Q61. (1 \frac{1}{4} ÷ 2 \frac{6}{7} का 5 \frac{3}{5} ) ÷ (6 \frac{2}{5} ÷ 4 \frac{1}{2} का 5 \frac{1}{3} ) × ( \frac{3}{4} × 2 \frac{2}{3} ÷ \frac{5}{9} का 1 \frac{1}{5} ) = 1 + k है तो k का मान किसके बीच आएगा ?
पहला भाग = 1 \frac{1}{3} ÷ 2 \frac{6}{7} का 5 \frac{3}{5}
= \frac{4}{3} ÷ \frac{20}{7} × \frac{28}{5}
= \frac{4}{3} ÷ 16
= \frac{1}{12}
दूसरा भाग = 6 \frac{2}{5} ÷ 4 \frac{1}{2} का 5 \frac{1}{3}
= \frac{32}{5} ÷ \frac{9}{2} × \frac{16}{3}
= \frac{32}{5} ÷ 24
= \frac{4}{15}
तीसरा भाग = \frac{3}{4} × 2 \frac{2}{3} ÷ \frac{5}{9} का 1 \frac{1}{5}
= \frac{3}{4} × \frac{8}{3} ÷ \frac{5}{9} × \frac{6}{5}
= \frac{3}{4} × \frac{8}{3} ÷ \frac{2}{3}
= 2 ÷ \frac{2}{3}
= 3
पहला भाग ÷ दूसरा भाग × तीसरा भाग = 1 + k
\frac{1}{12} ÷ \frac{4}{15} × 3 = 1 + k
\frac{1}{12} × \frac{15}{4} × 3 = 1 + k
\frac{15}{16} = 1 + k
k = – \frac{1}{16}
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