Question

Evaluate the expression \[ \frac{\left( 1\dfrac{29}{36} + 4\dfrac{1}{8} \times 1\dfrac{7}{11} \right) \div \left( 5\dfrac{1}{9} - 7\dfrac{7}{8} \div 9\dfrac{9}{20} \right)} {\, 3\dfrac{1}{5} \div \dfrac{9}{2} \ \text{ of } \ 5\dfrac{1}{3}} \]

• A. \(\dfrac{2}{15}\)
• B. \(2\)
• C. \(7\dfrac{1}{2}\)
• D. \(15\)

Hint:

Convert mixed numbers to improper fractions first. In school-order conventions, evaluate “of” (multiplication) before division, then proceed left to right for remaining \(\times,\div\).

Answer 1

The Correct Option is D

Step 1: Convert all mixed numbers to improper fractions.
\(1\dfrac{29}{36}=\dfrac{65}{36},\quad 4\dfrac{1}{8}=\dfrac{33}{8},\quad 1\dfrac{7}{11}=\dfrac{18}{11}\).
\(5\dfrac{1}{9}=\dfrac{46}{9},\quad 7\dfrac{7}{8}=\dfrac{63}{8},\quad 9\dfrac{9}{20}=\dfrac{189}{20}\).

Step 2: Simplify the left (numerator) block.
\(\dfrac{33}{8}\times\dfrac{18}{11}=\dfrac{27}{4}\Rightarrow \dfrac{65}{36}+\dfrac{27}{4}=\dfrac{65+243}{36}=\dfrac{308}{36}=\dfrac{77}{9}\).
Also \(\dfrac{63}{8}\div\dfrac{189}{20}=\dfrac{63}{8}\cdot\dfrac{20}{189}=\dfrac{5}{6}\). Hence \(\dfrac{46}{9}-\dfrac{5}{6}=\dfrac{92-15}{18}=\dfrac{77}{18}\).
Therefore,
\[ \left(\dfrac{77}{9}\right)\div\left(\dfrac{77}{18}\right)=\dfrac{77}{9}\cdot\dfrac{18}{77}=2. \]

Step 3: Simplify the right (denominator) block (of first).
\(\dfrac{9}{2}\ \text{ of }\ 5\dfrac{1}{3}=\dfrac{9}{2}\times\dfrac{16}{3}=24\).
So \(3\dfrac{1}{5}\div 24=\dfrac{16}{5}\div 24=\dfrac{16}{120}=\dfrac{2}{15}\).

Step 4: Combine numerator and denominator to get the final value.
\[ \dfrac{2}{\ \dfrac{2}{15}\ }=2\cdot\dfrac{15}{2}=15\Rightarrow \boxed{15}. \]