Option 3 : 10010

**Concept of Binary number system:**

- In the binary number system (base of 2), there are only two digits: 0 and 1,
- The place values are 20, 21, 22, 23, etc.
- Binary digits are abbreviated as bits. For example, 1101 is a binary number of 4 bits (is., it is a binary number containing four binary digits.)
- A binary number may have any number of bits. Consider the number 11001.01 1. Note the binary point (counterpart of the decimal point in decimal number system) in this number.
- The bit on the extreme right is called the least significant bit (LSB) and the bit on the extreme left is called the most significant bit (MSB). Each bit has its positional value as shown in Fig. given below.

__Calculation:__

Convert the given binary numbers into decimal equivalents.

Calculate the product of the two decimal equivalents.

Convert the decimal number into binary equivalent

Given binary numbers are 011 and 110

011 = (3)10

110 = (6)10

The product of the above two numbers = 3 × 6 = (18)10

= (10010)2

Option 2 : 1101

Given:

two numbers 1010 and 0011 expressed in binary format

Calculation:

**Method 1: **

1 + 0 = 1

1 + 1 = 0 (carry = 1)

0 + 0 = 0

(1010)_{2} + (0011)_{2} = (1101)_{2}

**Method 2:**

⇒ (1010)2 = 23 × 1 + 22 × 0 + 21 × 1 + 20 × 0 = (10)10

⇒ (0011)2 = 23 × 0 + 22 × 0 + 21 × 1 + 20 × 1 = (3)10

⇒ sum of two number = (10)10 + (3)10 = (13)10

∴ (13)10 = 2^{3} × 1 + 2^{2} × 1 + 2^{1} × 0 + 20 × 1 = (1101)2

Option 2 : 1000112

This problem can be solved in two ways

**Method 1:**

By following simple multiplication rule as shown

**Method 2:**

By converting binary to decimal, performing multiplication, and then again converting decimal to binary.

Here, (111)_{2} = 2^{2} + 2^{1} + 2^{0} = (7)_{10}

(101)_{2} = 2^{2} + 2^{0} = (5)_{10}

⇒ 5 × 7 = (35)_{10} = 2^{5} + 2^{1} + 2^{0} = **(100011) _{2}**

Option 1 : 110101

__Concept:__

**1. Signed magnitude representation **uses** the **most significant bit (MSB) a sign bit.

- If the sign bit is ‘0’ then the number is positive.
- If the sign bit is ‘1’ then the number is negative.

The remaining bits represent the magnitude of the binary number.

**2. 1’s complement** **representation:**

It is a representation of a binary number obtained by toggling all bits in it i.e. transforming the 0 bit to 1 and the 1 bit to 0.

**3. 2’s complement** **representation:**

It is obtained by simply adding 1 to the 1’s complement of that binary number.

__Calculation:__

Given: 4-bit binary number = 1100

**Signed magnitude representation (P):** - 4

**1’s complement** **representation (Q): **- 3

**2’s complement** **representation (R): **- 4

So,

P + Q + R = (-4) + (-3) + (-4) = -11

(-11)_{10} is represented in 2’s complement as:

-(11)_{10} = 10101

Since the Options are in 6 bits so, we copy sign bit once towards left.

So 6-bit representation of (-11)Option 3 : 5

__Concept:__

__Calculation:__

Given: (43)x = (y3)8

⇒ 3 * x^{0} + 4 * x^{1} = 3 * 80 + y * 8^{1}

⇒ 3 + 4x = 3 + 8y

⇒ 4x = 8y

∴ x = 2y

Now, y < 8 and x > 4 (because the number represented in base x is 43 or redix is always greater )

The following are possible solutions

y = 3, 4, 5, 6, 7 (y < 8)

x = 6, 8, 10, 12, 14

Hence number of possible solution is five.

Option 1 : 101

Let X = (A) (B)

Given that,

A = 101

X = 11001

By using decimal conversion of A & X, we get

(A)_{10 }= (2^{0} x 1) + (2^{1} x 0) + (2^{2} x 1) = 1 + 0 + 4

∴ (A)_{10} = 5

Now, (X)_{10} = (20 x 1) + (21 x 0) + (22 x 0) + (2^{3} x 1) + (2^{4} x 1)

(X)_{10} = 1 + 0 + 0 + 8 + 16

(X)10 = 25

So that, (B)_{10} = 25 / 5

(B)_{10} = 5

After decimal to binary conversion of (B)_{10 , }we get

**(B) _{2} = 101**

Option 3 : 11010100

1-1= 0 |

0-1= 1 (with borrow 1) |

1-0= 1 |

0-0= 0 |

1 1 1 0 0 1 1 1

__- 0 0 0 1 0 0 1 1__

1 1 0 1 0 1 0 0

Step 1: 1 – 1 = 0.

Step 2: 1 – 1 = 0.

Step 3: 1 – 0 = 1.

Step 4: 0 – 0 = 0.

Step 5: 0 – 1 = 1.Borrow to make 10 – 1 = 1.

Step 6: 0 – 0 = 0.

Step 7: 1 – 0 = 1.

Step 7: 1 – 0 = 1.

Hence **option (3)** is the correct answer.

__Remember:__ When zero takes 1 as carry from its left side number, '0' will become '10' which is equal to '2' (2-1=1) and if that '10' further gives carry then it will become '1' not '0'.

Option 1 : (1110)_{2} and (1)_{2}

(101011)_{2 }= 1 × 2^{5} + 0 × 24 + 1 × 23 + 0 × 22 + 1 × 21 + 1 × 20 =(43)10_{ }

(11)_{2 }=1 × 21 + 1 × 20 =(3)10

**Performing division 43 by 3**

3 | 43 | 1 |

14 |

(14)_{10} = (1110)_{2} **(quotient)**

(1)_{10 }= (1)_{2 }**(reminder)**

**Therefore Option 1 is correct**

Option 2 : 11001

**Concept:**

In Binary addition, the following rules obey:

- 1 + 1 = 10 (0 is sum value and 1 is carry)
- 1 + 0 = 1
- 0 + 1 = 1 and
- 0 + 0 = 0

__Calculation__:

1 1 0 1 (Binary number 1)

+**1 1 0 0** (Binary number 2)

--------------

1 1 0 0 1 (Sum values)

Hence **option (2)** is the correct answer

Option 1 : 211

__Key Points__

- A number with base 2 is known as a Binary Number, and numbers with base 10 are known as Decimal Numbers in the Number System.
- The decimal numbers are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, and so on.
- Binary numbers, on the other hand, are used in digital systems and are represented by just two digits – 0 and 1.

__Binary to Decimal Conversion:__

Given:

Given binary number = (11010011)2

Calculation:

(11010011)2 = (1 × 27 + 1 × 26 + 0 × 25 + 1 × 24 + 0 × 23 + 0 × 22 + 1 × 21 + 1 × 20)10

⇒ (128 + 64 + 16 + 2 + 1)10

⇒ (211)10

So, the decimal equivalent of the binary number (11010011)2 is (211)10.

∴ The number at the place of the question mark will be 211.