算数【入試】逆算

問１

解き方: \(7\displaystyle\frac{1}{7} \times (\displaystyle\frac{1}{8} + \boxed{\phantom{A}}\) ― \(\displaystyle\frac{1}{20}) + \displaystyle\frac{1}{28} = 1\displaystyle\frac{2}{7}\)

\(\dfrac{50}{7}\times(\dfrac{1}{8}+\boxed{\phantom{A}}-\dfrac{1}{20})=\dfrac{9}{7}-\dfrac{1}{28}\)

\(\dfrac{1}{8}+\boxed{\phantom{A}}-\dfrac{1}{20}=\dfrac{5}{4}\div\dfrac{50}{7}\)

\(\boxed{\phantom{A}}=\dfrac{7}{40}+\dfrac{1}{20}-\dfrac{1}{8}=\dfrac{1}{10}\)

解き方: \((2\displaystyle\frac{11}{40} + \boxed{\phantom{A}}\) \() \div 1.25\) ― \(\displaystyle\frac{7}{12} = 1\displaystyle\frac{44}{75}\)

\((\displaystyle\frac{91}{40} + \boxed{\phantom{A}}\) \() \div \dfrac{5}{4}= \displaystyle\frac{119}{75}+\displaystyle\frac{7}{12}\)

\(\displaystyle\frac{91}{40} + \boxed{\phantom{A}}= \displaystyle\frac{217}{100}\times\dfrac{5}{4}\)

\(\boxed{\phantom{A}}= \displaystyle\frac{217}{80}-\displaystyle\frac{91}{40}=\dfrac{7}{16}\)

解き方: \(\displaystyle\frac{2025 + \boxed{\phantom{A}}}{29} = 87 + 3 \div 0.0625\)

\(\displaystyle\frac{2025 + \boxed{\phantom{A}}}{29} = 87 + 3 \div \dfrac{5}{80}\)

\(\displaystyle\frac{2025 + \boxed{\phantom{A}}}{29} = 135\)

\(2025 + \boxed{\phantom{A}} = 135\times29\)

\(\boxed{\phantom{A}} = 3915-2025=1890\)

解き方: \(\{18 + (3 \times \boxed{\phantom{A}}\) ― \(\displaystyle\frac{1}{4}) \div \displaystyle\frac{5}{9}\} \times 0.8 = 27\)

\(\{18 + (3 \times \boxed{\phantom{A}}\) ― \(\displaystyle\frac{1}{4}) \div \displaystyle\frac{5}{9}\} = 27\div\dfrac{4}{5}\)

\((3 \times \boxed{\phantom{A}}\) ― \(\displaystyle\frac{1}{4}) \div \displaystyle\frac{5}{9} = \dfrac{135}{4}-18\)

\(3 \times \boxed{\phantom{A}}\) ― \(\displaystyle\frac{1}{4} = \dfrac{63}{4} \times\displaystyle\frac{5}{9} \)

\(3 \times \boxed{\phantom{A}}= \dfrac{35}{4} +\displaystyle\frac{1}{4}\)

\(\boxed{\phantom{A}}= 9\div3=3\)

解き方: \(4\) ― \(\{5\) ― \((\) \(\boxed{\phantom{A}}\) ― \(3) \times \displaystyle\frac{1}{3}\} \times \displaystyle\frac{3}{4} = 1\displaystyle\frac{1}{4}\)

\(\{5\) ― \((\) \(\boxed{\phantom{A}}\) ― \(3) \times \displaystyle\frac{1}{3}\} \times \displaystyle\frac{3}{4} = 4-1\displaystyle\frac{1}{4}\)

\(5\) ― \((\) \(\boxed{\phantom{A}}\) ― \(3) \times \displaystyle\frac{1}{3} = \displaystyle\frac{11}{4}\div \displaystyle\frac{3}{4}\)

\((\) \(\boxed{\phantom{A}}\) ― \(3) \times \displaystyle\frac{1}{3} = 5-\displaystyle\frac{11}{3}\)

\(\boxed{\phantom{A}}\) ― \(3 =\displaystyle\frac{4}{3}\div\displaystyle\frac{1}{3} \)

\(\boxed{\phantom{A}}=4+3=7\)

解き方: \((35\) ― \(7\times\boxed{\phantom{A}}\) \()\times6=84\)
\(35\) ― \(7\times\boxed{\phantom{A}}=84\div6\)
\(7\times\boxed{\phantom{A}}=35-14\)
\(\boxed{\phantom{A}}=21\div7=3\)

解き方: \(46-7\times(17+\boxed{\phantom{A}}\times3)\div7=11\)
\(7\times(17+\boxed{\phantom{A}}\times3)\div7=46-11\)
\(17+\boxed{\phantom{A}}\times3=35\)
\(\boxed{\phantom{A}}\times3=35-17\)
\(\boxed{\phantom{A}}=18\div3=6\)

解き方: \((1+\dfrac{1}{5})\times(1-\boxed{\phantom{A}}+\dfrac{1}{15})=1\dfrac{1}{25}\)

\(1-\boxed{\phantom{A}}+\dfrac{1}{15}=\dfrac{26}{25}\div\dfrac{6}{5}\)

\(\boxed{\phantom{A}}=1+\dfrac{1}{15}-\dfrac{13}{15}=\dfrac{1}{5}\)

解き方: \([\) \(9\div\{3\div(3\dfrac{8}{9}+3\dfrac{17}{18})\}\times\boxed{\phantom{A}}\) \(]\div(\dfrac{3}{4}\times0.75)=47\)

\(9\div(3\div\dfrac{47}{6})\times\boxed{\phantom{A}}=47\times\dfrac{9}{16}\)

\(9\times\boxed{\phantom{A}}=47\times\dfrac{9}{16}\times\dfrac{18}{47}\)

\(\boxed{\phantom{A}}=\dfrac{81}{8}\div9=1\dfrac{1}{8}\)

解き方: \((0.8\div0.26+\boxed{\phantom{A}}\times\dfrac{1}{13})\div0.3=20\)

\(\dfrac{40}{13}+\boxed{\phantom{A}}\times\dfrac{1}{13}=20\times0.3\)

\(\boxed{\phantom{A}}\times\dfrac{1}{13}=6-\dfrac{40}{13}\)

\(\boxed{\phantom{A}}=(6-\dfrac{40}{13})\times13=78-40=38\)