UGC NET CSE | June 2005 | Part 2 | Question: 6

Question

$\left (101011 \right)_{2} = \left (53 \right)_{b}$, then $’b’$ is equal to :

• A. $4$
• B. $8$
• C. $10$
• D. $16$

Answer 1

convert both the sides of equation to it's decimal values.

Therefore,

$32+8+2+1=3+5b$

$43=3+5b$

$40=5b$

$b=8$

Answer 2

When we convert both sides of the equation to decimal we get $b’s$ value.

$(101011)_2=(53)_b$

$= 1*2^0+1*2^1+0*2^2+1*2^3+0*2^4+1*2^5=3*b^0+5*b^1$

$=1+2+0+8+0+32=3+5b$

$=43=5b+3$

$=5b=40$

$=b=\frac{40}{5}$

$\therefore b=8$

Option $b$ is correct.