算数【入試】分数を含む四則計算（問題文に）

問１

\(\{\ \dfrac{2}{15}\ +\ \dfrac{2}{35}\ -\ (\ \dfrac{2}{63}\ +\ \dfrac{2}{99}\ )\ \}\ ×\ 1\dfrac{3}{4}\ =\)

答え

\(\dfrac{8}{33}\)

解き方

\(\{\dfrac{2}{15} + \dfrac{2}{35}\) ― \((\dfrac{2}{63} + \dfrac{2}{99})\} \times 1\dfrac{3}{4}\)

\(=[(\dfrac{1}{3}-\cancel{\dfrac{1}{5}})+(\cancel{\dfrac{1}{5}}-\dfrac{1}{7})-\{(\dfrac{1}{7}-\cancel{\dfrac{1}{9}})+(\cancel{\dfrac{1}{9}}-\dfrac{1}{11})\}]\times\dfrac{7}{4}\)

\(=\{(\dfrac{1}{3}-\dfrac{1}{7})-(\dfrac{1}{7}-\dfrac{1}{11})\}\times\dfrac{7}{4}\)

\(=(\dfrac{4}{21}-\dfrac{4}{77})\times\dfrac{7}{4}\)

\(=\dfrac{1}{3}-\dfrac{1}{11}=\dfrac{8}{33}\)

問２

\((\ \dfrac{3}{8}\ +\ \dfrac{1}{40}\ +\ \dfrac{6}{35}\ )\ ×\ (\ \dfrac{5}{6}\ -\ \dfrac{7}{12}\ +\ \dfrac{1}{20}\ )\ =\)

ヒント１

\(=\dfrac{1}{2}\times(\dfrac{3}{4}+\dfrac{1}{20}+\dfrac{12}{35})\times(\dfrac{5}{6}\) ― \(\dfrac{7}{12}+\dfrac{1}{20})\)

ヒント２

\(=\dfrac{1}{2}\times\{(\dfrac{1}{1}\) ― \(\cancel{\dfrac{1}{4}})+(\cancel{\dfrac{1}{4}}\) ― \(\cancel{\dfrac{1}{5}})+(\cancel{\dfrac{1}{5}}+\dfrac{1}{7})\}\times\{(\dfrac{1}{2}+\cancel{\dfrac{1}{3}})\) ― \((\cancel{\dfrac{1}{3}}+\cancel{\dfrac{1}{4}})+(\cancel{\dfrac{1}{4}}\) ― \(\dfrac{1}{5})\}\)

答え

\(\dfrac{6}{35}\)

解き方

\((\dfrac{3}{8} + \dfrac{1}{40} + \dfrac{6}{35}) \times (\dfrac{5}{6}\) ― \(\dfrac{7}{12} + \dfrac{1}{20})\)

\(=\dfrac{1}{2}\times(\dfrac{3}{4}+\dfrac{1}{20}+\dfrac{12}{35})\times(\dfrac{5}{6}\) ― \(\dfrac{7}{12}+\dfrac{1}{20})\)

\(=\dfrac{1}{2}\times\{(\dfrac{1}{1}\) ― \(\cancel{\dfrac{1}{4}})+(\cancel{\dfrac{1}{4}}\) ― \(\cancel{\dfrac{1}{5}})+(\cancel{\dfrac{1}{5}}+\dfrac{1}{7})\}\times\{(\dfrac{1}{2}+\cancel{\dfrac{1}{3}})\) ― \((\cancel{\dfrac{1}{3}}+\cancel{\dfrac{1}{4}})+(\cancel{\dfrac{1}{4}}\) ― \(\dfrac{1}{5})\}\)

\(=\dfrac{1}{2}\times(1+\dfrac{1}{7})\times(\dfrac{1}{2}\) ― \(\dfrac{1}{5})\)

\(=\dfrac{1}{2}\times\dfrac{8}{7}\times\dfrac{3}{10}=\dfrac{6}{35}\)

問３

\((\ \dfrac{7}{24}\ -\ \dfrac{9}{40}\ +\ \dfrac{1}{42}\ )\ ÷\ \dfrac{57}{70}\ =\)

答え

\(\dfrac{1}{9}\)

解き方

\((\dfrac{7}{24}\) ― \(\dfrac{9}{40} + \dfrac{1}{42}) \div \dfrac{57}{70}\)

\(=\dfrac{1}{2}\times(\dfrac{7}{12}\) ― \(\dfrac{9}{20} + \dfrac{1}{21})\times \dfrac{70}{57}\)

\(=\dfrac{1}{2}\times\{(\dfrac{1}{3}+\cancel{\dfrac{1}{4}})\) ― \((\cancel{\dfrac{1}{4}}+\dfrac{1}{5}) + \dfrac{1}{21}\}\times \dfrac{70}{57}\)

\(=\dfrac{1}{2}\times(\dfrac{1}{3}\) ― \(\dfrac{1}{5} + \dfrac{1}{21})\times \dfrac{70}{57}\)

\(=\dfrac{1}{2}\times\dfrac{19}{105}\times \dfrac{70}{57}=\dfrac{1}{9}\)

問４

\(3\dfrac{1}{56}\ +\ 5\dfrac{1}{72}\ -\ 7\dfrac{2}{63}\ =\)

答え

1

解き方

\(3\dfrac{1}{56} + 5\dfrac{1}{72}\) ― \(7\dfrac{2}{63}\)

\(=(3+\dfrac{1}{56}) + (5+\dfrac{1}{72})\) ― \((7+\dfrac{2}{63})\)

\(=3+5\) ― \(7+\dfrac{1}{56}+\dfrac{1}{72}\) ― \(\dfrac{2}{63}\)

\(=1+(\cancel{\dfrac{1}{7}}\) ― \(\cancel{\dfrac{1}{8}})+(\cancel{\dfrac{1}{8}}\) ― \(\cancel{\dfrac{1}{9}})\) ― \((\cancel{\dfrac{1}{7}}\) ― \(\cancel{\dfrac{1}{9}})=1\)

問５

\(\dfrac{2}{1\ ×\ 3}\ +\ \dfrac{2}{3\ ×\ 5}\ +\ \dfrac{2}{5\ ×\ 7}\ +\ \dfrac{2}{7\ ×\ 9}\ =\)

答え

\(\dfrac{8}{9}\)

解き方

\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}\)

\(=(\dfrac{1}{1}-\cancel{\dfrac{1}{3}})+(\cancel{\dfrac{1}{3}}-\cancel{\dfrac{1}{5}})+(\cancel{\dfrac{1}{5}}-\cancel{\dfrac{1}{7}})+(\cancel{\dfrac{1}{7}}-\dfrac{1}{9})\)

\(=\dfrac{8}{9}\)

問６

\(\dfrac{1}{3}\ ×\ \dfrac{7}{4}\ -\ \dfrac{5}{9}\ ÷\ \dfrac{4}{3}\ +\ \dfrac{1}{6}\ ×\ \dfrac{3}{5}\ =\)

答え

\(\dfrac{4}{15}\)

解き方

\(\dfrac{1}{3}\times\dfrac{7}{4}\) ― \(\dfrac{5}{9}\div\dfrac{4}{3}+\dfrac{1}{6}\times\dfrac{3}{5}\)

\(=\dfrac{1}{3}\times\dfrac{7}{4}\) ― \(\dfrac{5}{9}\times\dfrac{3}{4}+\dfrac{1}{6}\times\dfrac{3}{5}\)

\(=\dfrac{7}{12}-\dfrac{5}{12}+\dfrac{1}{10}\)

\(=\dfrac{1}{6}+\dfrac{1}{10}\)

\(=\dfrac{5+3}{30}=\dfrac{8}{30}=\dfrac{4}{15}\)