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

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

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

= { ( \(\dfrac{1}{3}\) – \(\dfrac{1}{7}\) ) – ( \(\dfrac{1}{7}\) – \(\dfrac{1}{11}\) ) } × \(\dfrac{7}{4}\)

= ( \(\dfrac{4}{21}\) – \(\dfrac{4}{77}\) ) × \(\dfrac{7}{4}\)

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

問２

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

答え

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

解き方

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

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

= \(\dfrac{1}{2}\) × { ( \(\dfrac{1}{1}\) – \(\cancel{\dfrac{1}{4}}\) ) + ( \(\cancel{\dfrac{1}{4}}\) – \(\cancel{\dfrac{1}{5}}\) ) + ( \(\cancel{\dfrac{1}{5}}\) + \(\dfrac{1}{7}\) ) } × { ( \(\dfrac{1}{2}\) + \(\cancel{\dfrac{1}{3}}\) ) – ( \(\cancel{\dfrac{1}{3}}\) + \(\cancel{\dfrac{1}{4}}\) ) + ( \(\cancel{\dfrac{1}{4}}\) – \(\dfrac{1}{5}\) ) }

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

= \(\dfrac{1}{\cancel{2}}\) × \(\dfrac{\cancelto{\cancelto{2}{4}}{8}}{7}\) × \(\dfrac{3}{\cancelto{5}{10}}\) = \(\dfrac{6}{35}\)

問３

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

答え

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

解き方

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

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

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

= \(\dfrac{1}{2}\) × ( \(\dfrac{1}{3}\) – \(\dfrac{1}{5}\) + \(\dfrac{1}{21}\) ) × \(\dfrac{70}{57}\)

= \(\dfrac{1}{\cancel{2}}\) × \(\dfrac{\cancel{19}}{\cancelto{3}{105}}\) × \(\dfrac{\cancelto{\cancel{35}}{70}}{\cancelto{3}{57}}\) = \(\dfrac{1}{9}\)

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

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

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

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

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

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

= \(\dfrac{1}{3}\) × \(\dfrac{7}{4}\) – \(\dfrac{5}{\cancelto{3}{9}}\) × \(\dfrac{\cancel{3}}{4}\) + \(\dfrac{1}{\cancelto{2}{9}}\) × \(\dfrac{\cancel{3}}{5}\)

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

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

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