Simplify (\(\frac{3}{4} \div 2\frac{1}{4}\)) of 1 \(\frac{7}{11}\) (3 \(\frac{2}{3}\) - \(\frac{15}{6}\))

To simplify this expression, let's break it down step by step. The problem is:

\[ \left( \frac{3}{4} \div 2\frac{1}{4} \right) \text{ of } 1\frac{7}{11} \left( 3\frac{2}{3} - \frac{15}{6} \right) \] Remember that "of" in mathematics means multiplication.

• Step 1: Simplify \(2\frac{1}{4}\) and \(3\frac{2}{3}\) to improper fractions.
  \[ 2\frac{1}{4} = \frac{2 \times 4 + 1}{4} = \frac{9}{4} \] \[ 3\frac{2}{3} = \frac{3 \times 3 + 2}{3} = \frac{11}{3} \]


• Step 2: Simplify \(\frac{3}{4} \div \frac{9}{4}\).
  When dividing by a fraction, multiply by its reciprocal: \[ \frac{3}{4} \div \frac{9}{4} = \frac{3}{4} \times \frac{4}{9} \] Cancel out \(4\) in numerator and denominator: \[ = \frac{3}{1} \times \frac{1}{9} = \frac{3}{9} = \frac{1}{3} \]


• Step 3: Simplify inside the parenthesis: \(3\frac{2}{3} - \frac{15}{6}\).
  We already have \(3\frac{2}{3} = \frac{11}{3}\). Let's convert \(\frac{11}{3}\) to have the same denominator as \(\frac{15}{6}\). \[ \frac{11}{3} = \frac{22}{6} \] So, \[ \frac{22}{6} - \frac{15}{6} = \frac{7}{6} \]


• Step 4: Next, we have: \[ \left( \frac{1}{3} \right) \text{ of } 1\frac{7}{11} \left( \frac{7}{6} \right) \] Let's also convert \(1\frac{7}{11}\) to an improper fraction: \[ 1\frac{7}{11} = \frac{18}{11} \]


• Step 5: Now, the entire expression is: \[ \frac{1}{3} \times \frac{18}{11} \times \frac{7}{6} \] Multiply all numerators together and all denominators together: \[ = \frac{1 \times 18 \times 7}{3 \times 11 \times 6} \] \[ = \frac{126}{198} \]
  Now, simplify \(\frac{126}{198}\). Both numbers are divisible by \(18\): \[ 126 \div 18 = 7 \] \[ 198 \div 18 = 11 \] So, \[ \frac{126}{198} = \frac{7}{11} \]

The correct simplified value is \(\frac{7}{11}\).

Underlying concept: This question tests your ability to work with mixed numbers (converting them to improper fractions), apply the order of operations, multiply and divide fractions, and simplify fractions to their lowest terms.