If the average of a, b and c is 11, the average of c, d and e is 17, the average of e and f is 22 and the average of e and c is 17, then the average of a, b, c, d, e and f will be:

$15 \frac{2}{3}$

average a, b and c = a+b+c/3 = 11

a+b+c = 33

the average of c, d and e is 17, c+d+e/3 = 17

c+d+e = 51

the average of e and f = e+f/2 = 22

e+f = 44

average of e and c = e+c/2 = 17

e+c = 34

average a,b,c,d,e and f = (a+b+c+d+e+f)/6 = [(a+b+c)+(c+d+e)+(e+f)-(e+c)] = 33+51+44-34 = 94/6

The correct answer is Option (1) → $15 \frac{2}{3}$