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

Soalan 6:

$1 \frac{1}{2} \div \frac{1}{4} \times \frac{4}{5} =$

Penyelesaian:

$\begin{array}{l} 1 \frac{1}{2} \div \frac{1}{4} \times \frac{4}{5} \\ = \frac{3}{2} \times \frac{4^{2}}{1} \times \frac{4}{5} \\ = \frac{24}{5} \\ = 4 \frac{4}{5} \end{array}$

Soalan 7:

$4 \frac{1}{5} \div 1 \frac{1}{6} \times 2 \frac{2}{9} =$

Penyelesaian:

$\begin{array}{l} 4 \frac{1}{5} \div 1 \frac{1}{6} \times 2 \frac{2}{9} \\ = \frac{21^{3}}{5} \times \frac{6^{2}}{7} \times \frac{20^{4}}{9^{3}} \\ = 8 \end{array}$

Soalan 8:

$2 \frac{3}{4} \times 1 \frac{1}{3} \div 7 \frac{1}{3} =$

Penyelesaian:

$\begin{array}{l} 2 \frac{3}{4} \times 1 \frac{1}{3} \div 7 \frac{1}{3} \\ = \frac{11}{4} \times \frac{4}{3} \div \frac{22}{3} \\ = \frac{11}{4} \times \frac{4}{3} \times \frac{3}{22^{2}} \\ = \frac{1}{2} \end{array}$

Soalan 9:

$4 \frac{5}{6} - 2 \frac{1}{3} \div \frac{5}{9} =$

Penyelesaian:

$\begin{array}{l} 4 \frac{5}{6} - 2 \frac{1}{3} \div \frac{5}{9} \\ = 4 \frac{5}{6} - (\frac{7}{3} \div \frac{5}{9}) \\ = 4 \frac{5}{6} - (\frac{7}{3} \times \frac{9^{3}}{5}) \\ = 4 \frac{5}{6} - \frac{21}{5} \\ = 4 \frac{5}{6} - 4 \frac{1}{5} \\ = \frac{25}{30} - \frac{6}{30} \\ = \frac{19}{30} \end{array}$

Soalan 10:

$(\frac{7}{9} + \frac{11}{15}) \div (7 \frac{8}{15} - 4 \frac{1}{5}) =$

Penyelesaian:

$\begin{array}{l} (\frac{7}{9} + \frac{11}{15}) \div (4 \frac{8}{15} - 3 \frac{1}{5}) \\ = (\frac{35}{45} + \frac{33}{45}) \div (4 \frac{8}{15} - 3 \frac{3}{15}) \\ = \frac{68}{45} \div 1 \frac{5}{15} \\ = \frac{68}{45} \div 1 \frac{1}{3} \\ = \frac{68}{45} \div \frac{4}{3} \\ = \frac{68^{17}}{45^{15}} \times \frac{3}{4} \\ = \frac{17}{15} \\ = 1 \frac{2}{15} \end{array}$

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