1.4.1 Ulangkaji Pecahan, Praktis Berformat PT3 - Matematik Tingkatan 1, 2 dan 3

Soalan 1:

3 \frac{2}{5} + 1 \frac{5}{6} - \frac{1}{5} =

Penyelesaian:

\begin{array}{l} 3 \frac{2}{5} + 1 \frac{5}{6} - \frac{1}{5} \\ = 3 \frac{12}{30} + 1 \frac{25}{30} - \frac{6}{30} \\ = 4 \frac{37}{30} - \frac{6}{30} \\ = 4 \frac{31}{30} \\ = 5 \frac{1}{30} \end{array}

Soalan 2:

6 \frac{1}{12} - 3 \frac{2}{3} - 1 \frac{5}{6} =

Penyelesaian:

\begin{array}{l} 6 \frac{1}{12} - 3 \frac{2}{3} - 1 \frac{5}{6} \\ = 6 \frac{1}{12} - 3 \frac{8}{12} - 1 \frac{10}{12} \\ = (5 \frac{13}{12} - 3 \frac{8}{12}) - 1 \frac{10}{12} \\ = 2 \frac{5}{12} - 1 \frac{10}{12} \\ = 1 \frac{17}{12} - 1 \frac{10}{12} \\ = \frac{7}{12} \end{array}

Soalan 3:

30 - (\frac{2}{7} \div \frac{2}{21}) =

Penyelesaian:

\begin{array}{l} 30 - (\frac{2}{7} \div \frac{2}{21}) \\ = 30 - (\frac{2}{7} \times \frac{21^{3}}{2}) \\ = 30 - 3 \\ = 27 \end{array}

Soalan 4:

(7 \frac{1}{2} - 3 \frac{2}{3}) \times \frac{12}{69} =

Penyelesaian:

\begin{array}{l} (7 \frac{1}{2} - 3 \frac{2}{3}) \times \frac{12}{69} \\ = (7 \frac{3}{6} - 3 \frac{4}{6}) \times \frac{12}{69} \\ = (6 \frac{9}{6} - 3 \frac{4}{6}) \times \frac{12}{69} \\ = 3 \frac{5}{6} \times \frac{12}{69} \\ = \frac{23}{6} \times \frac{12^{2}}{69^{3}} \\ = \frac{2}{3} \end{array}

Soalan 5:

1 \frac{1}{5} \div (4 - 2 \frac{1}{2}) =

Penyelesaian:

\begin{array}{l} 1 \frac{1}{5} \div (4 - 2 \frac{1}{2}) \\ = 1 \frac{1}{5} \div (3 \frac{2}{2} - 2 \frac{1}{2}) \\ = \frac{6}{5} \div (1 \frac{1}{2}) \\ = \frac{6}{5} \div \frac{3}{2} \\ = \frac{6^{2}}{5} \times \frac{2}{3} \\ = \frac{4}{5} \end{array}