SOLUTION: Practice exercises on number bases 1 - Studypool - Free Printable

Problem: Convert numbers from different bases to base 10 (denary)

We will solve each problem step by step.

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#### Problem 1: Convert \( 11011_2 \) to a number in base ten.

Solution:
To convert a binary number (\( base \, 2 \)) to decimal (\( base \, 10 \)), we use the positional value method. Each digit is multiplied by \( 2^n \), where \( n \) is the position of the digit from right to left, starting at 0.

The binary number \( 11011_2 \) can be expanded as:
\[
11011_2 = 1 \cdot 2^4 + 1 \cdot 2^3 + 0 \cdot 2^2 + 1 \cdot 2^1 + 1 \cdot 2^0
\]

Now, calculate each term:
\[
1 \cdot 2^4 = 1 \cdot 16 = 16
\]
\[
1 \cdot 2^3 = 1 \cdot 8 = 8
\]
\[
0 \cdot 2^2 = 0 \cdot 4 = 0
\]
\[
1 \cdot 2^1 = 1 \cdot 2 = 2
\]
\[
1 \cdot 2^0 = 1 \cdot 1 = 1
\]

Add these values together:
\[
16 + 8 + 0 + 2 + 1 = 27
\]

Thus, the decimal equivalent of \( 11011_2 \) is \( 27 \).

Answer:
\[
\boxed{C}
\]

---

#### Problem 2: Convert \( 100011_2 \) to a number in base ten.

Solution:
Similarly, expand the binary number \( 100011_2 \):
\[
100011_2 = 1 \cdot 2^5 + 0 \cdot 2^4 + 0 \cdot 2^3 + 0 \cdot 2^2 + 1 \cdot 2^1 + 1 \cdot 2^0
\]

Calculate each term:
\[
1 \cdot 2^5 = 1 \cdot 32 = 32
\]
\[
0 \cdot 2^4 = 0 \cdot 16 = 0
\]
\[
0 \cdot 2^3 = 0 \cdot 8 = 0
\]
\[
0 \cdot 2^2 = 0 \cdot 4 = 0
\]
\[
1 \cdot 2^1 = 1 \cdot 2 = 2
\]
\[
1 \cdot 2^0 = 1 \cdot 1 = 1
\]

Add these values together:
\[
32 + 0 + 0 + 0 + 2 + 1 = 35
\]

Thus, the decimal equivalent of \( 100011_2 \) is \( 35 \).

Answer:
\[
\boxed{A}
\]

---

#### Problem 3: Convert \( 211_3 \) to a number in base ten.

Solution:
To convert a ternary number (\( base \, 3 \)) to decimal (\( base \, 10 \)), we use the positional value method. Each digit is multiplied by \( 3^n \), where \( n \) is the position of the digit from right to left, starting at 0.

The ternary number \( 211_3 \) can be expanded as:
\[
211_3 = 2 \cdot 3^2 + 1 \cdot 3^1 + 1 \cdot 3^0
\]

Now, calculate each term:
\[
2 \cdot 3^2 = 2 \cdot 9 = 18
\]
\[
1 \cdot 3^1 = 1 \cdot 3 = 3
\]
\[
1 \cdot 3^0 = 1 \cdot 1 = 1
\]

Add these values together:
\[
18 + 3 + 1 = 22
\]

Thus, the decimal equivalent of \( 211_3 \) is \( 22 \).

Answer:
\[
\boxed{B}
\]

---

#### Problem 4: Convert \( 1102_3 \) to a number in base ten.

Solution:
Expand the ternary number \( 1102_3 \):
\[
1102_3 = 1 \cdot 3^3 + 1 \cdot 3^2 + 0 \cdot 3^1 + 2 \cdot 3^0
\]

Calculate each term:
\[
1 \cdot 3^3 = 1 \cdot 27 = 27
\]
\[
1 \cdot 3^2 = 1 \cdot 9 = 9
\]
\[
0 \cdot 3^1 = 0 \cdot 3 = 0
\]
\[
2 \cdot 3^0 = 2 \cdot 1 = 2
\]

Add these values together:
\[
27 + 9 + 0 + 2 = 38
\]

Thus, the decimal equivalent of \( 1102_3 \) is \( 38 \).

Answer:
\[
\boxed{D}
\]

---

#### Problem 5: Convert \( 111.11_2 \) to its denary equivalent.

Solution:
For binary numbers with a fractional part, the integer part is converted as usual, and the fractional part is converted by multiplying each digit by \( 2^{-n} \), where \( n \) is the position of the digit after the radix point (starting at 1).

The binary number \( 111.11_2 \) can be split into:
- Integer part: \( 111_2 \)
- Fractional part: \( .11_2 \)

Integer part:
\[
111_2 = 1 \cdot 2^2 + 1 \cdot 2^1 + 1 \cdot 2^0
\]
\[
= 1 \cdot 4 + 1 \cdot 2 + 1 \cdot 1 = 4 + 2 + 1 = 7
\]

Fractional part:
\[
.11_2 = 1 \cdot 2^{-1} + 1 \cdot 2^{-2}
\]
\[
= 1 \cdot \frac{1}{2} + 1 \cdot \frac{1}{4} = 0.5 + 0.25 = 0.75
\]

Combine the integer and fractional parts:
\[
7 + 0.75 = 7.75
\]

Thus, the denary equivalent of \( 111.11_2 \) is \( 7.75 \).

Answer:
\[
\boxed{A}
\]

---

#### Problem 6: Convert \( 101.11_2 \) to its denary equivalent.

Solution:
Similar to Problem 5, split the binary number \( 101.11_2 \) into:
- Integer part: \( 101_2 \)
- Fractional part: \( .11_2 \)

Integer part:
\[
101_2 = 1 \cdot 2^2 + 0 \cdot 2^1 + 1 \cdot 2^0
\]
\[
= 1 \cdot 4 + 0 \cdot 2 + 1 \cdot 1 = 4 + 0 + 1 = 5
\]

Fractional part:
\[
.11_2 = 1 \cdot 2^{-1} + 1 \cdot 2^{-2}
\]
\[
= 1 \cdot \frac{1}{2} + 1 \cdot \frac{1}{4} = 0.5 + 0.25 = 0.75
\]

Combine the integer and fractional parts:
\[
5 + 0.75 = 5.75
\]

Thus, the denary equivalent of \( 101.11_2 \) is \( 5.75 \).

Answer:
\[
\boxed{C}
\]

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Final Answers:

1. \(\boxed{C}\)
2. \(\boxed{A}\)
3. \(\boxed{B}\)
4. \(\boxed{D}\)
5. \(\boxed{A}\)
6. \(\boxed{C}\)