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Simplify \(\left\{\left(\dfrac{20}{36}\right)\div\left(\dfrac{20}{5}\right)\right\}\div\left(\dfrac{5}{11}\times\dfrac{22}{15}+\dfrac{3}{12}\right)+\dfrac{5}{5}\div\dfrac{33}{21}\text{ of }\dfrac{21}{5}\)

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RRB NTPC 2025 Under Graduate CBT 1 Question Paper PDF (20-Jun-2026) (Shift 3)
The correct answer is

\(\tfrac{10}{33}\)

\(\left(\tfrac{20}{36}\right)\div\left(\tfrac{20}{5}\right) = \tfrac{20}{36}\times\tfrac{5}{20} = \tfrac{5}{36}\).

Inside the bracket: \(\tfrac{5}{11}\times\tfrac{22}{15} = \tfrac23\), plus \(\tfrac{3}{12}=\tfrac14\) gives \(\tfrac23+\tfrac14 = \tfrac{11}{12}\).

First part: \(\tfrac{5}{36}\div\tfrac{11}{12} = \tfrac{5}{36}\times\tfrac{12}{11} = \tfrac{5}{33}\).

For the second part, \(\tfrac55=1\), and \(\tfrac{33}{21}\text{ of }\tfrac{21}{5} = \tfrac{33}{5}\), so \(1\div\tfrac{33}{5} = \tfrac{5}{33}\).

Adding both parts: \(\tfrac{5}{33}+\tfrac{5}{33} = \tfrac{10}{33}\).

Hence, the simplified value of the expression is \(\tfrac{10}{33}\).

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Important Questions from Bodmas Rule

  1. The value of 90 ÷ 20 of 6 × [11 ÷ 4 of {3 × 2 - (3 - 8)}] ÷ (9 ÷ 3 × 2) is:

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  3. The value of 20 ÷ 5 of 8 × [9 ÷ 6 × (6 - 3)] - (10 ÷ 2 of 20) is:

  4. The value of \(\left( {18 \div 2\;of\frac{1}{4}} \right)\; \times \;\left( {\frac{2}{3} \div \frac{3}{4}\; \times \;\frac{5}{8}} \right) \div \left( {\frac{2}{3} \div \frac{3}{4}of\frac{3}{4}} \right)\) is:

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