1.prac

1. Convert 0b00110011 to Decimal

numerical value of position	128	64	32	16	8	4	2	1
Bits	0	0	1	1	0	0	1	1

$32+16+2+1=51$

Ans → 51

2. Convert 0o7654 to Hexadecimal

Step 1: Convert to binary

Octal (0o)	7	6	5	4
Binary (0b)	111	110	101	100

Step 2: Regroup into fours, starting from the least significant bit

From this → 111 110 101 100

To this → {111 1} {10 10} {1 100}

Step 3: Convert to Hexadecimal

Binary (0b)	1111	1010	1100
Hex (0x)	F	A	C

Ans → 0xFAC

3. Convert -5 to 2’s Complement. Express the answer as an 8 bit binary

5 expressed as an 8-bit binary → 0b00000101

Now, we convert it to -5 expressed as a 2’s Complement binary:

1. Invert: 0b00000101 → 0b11111010
2. Add 1: 0b11111010 + 1 = 0b11111011
3. Yuhhhhhh

4. Convert 0x8F00 to Decimal

Step 1: Break into Binary

Hex (0x)	8	F	0	0
Binary (0b)	1000	1111	0000	0000

Step 2: Count! just like in qn 1!

You should b able to get this: $32,768+2,048+1,024+512+256=36,608$

BUT WAITTT, there’s a trick for this. Getting “32,768” should be fairly straight forward (u should have this memorised). But the addition of the rest is very tedious. So, what you could do instead, is:

Original Binary (0b)	1000	1111	0000	0000
“Added” Binary	0000	0001	0000	0000
“Rounded Up” Binary	1001	0000	0000	0000
So, we’ve added another binary number to the equation to round up the original number.

This is rounded up binary is MUCH easier to calculate! $32,768+4096=36,864$

But we have to remove our rounding errors… We added 0b 0001 0000 0000 to round up our original binary, so we have to subtract it from $36,864$!

So the full equation is → $32,768+4096-256=36,608$

Multiplication Method

Hex (0x)	8	F	0	0
Decimal Value	$8\times 16^3$	$15\times 16^2$	0000	0000

ANS → $8\times 16^3+15\times 16^2=36,608$

This might b faster since ya’ll are allowed scrap paper to do multiplication! Whichever floats ur boat

5. Solve NOT((7 OR 5) XOR 10)

Also can be expressed as → ~((7 | 5)^10)

Step 1: Convert to binary

~((0b111 | 0b101) ^ 0b1010)

Step 2: solve inner statement (0b111 | 0b101)

Bitwise OR
0b111 →	1	1	1
0b101 →	1	0	1
Output	1	1	1

(0b111 | 0b101) = 0b111

Step 3: solve remaining statement ~(0b111 ^ 0b1010)

Bitwise XOR
0b111 →	0	1	1	1
0b1010 →	1	0	1	0
Output	1	1	0	1

~(0b111 ^ 0b1010) = ~0b1101

= 0b0010

= 2 ← ANS