Real numbers a and b satisfy the equations 3a=81b+2 and 125b=5a−3 The value of ab is
A
17
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B
9
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C
12
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D
60
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Solution
The correct option is D 60 3a=81b+2 3a=(34)b+2 3a=34b+8 Base are same so a=4b+8 a−4b=8.............eq1 125b=5a−3 (53)b=5a−3 53b=5a−3 base are same so 3b=a−3 a−3b=3..............eq2 subtract eq1 and eq2 b=−5 Substitute x=-5 in eq1 a+−4(−5)=8 a+20=8 a=−12 Hence ab=−12×−5=60