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CAT 2001 | Question: 19

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The product of $44*11$ in base-10 is $484$. now if the base is $b$ then

$(484)_{10}=(3414)_b\implies 4+b+4b^2+3b^3=484\implies 3b^3+4b^2+b=480$

now put $b=5$ it will satisfy the equation.

in decimal number system $(3111)_5\implies 1+5+25+375=406$

Option (A) is correct.
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