Q.1 Find the sum:
(i) \(\frac{5}{4} \) +( \(\frac{-11}{4} \))
(ii) \(\frac{5}{3} \) + \(\frac{3}{5} \)
(iii) \(\frac{-9}{10} \) + \(\frac{22}{15} \)
(iv) \(\frac{-3}{-11} \) + \(\frac{5}{9} \)
(v) \(\frac{-8}{-19} \) + \(\frac{(-2)}{57} \)
(vi) \(\frac{-2}{3} \) + 0
(vii) \(-2\frac{1}{3} \) + \(4\frac{3}{5} \)


Answer :

(i) \(\frac{5}{4} \) +( \(\frac{-11}{4} \)) =\(\frac{5}{4} \) -\(\frac{11}{4} \) = \(\frac{5-11}{4} \)= \(\frac{-6}{4} \)
(ii) \(\frac{5}{3} \) + \(\frac{3}{5} \) = \(\frac{5×5}{3×5} \) + \(\frac{3×3}{5×3} \)
= \(\frac{25}{15} \) + \(\frac{9}{15} \) = \(\frac{25+9}{15} \) =\(\frac{25+9}{15} \) =\(\frac{34 }{15} \) = \(2\frac{4}{15} \)
(iii) \(\frac{-9}{10} \) + \(\frac{22}{15} \) = \(\frac{-9×3}{10×3} \) + \(\frac{22×2}{15×2} \)=\(\frac{-27}{30} \) + \(\frac{44}{30} \) = \(\frac{-27+44}{30} \) =\(\frac{17}{30} \)
(iv) \(\frac{-3}{-11} \) + \(\frac{5}{9} \) = \(\frac{3×9}{11×9} \) + \(\frac{5×11}{9×11} \) =\(\frac{27}{99} \) + \(\frac{55}{99} \) = \(\frac{27+55}{99} \) =\(\frac{82}{99} \)
(v) \(\frac{-8}{-19} \) + \(\frac{(-2)}{57} \) =\(\frac{8}{19} \) - \(\frac{2}{57} \) = \(\frac{8×3}{19×3} \) - \(\frac{2}{57} \)=\(\frac{24-2}{57} \) =\(\frac{22}{57} \)
(vi) \(\frac{-2}{3} \) + 0 = \(\frac{-2}{3} \)
(vii) \(-2\frac{1}{3} \) + \(4\frac{3}{5} \)= \(\frac{-7}{3} \) + \(\frac{23}{5} \)
= \(\frac{-7×5}{3×5} \) + \(\frac{23×3}{5×3} \) = \(\frac{-35}{15} \) + \(\frac{69}{15} \)
= \(\frac{-35+69}{15} \)=\(\frac{34}{15} \)=\(2\frac{4}{15} \)

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