CUET Preparation Today
If \(\frac{a}{4}\) = \(\frac{b}{9}\) = \(\frac{c}{1}\) then find (\(\frac{a + b + c}{b + c }\)). |
\(\frac{8}{5}\) \(\frac{7}{5}\) \(\frac{7}{6}\) \(\frac{9}{5}\) |
\(\frac{7}{5}\) |
Given, \(\frac{a}{4}\) = \(\frac{b}{9}\) = \(\frac{c}{1}\) Here we can directly conclude that a = 4, b = 9, c = 1, hence ⇒ (\(\frac{a + b + c}{b + c }\)) = (\(\frac{4 + 9 + 1}{ 9 + 1 }\)) = \(\frac{7}{5}\) |
