$\frac{11}{5}$ of a number A is 22% of a number B. The number B is equal to 2.5% of a third number C. If the value of C is 5500, then the sum of 80% of A and 40% of B is :

66

\(\frac{11}{5}\) × A =\(\frac{22}{100}\) × B

 A =\(\frac{B}{10}\)

 \(\frac{A}{B}\) = \(\frac{1}{10}\)

Also, B =\(\frac{2.5}{100}\) × C

 B = \(\frac{C}{40}\)   [∵ 2.5% = \(\frac{1}{40}\) ]

\(\frac{B}{C}\) = \(\frac{1}{40}\)

\(\frac{B}{C}\) = \(\frac{10}{400}\)   [∵\(\frac{A}{B}\) = \(\frac{1}{0}\)]

Here, 400 units = 5500

⇒ 1 unit = \(\frac{5500}{400}\) = \(\frac{55}{4}\)

The sum of 80% of A and 40% of B = [ \(\frac{4}{5}\) × \(\frac{55}{4}\)] + [\(\frac{2}{5}\) ×\(\frac{550}{4}\)]  [∵ 80% = \(\frac{4}{5}\) and 40% = \(\frac{2}{5}\)]

= 11 + 55

= 66