Question: Find the sum: (A) 25+310\frac{2}{5} + \frac{3}{10}52+103 (B) 712+58\frac{7}{12} + \frac{5}{8}127+85 (C) −47+314\frac{-4}{7} + \frac{3}{14}7−4+143 [Page No. 49] Answer: Verified 10. (A) 25+310:410+310=71010. \text{ (A) } \frac{2}{5} + \frac{3}{10}: \frac{4}{10} + \frac{3}{10} = \frac{7}{10}10. (A) 52+103:104+103=107 (B)712+58:1424+1524=2924(\mathbb{B})\frac{7}{12}+\frac{5}{8}:\frac{14}{24}+\frac{15}{24}=\frac{29}{24}(B)127+85:2414+2415=2429 (C)−47+314:−814+314=−514(C)\frac{-4}{7}+\frac{3}{14}:\frac{-8}{14}+\frac{3}{14}=\frac{-5}{14}(C)7−4+143:14−8+143=14−5