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:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(15\frac{2}{3}\)
Turn each average into a total (average \(\times\) number of letters). The groups overlap, so they cannot just be added: c and e appear in two groups each. The sum of all six letters is built from the groups and the overlap c + e is taken away.
Average 11 of a, b, c: \(a+b+c=33\) (1)
Average 17 of c, d, e: \(c+d+e=51\) (2)
Average 22 of e, f: \(e+f=44\) (3)
Average 17 of e, c: \(c+e=34\) (4)
Adding (1), (2) and (3) counts c and e twice, so (4) is subtracted once:
Sum of all six: \(a+b+c+d+e+f=33+51+44-34=94\)
Average of the six: \(\displaystyle \frac{94}{6}=15\frac{2}{3}\)