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Find the value of

[(48÷8)×[497+404×(7−3)]][(48 \div 8)\times[\frac{49}{7}+\frac{40}{4}\times(7-3)]]

A282

B286

C284

D273

Answer:

A. 282

Read Explanation:

Let’s solve step by step:
(48÷8)×[497+404×(7−3)]](48 \div 8)\times[\frac{49}{7}+\frac{40}{4}\times(7-3)]]


48÷8=648 \div 8 = 6
497=7\frac{49}{7} = 7
404=10\frac{40}{4} = 10
(7−3)=4(7 - 3) = 4

So,
10×4=4010 \times 4 = 40
7+40=477 + 40 = 47
6×47=2826 \times 47 = \boxed{282}
Final Answer: 282


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