算数【応用】分数を含む四則計算（問題文に）

問１

( \(\dfrac{7}{9}\) – \(\dfrac{5}{12}\) ) × ( \(2\dfrac{4}{5}\) – \(2\dfrac{2}{7}\) ) =

答え

\(\dfrac{13}{70}\)

解き方

( \(\dfrac{7}{9}\) – \(\dfrac{5}{12}\) ) × ( \(2\dfrac{4}{5}\) – \(2\dfrac{2}{7}\) )
= \(\dfrac{28\ -\ 15}{36}\) × ( \(\dfrac{14}{5}\) – \(\dfrac{16}{7}\) )
= \(\dfrac{13}{36}\) × \(\dfrac{98\ -\ 80}{35}\)
= \(\dfrac{13}{\cancelto{2}{36}}\) × \(\dfrac{\cancel{18}}{35}\) = \(\dfrac{13}{70}\)

問２

\(\dfrac{1\ ×\ 2\ ×\ 3\ ×\ 4\ ×\ 5\ ×\ 6\ ×\ 7}{3\ ×\ 4\ ×\ 5\ ×\ 6\ ×\ 7\ ×\ 8\ ×\ 9}\) =

答え

\(\dfrac{1}{36}\)

解き方

\(\dfrac{1\ ×\ 2\ ×\ \cancel{3}\ ×\ \cancel{4}\ ×\ \cancel{5}\ ×\ \cancel{6}\ ×\ \cancel{7}}{\cancel{3}\ ×\ \cancel{4}\ ×\ \cancel{5}\ ×\ \cancel{6}\ ×\ \cancel{7}\ ×\ 8\ ×\ 9}\)
= \(\dfrac{1\ ×\ \cancel{2}}{\cancelto{4}{8}\ ×\ 9}\) = \(\dfrac{1}{36}\)

問３

( \(8\dfrac{1}{3}\) – \(1\dfrac{8}{15}\) ) ÷ 4 ÷ \(1\dfrac{1}{2}\) + \(\dfrac{2}{3}\) =

答え

\(1\dfrac{4}{5}\)

解き方

( \(8\dfrac{1}{3}\) – \(1\dfrac{8}{15}\) ) ÷ 4 ÷ \(1\dfrac{1}{2}\) + \(\dfrac{2}{3}\)
= ( \(\dfrac{25}{3}\) – \(\dfrac{23}{15}\) ) ÷ 4 ÷ \(\dfrac{3}{2}\) + \(\dfrac{2}{3}\)
= \(\dfrac{125\ -\ 23}{15}\) × \(\dfrac{1}{\cancelto{2}{4}}\) × \(\dfrac{\cancel{2}}{3}\) + \(\dfrac{2}{3}\)
= \(\dfrac{\cancelto{17}{102}}{15}\) × \(\dfrac{1}{\cancel{6}}\) + \(\dfrac{2}{3}\)
= \(\dfrac{17}{15}\) + \(\dfrac{10}{15}\) = \(\dfrac{27}{15}\) = \(\dfrac{9}{5}\) = \(1\dfrac{4}{5}\)

問４

\(\dfrac{4}{5}\) × ( \(\dfrac{1}{4}\) + \(\dfrac{1}{3}\) ) ÷ \(2\dfrac{1}{3}\) =

答え

\(\dfrac{1}{5}\)

解き方

\(\dfrac{4}{5}\) × ( \(\dfrac{1}{4}\) + \(\dfrac{1}{3}\) ) ÷ \(2\dfrac{1}{3}\)
= ( \(\dfrac{\cancel{4}}{5}\) × \(\dfrac{1}{\cancel{4}}\) + \(\dfrac{4}{5}\) × \(\dfrac{1}{3}\) ) ÷ \(\dfrac{7}{3}\)
= ( \(\dfrac{1}{5}\) + \(\dfrac{4}{15}\) ) × \(\dfrac{3}{7}\)
= ( \(\dfrac{3}{15}\) + \(\dfrac{4}{15}\) ) × \(\dfrac{3}{7}\)
= \(\dfrac{\cancel{7}}{\cancelto{5}{15}}\) × \(\dfrac{\cancel{3}}{\cancel{7}}\) = \(\dfrac{1}{5}\)

問５

81 ÷ \(2\dfrac{1}{4}\) + 52 × ( \(\dfrac{11}{26}\) – \(\dfrac{5}{13}\) ) =

答え

38

解き方

81 ÷ \(2\dfrac{1}{4}\) + 52 × ( \(\dfrac{11}{26}\) – \(\dfrac{5}{13}\) )
= 81 ÷ \(\dfrac{9}{4}\) + ( \(\cancelto{2}{52}\) × \(\dfrac{11}{\cancel{26}}\) – \(\cancelto{4}{52}\) × \(\dfrac{5}{\cancel{13}}\) )
= \(\cancelto{9}{81}\) × \(\dfrac{4}{\cancel{9}}\) + ( 22 – 20 )
= 36 + 2 = 38

問６

\(2\dfrac{7}{9}\) ÷ \(\dfrac{10}{27}\) – \(1\dfrac{2}{7}\) ÷ \(2\dfrac{4}{7}\) =

答え

7

解き方

\(2\dfrac{7}{9}\) ÷ \(\dfrac{10}{27}\) – \(1\dfrac{2}{7}\) ÷ \(2\dfrac{4}{7}\)
= \(\dfrac{25}{9}\) ÷ \(\dfrac{10}{27}\) – \(\dfrac{9}{7}\) ÷ \(\dfrac{18}{7}\)
= \(\dfrac{\cancelto{5}{25}}{\cancel{9}}\) × \(\dfrac{\cancelto{3}{27}}{\cancelto{2}{10}}\) – \(\dfrac{\cancel{9}}{\cancel{7}}\) × \(\dfrac{\cancel{7}}{\cancelto{2}{18}}\)
= \(\dfrac{15}{2}\) – \(\dfrac{1}{2}\) = \(\dfrac{14}{2}\) = 7

問７

\(3\dfrac{13}{24}\) ÷ ( 7 – \(1\dfrac{3}{8}\) ) – ( \(3\dfrac{2}{3}\) – \(\dfrac{5}{6}\) ) × \(\dfrac{2}{9}\) =

答え

0

解き方

\(3\dfrac{13}{24}\) ÷ ( 7 – \(1\dfrac{3}{8}\) ) – ( \(3\dfrac{2}{3}\) – \(\dfrac{5}{6}\) ) × \(\dfrac{2}{9}\)
= \(\dfrac{85}{24}\) ÷ ( \(\dfrac{56}{8}\) – \(\dfrac{11}{8}\) ) – ( \(\dfrac{11}{3}\) – \(\dfrac{5}{6}\) ) × \(\dfrac{2}{9}\)
= \(\dfrac{85}{24}\) ÷ \(\dfrac{45}{8}\) – ( \(\dfrac{22}{6}\) – \(\dfrac{5}{6}\) ) × \(\dfrac{2}{9}\)
= \(\dfrac{\cancelto{17}{85}}{\cancelto{3}{24}}\) × \(\dfrac{\cancel{8}}{\cancelto{9}{45}}\) – \(\dfrac{17}{\cancelto{3}{6}}\) × \(\dfrac{\cancel{2}}{9}\)
= \(\dfrac{17}{27}\) – \(\dfrac{17}{27}\) = 0

問８

( 1 – \(\dfrac{1}{2}\) ) × ( 1 – \(\dfrac{1}{3}\) ) × ( 1 – \(\dfrac{1}{4}\) ) × ・・・ × ( 1 – \(\dfrac{1}{9}\) ) × ( 1 – \(\dfrac{1}{10}\) ) =

答え

\(\dfrac{1}{10}\) ( または 0.1 )

解き方

( 1 – \(\dfrac{1}{2}\) ) × ( 1 – \(\dfrac{1}{3}\) ) × ( 1 – \(\dfrac{1}{4}\) ) × ・・・ × ( 1 – \(\dfrac{1}{9}\) ) × ( 1 – \(\dfrac{1}{10}\) )
= \(\dfrac{1}{\cancel{2}}\) × \(\dfrac{\cancel{2}}{\cancel{3}}\) × \(\dfrac{\cancel{3}}{\cancel{4}}\) × ・・・ × \(\dfrac{\cancel{8}}{\cancel{9}}\) × \(\dfrac{\cancel{9}}{10}\)
= \(\dfrac{1}{10}\)

問９

\(4\dfrac{3}{4}\) + ( \(4\dfrac{5}{6}\) – \(\dfrac{2}{3}\) ) ÷ \(3\dfrac{1}{3}\) =

答え

6

解き方

\(4\dfrac{3}{4}\) + ( \(4\dfrac{5}{6}\) – \(\dfrac{2}{3}\) ) ÷ \(3\dfrac{1}{3}\)
= \(\dfrac{19}{4}\) + ( \(\dfrac{29}{6}\) – \(\dfrac{4}{6}\) ) ÷ \(\dfrac{10}{3}\)
= \(\dfrac{19}{4}\) + \(\dfrac{\cancelto{5}{25}}{\cancelto{2}{6}}\) × \(\dfrac{\cancel{3}}{\cancelto{2}{10}}\)
= \(\dfrac{19}{4}\) + \(\dfrac{5}{4}\) = \(\dfrac{24}{4}\) = 6

問１０

\(5\dfrac{4}{9}\) ÷ \(5\dfrac{3}{5}\) – \(\dfrac{2}{3}\) × \(\dfrac{5}{6}\) =

答え

\(\dfrac{5}{12}\)

解き方

\(5\dfrac{4}{9}\) ÷ \(5\dfrac{3}{5}\) – \(\dfrac{2}{3}\) × \(\dfrac{5}{6}\)
= \(\dfrac{49}{9}\) ÷ \(\dfrac{28}{5}\) – \(\dfrac{\cancel{2}}{3}\) × \(\dfrac{5}{\cancelto{3}{6}}\)
= \(\dfrac{\cancelto{7}{49}}{9}\) × \(\dfrac{5}{\cancelto{4}{28}}\) – \(\dfrac{5}{9}\)
= \(\dfrac{35}{36}\) – \(\dfrac{20}{36}\) = \(\dfrac{15}{36}\) = \(\dfrac{5}{12}\)

問１１

( \(2\dfrac{1}{12}\) – \(1\dfrac{8}{9}\) ) ÷ \(\dfrac{7}{18}\) × \(3\dfrac{3}{5}\) =

答え

\(1\dfrac{4}{5}\) ( または 1.8 )

解き方

( \(2\dfrac{1}{12}\) – \(1\dfrac{8}{9}\) ) ÷ \(\dfrac{7}{18}\) × \(3\dfrac{3}{5}\)
= ( \(\dfrac{25}{12}\) – \(\dfrac{17}{9}\) ) × \(\dfrac{18}{7}\) × \(\dfrac{18}{5}\)
= ( \(\dfrac{25}{\cancelto{2}{12}}\) × \(\cancelto{3}{18}\) – \(\dfrac{17}{\cancel{9}}\) × \(\cancelto{2}{18}\) ) × \(\dfrac{1}{7}\) × \(\dfrac{18}{5}\)
= ( \(\dfrac{75}{2}\) – 34 ) × \(\dfrac{18}{35}\)
= \(\dfrac{75\ -\ 68}{2}\) × \(\dfrac{18}{35}\)
= \(\dfrac{\cancel{7}}{\cancel{2}}\) × \(\dfrac{\cancelto{9}{18}}{\cancelto{5}{35}}\) = \(\dfrac{9}{5}\) = \(1\dfrac{4}{5}\)

問１２

( \(\dfrac{5}{2}\) – \(\dfrac{5}{3}\) + \(\dfrac{5}{9}\) ) ÷ \(\dfrac{7}{18}\) =

答え

\(3\dfrac{4}{7}\)

解き方

( \(\dfrac{5}{2}\) – \(\dfrac{5}{3}\) + \(\dfrac{5}{9}\) ) ÷ \(\dfrac{7}{18}\)
= ( \(\dfrac{5}{2}\) – \(\dfrac{5}{3}\) + \(\dfrac{5}{9}\) ) × \(\dfrac{18}{7}\)
= ( \(\dfrac{5}{\cancel{2}}\) × \(\cancelto{9}{18}\) – \(\dfrac{5}{\cancel{3}}\) × \(\cancelto{6}{18}\) + \(\dfrac{5}{9}\) × \(\cancelto{2}{18}\) ) × \(\dfrac{1}{7}\)
= ( 45 – 30 + 10 ) × \(\dfrac{1}{7}\)
= \(\dfrac{25}{7}\) = \(3\dfrac{4}{7}\)

解き方

( \(2\dfrac{1}{12}\) – \(1\dfrac{8}{9}\) ) ÷ \(\dfrac{7}{18}\) × \(3\dfrac{3}{5}\)
= ( \(\dfrac{25}{12}\) – \(\dfrac{17}{9}\) ) × \(\dfrac{18}{7}\) × \(\dfrac{18}{5}\)
= ( \(\dfrac{25}{\cancelto{2}{12}}\) × \(\cancelto{3}{18}\) – \(\dfrac{17}{\cancel{9}}\) × \(\cancelto{2}{18}\) ) × \(\dfrac{1}{7}\) × \(\dfrac{18}{5}\)
= ( \(\dfrac{75}{2}\) – 34 ) × \(\dfrac{18}{35}\)
= \(\dfrac{75\ -\ 68}{2}\) × \(\dfrac{18}{35}\)
= \(\dfrac{\cancel{7}}{\cancel{2}}\) × \(\dfrac{\cancelto{9}{18}}{\cancelto{5}{35}}\) = \(\dfrac{9}{5}\) = \(1\dfrac{4}{5}\)

問１３

23 + 2 × 14 ÷ \(\dfrac{1}{10}\) + 20 × 10 + 5 ÷ \(\dfrac{1}{4}\) ÷ \(\dfrac{7}{17}\) =

答え

\(551\dfrac{4}{7}\)

解き方

23 + 2 × 14 ÷ \(\dfrac{1}{10}\) + 20 × 10 + 5 ÷ \(\dfrac{1}{4}\) ÷ \(\dfrac{7}{17}\)
= 23 + 2 × 14 × 10 + 20 × 10 + 5 × 4 × \(\dfrac{17}{7}\)
= 23 + 280 + 200 + \(\dfrac{340}{7}\)
= 503 + \(48\dfrac{4}{7}\) = \(551\dfrac{4}{7}\)

問１４

( \(1\dfrac{5}{12}\) – \(1\dfrac{1}{9}\) ) ÷ \(3\dfrac{2}{3}\) × \(1\dfrac{1}{2}\) =

答え

\(\dfrac{1}{8}\)

解き方

( \(1\dfrac{5}{12}\) – \(1\dfrac{1}{9}\) ) ÷ \(3\dfrac{2}{3}\) × \(1\dfrac{1}{2}\)
= ( \(\dfrac{17}{12}\) – \(\dfrac{10}{9}\) ) ÷ \(\dfrac{11}{3}\) × \(\dfrac{3}{2}\)
= ( \(\dfrac{17}{\cancelto{4}{12}}\) × \(\dfrac{\cancel{3}}{2}\) – \(\dfrac{\cancelto{5}{10}}{\cancelto{3}{9}}\) × \(\dfrac{\cancel{3}}{\cancel{2}}\) ) × \(\dfrac{3}{11}\)
= ( \(\dfrac{17}{8}\) – \(\dfrac{5}{3}\) ) × \(\dfrac{3}{11}\)
= ( \(\dfrac{51}{24}\) – \(\dfrac{40}{24}\) ) × \(\dfrac{3}{11}\)
= \(\dfrac{\cancel{11}}{\cancelto{8}{24}}\) × \(\dfrac{\cancel{3}}{\cancel{11}}\) = \(\dfrac{1}{8}\)

解き方

( \(2\dfrac{1}{12}\) – \(1\dfrac{8}{9}\) ) ÷ \(\dfrac{7}{18}\) × \(3\dfrac{3}{5}\)
= ( \(\dfrac{25}{12}\) – \(\dfrac{17}{9}\) ) × \(\dfrac{18}{7}\) × \(\dfrac{18}{5}\)
= ( \(\dfrac{25}{\cancelto{2}{12}}\) × \(\cancelto{3}{18}\) – \(\dfrac{17}{\cancel{9}}\) × \(\cancelto{2}{18}\) ) × \(\dfrac{1}{7}\) × \(\dfrac{18}{5}\)
= ( \(\dfrac{75}{2}\) – 34 ) × \(\dfrac{18}{35}\)
= \(\dfrac{75\ -\ 68}{2}\) × \(\dfrac{18}{35}\)
= \(\dfrac{\cancel{7}}{\cancel{2}}\) × \(\dfrac{\cancelto{9}{18}}{\cancelto{5}{35}}\) = \(\dfrac{9}{5}\) = \(1\dfrac{4}{5}\)

問１５

( 1 ÷ \(\dfrac{1}{5}\) + \(\dfrac{4}{5}\) ÷ \(\dfrac{1}{10}\) ) + ( \(\dfrac{1}{4}\) ÷ \(\dfrac{1}{7}\) + \(\dfrac{1}{5}\) ÷ \(\dfrac{1}{10}\) ) – ( \(\dfrac{1}{4}\) ÷ \(\dfrac{1}{7}\) + \(\dfrac{1}{5}\) ÷ \(\dfrac{1}{10}\) ) × \(\dfrac{1}{5}\) =

答え

16

解き方

( 1 ÷ \(\dfrac{1}{5}\) + \(\dfrac{4}{5}\) ÷ \(\dfrac{1}{10}\) ) + ( \(\dfrac{1}{4}\) ÷ \(\dfrac{1}{7}\) + \(\dfrac{1}{5}\) ÷ \(\dfrac{1}{10}\) ) – ( \(\dfrac{1}{4}\) ÷ \(\dfrac{1}{7}\) + \(\dfrac{1}{5}\) ÷ \(\dfrac{1}{10}\) ) × \(\dfrac{1}{5}\)
= ( 1 × 5 + \(\dfrac{4}{\cancel{5}}\) × \(\cancelto{2}{10}\) ) + ( \(\dfrac{1}{4}\) ÷ \(\dfrac{1}{7}\) + \(\dfrac{1}{5}\) ÷ \(\dfrac{1}{10}\) ) × ( 1 – \(\dfrac{1}{5}\) )
= ( 5 + 8 ) + ( \(\dfrac{1}{4}\) × 7 + \(\dfrac{1}{\cancel{5}}\) × \(\cancelto{2}{10}\) ) × \(\dfrac{4}{5}\)
= 13 + ( \(\dfrac{7}{4}\) + 2 ) × \(\dfrac{4}{5}\)
= 13 + \(\dfrac{\cancelto{3}{15}}{\cancel{4}}\) × \(\dfrac{\cancel{4}}{\cancel{5}}\)
= 13 + 3 = 16

問１６

1 – { 5 – ( \(1\dfrac{2}{3}\) + \(\dfrac{3}{8}\) ) ÷ \(\dfrac{5}{12}\) } =

答え

\(\dfrac{9}{10}\) ( または 0.9 )

解き方

1 – { 5 – ( \(1\dfrac{2}{3}\) + \(\dfrac{3}{8}\) ) ÷ \(\dfrac{5}{12}\) }
= 1 – { 5 – ( \(\dfrac{5}{3}\) + \(\dfrac{3}{8}\) ) × \(\dfrac{12}{5}\) }
= 1 – { 5 – ( \(\dfrac{\cancel{5}}{\cancel{3}}\) × \(\dfrac{\cancelto{4}{12}}{\cancel{5}}\) + \(\dfrac{3}{\cancelto{2}{8}}\) × \(\dfrac{\cancelto{3}{12}}{5}\) ) }
= 1 – { 5 – ( 4 + \(\dfrac{9}{10}\) ) }
= 1 – { 5 – ( 4 + 0.9 ) }
= 1 – ( 5 – 4.9 )
= 1 – 0.1 = 0.9 = \(\dfrac{9}{10}\)

解き方

( \(2\dfrac{1}{12}\) – \(1\dfrac{8}{9}\) ) ÷ \(\dfrac{7}{18}\) × \(3\dfrac{3}{5}\)
= ( \(\dfrac{25}{12}\) – \(\dfrac{17}{9}\) ) × \(\dfrac{18}{7}\) × \(\dfrac{18}{5}\)
= ( \(\dfrac{25}{\cancelto{2}{12}}\) × \(\cancelto{3}{18}\) – \(\dfrac{17}{\cancel{9}}\) × \(\cancelto{2}{18}\) ) × \(\dfrac{1}{7}\) × \(\dfrac{18}{5}\)
= ( \(\dfrac{75}{2}\) – 34 ) × \(\dfrac{18}{35}\)
= \(\dfrac{75\ -\ 68}{2}\) × \(\dfrac{18}{35}\)
= \(\dfrac{\cancel{7}}{\cancel{2}}\) × \(\dfrac{\cancelto{9}{18}}{\cancelto{5}{35}}\) = \(\dfrac{9}{5}\) = \(1\dfrac{4}{5}\)

問１７

\(\dfrac{5}{7}\) ÷ { ( \(\dfrac{4}{5}\) – \(\dfrac{2}{7}\) ) × \(2\dfrac{5}{6}\) } =

答え

\(\dfrac{25}{51}\)

解き方

\(\dfrac{5}{7}\) ÷ { ( \(\dfrac{4}{5}\) – \(\dfrac{2}{7}\) ) × \(2\dfrac{5}{6}\) }
= \(\dfrac{5}{7}\) ÷ { ( \(\dfrac{28}{35}\) – \(\dfrac{10}{35}\) ) × \(\dfrac{17}{6}\) }
= \(\dfrac{5}{7}\) ÷ ( \(\dfrac{\cancelto{3}{18}}{35}\) × \(\dfrac{17}{\cancel{6}}\) )
= \(\dfrac{5}{7}\) ÷ \(\dfrac{51}{35}\)
= \(\dfrac{5}{\cancel{7}}\) × \(\dfrac{\cancelto{5}{35}}{51}\) = \(\dfrac{25}{51}\)

問１８

\(\dfrac{1}{12}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) – \(\dfrac{1}{56}\) – \(\dfrac{1}{7}\) – \(\dfrac{1}{8}\) + \(\dfrac{1}{9}\) =

答え

\(\dfrac{31}{63}\)

解き方

\(\dfrac{1}{12}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) – \(\dfrac{1}{56}\) – \(\dfrac{1}{7}\) – \(\dfrac{1}{8}\) + \(\dfrac{1}{9}\)
= ( \(\dfrac{1}{3}\) – \(\dfrac{1}{4}\) ) + \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) – ( \(\dfrac{1}{7}\) – \(\dfrac{1}{8}\) ) – \(\dfrac{1}{7}\) – \(\dfrac{1}{8}\) + \(\dfrac{1}{9}\)
= \(\dfrac{1}{3}\) – \(\cancel{\dfrac{1}{4}}\) + \(\dfrac{1}{3}\) + \(\cancel{\dfrac{1}{4}}\) – \(\dfrac{1}{7}\) + \(\cancel{\dfrac{1}{8}}\) – \(\dfrac{1}{7}\) – \(\cancel{\dfrac{1}{8}}\) + \(\dfrac{1}{9}\)
= \(\dfrac{2}{3}\) – \(\dfrac{1}{7}\) – \(\dfrac{1}{7}\) + \(\dfrac{1}{9}\)
= \(\dfrac{42\ -\ 9\ -\ 9\ +\ 7}{63}\) = \(\dfrac{31}{63}\)

解き方

( \(2\dfrac{1}{12}\) – \(1\dfrac{8}{9}\) ) ÷ \(\dfrac{7}{18}\) × \(3\dfrac{3}{5}\)
= ( \(\dfrac{25}{12}\) – \(\dfrac{17}{9}\) ) × \(\dfrac{18}{7}\) × \(\dfrac{18}{5}\)
= ( \(\dfrac{25}{\cancelto{2}{12}}\) × \(\cancelto{3}{18}\) – \(\dfrac{17}{\cancel{9}}\) × \(\cancelto{2}{18}\) ) × \(\dfrac{1}{7}\) × \(\dfrac{18}{5}\)
= ( \(\dfrac{75}{2}\) – 34 ) × \(\dfrac{18}{35}\)
= \(\dfrac{75\ -\ 68}{2}\) × \(\dfrac{18}{35}\)
= \(\dfrac{\cancel{7}}{\cancel{2}}\) × \(\dfrac{\cancelto{9}{18}}{\cancelto{5}{35}}\) = \(\dfrac{9}{5}\) = \(1\dfrac{4}{5}\)

問１９

\(\dfrac{4}{3}\) + \(\dfrac{3}{4}\) + \(\dfrac{11}{9}\) + \(\dfrac{11}{12}\) =

答え

\(4\dfrac{2}{9}\)

解き方

\(\dfrac{4}{3}\) + \(\dfrac{3}{4}\) + \(\dfrac{11}{9}\) + \(\dfrac{11}{12}\)
= \(\dfrac{16}{12}\) + \(\dfrac{9}{12}\) + \(\dfrac{11}{9}\) + \(\dfrac{11}{12}\)
= \(\dfrac{36}{12}\) + \(\dfrac{11}{9}\)
= 3 + \(1\dfrac{2}{9}\) = \(4\dfrac{2}{9}\)

問２０

( \(4\dfrac{1}{3}\) + \(17\dfrac{4}{5}\) – \(19\dfrac{5}{6}\) ) × 10 =

答え

23

解き方

( \(4\dfrac{1}{3}\) + \(17\dfrac{4}{5}\) – \(19\dfrac{5}{6}\) ) × 10
= { ( 4 + \(\dfrac{1}{3}\) ) + ( 17 + \(\dfrac{4}{5}\) ) – ( 19 + \(\dfrac{5}{6}\) ) } × 10
= { ( 4 + 17 – 19 ) + ( \(\dfrac{1}{3}\) + \(\dfrac{4}{5}\) – \(\dfrac{5}{6}\) ) } × 10
= { 2 + ( \(\dfrac{10}{30}\) + \(\dfrac{24}{30}\) – \(\dfrac{25}{30}\) ) } × 10
= ( 2 + \(\dfrac{9}{30}\) ) × 10
= ( 2 + \(\dfrac{3}{10}\) ) × 10 = 20 + 3 = 23