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Best answer
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4 votes

The key thing to notice here is : The given expansion is in octal system i.e. base 8..

Given expansion  :   3 × 512   +   7 × 64   +   5 × 8   +   3

                      =      3  × 83   +   7  × 82  +   5  × 8   +  3

                      =     (3753)8

                      =    (011 111 101 011)2

Hence    number of 1's in given expansion   =   9.

Hence B) should be correct option.

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The given no can be written as 

$3 * 8^{3} + 7 * 8^{2} + 5 * 8^{1} + 3 * 8^{0} = (3753){_{8}}$

We can write 

$(3753){_{8}} = (011111101011){_{2}}$

No of 1's present are 9.

Hence option b) is correct

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