UGC NET Computer Science Number System Previous Year Questions (PYQs)

Given equation is

$(146)b+(313){b-2}=(246)_8$

Now,

$(146)_b=1\cdot b^2+4\cdot b+6$

$=b^2+4b+6$

Also,

$(313)_{b-2}=3(b-2)^2+1(b-2)+3$

$=3(b^2-4b+4)+b-2+3$

$=3b^2-12b+12+b+1$

$=3b^2-11b+13$

Now,

$(246)_8=2\cdot 8^2+4\cdot 8+6$

$=128+32+6$

$=166$

So,

$b^2+4b+6+3b^2-11b+13=166$

$4b^2-7b+19=166$

$4b^2-7b-147=0$

Now factorizing,

$4b^2-28b+21b-147=0$

$4b(b-7)+21(b-7)=0$

$(b-7)(4b+21)=0$

So,

$b=7$

Convert each number into decimal form.

$(142)_b=1\cdot b^2+4\cdot b+2$

$=b^2+4b+2$

Now,

$(112)_{b-2}=1\cdot (b-2)^2+1\cdot (b-2)+2$

$=(b-2)^2+(b-2)+2$

$=b^2-4b+4+b-2+2$

$=b^2-3b+4$

Also,

$(75)_8=7\cdot 8+5=56+5=61$

Now,

$(142)b+(112){b-2}=(75)_8$

$(b^2+4b+2)+(b^2-3b+4)=61$

$2b^2+b+6=61$

$2b^2+b-55=0$

$2b^2+11b-10b-55=0$

$b(2b+11)-5(2b+11)=0$

$(b-5)(2b+11)=0$

So,

$b=5$