If \(\frac{a}{7}\) = \(\frac{b}{7}\) = \(\frac{c}{8}\) then find (\(\frac{a + b }{a + b + c }\)).

\(\frac{7}{11}\) \(\frac{8}{11}\) \(\frac{3}{11}\) \(\frac{6}{11}\)

\(\frac{7}{11}\)

Given, \(\frac{a}{7}\) = \(\frac{b}{7}\) = \(\frac{c}{8}\) Here we can directly conclude that  a = 7, b = 7, c = 8, hence ⇒ (\(\frac{a + b }{a + b + c }\)) = (\(\frac{7 + 7 }{7 + 7 + 8 }\)) = \(\frac{7}{11}\)