frac{18}{100}\times6000\right)$ = 1080
E → $\left(\frac{29}{100}\times6000\right)$ = 1740
F → $\left(\frac{6}{100}\times6000\right)$ = 360
Number of boys in :
A → 500, B → 400, C → 900, D → 600, E → 1200, F → 100.
Number of girls in:
A → (720 - 500) = 220; B → (540 - 400) 140; C → (1560 - 900) = 660
D → (1080 - 600) = 480; E → (1740 - 1200) = 540; F → (360 - 100) = 260.
Required sum = (660 + 540 + 600) = 1800
