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If \(\frac{a}{7}\) = \(\frac{b}{3}\) = \(\frac{c}{5}\) then find (\(\frac{a + b + c}{b + c }\)). |
\(\frac{15}{8}\) \(\frac{17}{8}\) \(\frac{19}{8}\) \(\frac{21}{8}\) |
\(\frac{15}{8}\) |
Given, \(\frac{a}{7}\) = \(\frac{b}{3}\) = \(\frac{c}{5}\) Here we can directly conclude that a = 7, b = 3, c = 5, hence ⇒ (\(\frac{a + b + c}{b + c }\)) = (\(\frac{7 + 3 + 5}{ 3 + 5 }\)) = \(\frac{15}{8}\) |
