Quiz 1 — Number Systems & Conversions

Question 1 of 12 · easy · Base Conversions

The binary equivalent of the decimal number 25 is:

$25 ÷ 2 = 12$ R$1$, $12 ÷ 2 = 6$ R$0$, $6 ÷ 2 = 3$ R$0$, $3 ÷ 2 = 1$ R$1$, $1 ÷ 2 = 0$ R$1$. Read bottom to top: $11001$.

Question 2 of 12 · easy · Base Conversions

The hexadecimal equivalent of the binary number 11011010 is:

Group by 4 from right: $1101 \ 1010$. $1010=A$, $1101=D$. Result: $DA$.

Question 3 of 12 · easy · Binary Codes

The decimal number 178 in BCD code is:

BCD: each decimal digit converted separately. $1→0001$, $7→0111$, $8→1000$. So $178 = 0001 \ 0111 \ 1000$.

Question 4 of 12 · easy · Complements

The 1's complement of the binary number 1010110 is:

1's complement: flip every bit. $1010110 → 0101001$.

Question 5 of 12 · easy · Complements

The 2's complement of 1010 is:

1's complement of $1010 = 0101$. Add 1: $0101 + 1 = 0110$.

Question 6 of 12 · medium · Base Conversions

The octal equivalent of binary 110101011 is:

Group by 3 from right: $110 \ 101 \ 011$. $011=3$, $101=5$, $110=6$. Result: $653$.

Question 7 of 12 · easy · Binary Codes

Which of the following is NOT a valid BCD code?

BCD only uses codes 0000-1001 (decimal 0-9). $1010 = 10$, which is NOT a valid BCD digit.

Question 8 of 12 · medium · Binary Codes

The Gray code for binary 1010 is 1111.

$G_3=B_3=1$, $G_2=1⊕0=1$, $G_1=0⊕1=1$, $G_0=1⊕0=1$. Gray = $1111$. ✓

Question 9 of 12 · easy · Number Systems

Convert $(3F)_{16}$ to decimal.

$3 × 16^1 + 15 × 16^0 = 48 + 15 = 63$.

Question 10 of 12 · medium · Complements

Perform binary subtraction using 2's complement: $1001 - 0100$

2's complement of $0100$: flip → $1011$, add 1 → $1100$. Add: $1001 + 1100 = 10101$. Discard carry: $0101$. Verify: $9 - 4 = 5$ ✓.

Question 11 of 12 · easy · Binary Codes

The Excess-3 code for decimal digit 4 is:

Excess-3: add 3 to digit then convert to binary. $4 + 3 = 7 = 0111$.

Question 12 of 12 · medium · Number Systems

How many bits are needed to represent the decimal number 200 in binary?

$2^7 = 128 < 200 < 256 = 2^8$. So 8 bits are needed.