Electronics - Number Systems and Codes - Discussion

Not getting this, please anyone explain clearly.

28, 64, 32, 16, 8, 4,2,1 ---- 0.5, 0.25, 0.125.
So, 1 0 0 1 0 1 1 1 ---- 1 1 = Ans# 10010111.11.

Add 1 to odd numbers, Add 0 to even numbers.
151 (odd)=1,,
151/2=75 (odd)=1,
75/2=37(odd) =1,
37/2=18(even)=0,
18/2=9=1,
9/2=4=0,
4/2=2=0,
2/2=1=1 then in order=10010111.

The system work as below ( x2 increment -- divede 2 decrement).
0.25/2= 0.125
0.5/2 = 0.25
1/2 = 0.5
1
1x2 =2
2x2 = 4
4x2 = 8
8x2 = 16
16x2 = 32
32x2= 64
64x2=128.

128, 64, 32, 16, 8, 4,2,1 ---- 0.5, 0.25, 0.125.
So, 1 0 0 1 0 1 1 1 ---- 1 1 = Ans# 10010111.11.

128 64 32 16 8 4 2 1
151:- 1 0 0 1 0 1 1 1 =151.

NXT IS DECIMAL VALUE
0.75 * 2 = 1.5.

WE WANT TO TAKE ONLY DECIMAL VALUEOF THE ANSWER
eqivalent value for 0.75 =0.11.

The final answer is 10010111.11.

151 = 10010111.
.75 * 2 = 1.50.
.50 * 2 = 1.00.

i.e lower to upper count no. point before =11.
i.e (10010111.11).

151 = 10010111.
.75 * 2 = 1.50.
.50 * 2 = 1.00.

i.e lower to upper count no. point before =11.
i.e (10010111.11).

Convert 151 to hexadecimal (base = 16) first and then the hexadecimal can easily be converted to binary.

First divide 151 by 16. Remainder is 7 and quotient is 9. Preserve the remainder (i.e. 7). Divide 9 by 16 (i.e. the quotient by 16). Remainder is 9 and quotient is 0.

Preserve the remainder and continue the division until the quotient is 0. So the equivalent hexadecimal no. is (97) base 16.

Now make pair of 4 bits for each digit. We know 7 is written as 0111 and 9 as 1001 in binary number system using 4 bits. So the equivalent binary number is (10010111) base 2.

@Lavanya, can you explain the conversion more clearly please?

actually we have to do this in decimal to binary means 151 is divided by 75 now binary no is 1 next 75 is divided by 37 binary n0 is 1 upto 1-0 we will do like that so the ans is D.

The decimal form of 10010111 is 151 and .11 is 0.75

10010111=1*2^7+0+0+1*2^4+0+1*2^2+1*2^1+1*2^0=128+16+4+2+1=151 and

.11=1*2^-1+1*2^-2=0.5+0.25=0.75

In this question 1st check the answer and find easily the answer.

best of luck.