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Find the value of

[(44÷4)×[488+215×(7−2)]][(44 \div 4) \times [\frac{48}{8}+\frac{21}{5}\times(7-2)]]

A296

B299

C313

D297

Answer:

D. 297

Read Explanation:

[(44÷4)×(488+215×(7−2))]\Big[ (44 \div 4) \times \Big( \frac{48}{8} + \frac{21}{5} \times (7-2) \Big) \Big]

  1. (44÷4=1144 \div 4 = 11)

  2. (48÷8=648 \div 8 = 6)

  3. (7 - 2 = 5)

  4. (215×5=21\frac{21}{5} \times 5 = 21)

Now the inner bracket becomes:

488+215×(7−2)=6+21=27\frac{48}{8} + \frac{21}{5} \times (7-2) = 6 + 21 = 27

Step 2: Multiply by 11
11×27=29711 \times 27 = 297


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