(4×103)+(2×102)+(0×101)+(2×100)+(1×10−1)+(2×10−2) is the expanded form of the number ___.
A
4202.12
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B
4220.12
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C
4202.02
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D
4202.21
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Solution
The correct option is A4202.12 (4×103)+(2×102)+(0×101)+(2×100)+(1×10−1)+(2×10−2) =4000+200+0+(2×1)+(1×110)+(2×1100)(∵a0=1) =4000+200+0+2+0.1+0.02 =4202.12