GATE CSE 2025 | Set 2 | Question: 39

Question

Three floating point numbers $X, Y$, and $Z$ are stored in three registers $\mathrm{R}_{\mathrm{X}}, \mathrm{R}_{\mathrm{Y}}$, and $\mathrm{R}_{\mathrm{Z}}$, respectively in $IEEE$ $754$ single precision format as given below in hexadecimal:

$$\mathrm{R}_{\mathrm{X}}=0 \mathrm{xC} 1100000, \mathrm{R}_{\mathrm{Y}}=0 \mathrm{x}  40 \mathrm{C} 00000, \text { and } \mathrm{R}_{\mathrm{Z}}=0 \mathrm{x} 41400000$$

Which of the following option(s) is/are CORRECT?

• A. $4(X+Y)+Z=0$
• B. $2 Y-Z=0$
• C. $4 X+3 Z=0$
• D. $X+Y+Z=0$

Comments

Similar questions:
Digital Logic: GATE CSE 2020 | Question: 29
Digital Logic: GATE CSE 2022 | Question: 31

Digital Logic: GATE CSE 2023 | Question: 35

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X = -9,  Y =  6,  Z = 12

Option A,B,C is satisfied the values

Answer 1

Given Values :-

 

                   RX = 0×C1100000 = 1100 0001 0001 0000 0000 0000 0000 0000

                                IEEE 754  Single  Precision  Format (32-bit) –

Sign (1 bit )

Biased   Exponent ( 8 bits )

Mantissa ( 23 bits )

1

100 0001 0

001 0000 0000 0000 0000 0000

 

                 Biased  Exponent (E) =  (10000010)₂ = 130

       Actual  Exponent (e) = Biased  Exponent (E)  –  Bias (127 for single precision) = 130 - 127 = 3

                Mantissa (M) = 001 0000 0000 0000 0000 0000

                          

                             We Know that :-

Actual number( N ) = (-1)S  X (1.M) X 2e

       

                                            Actual Number = (-1)¹ X 1.00100000000000000000000 X 2³ =  -9

RX = -9

 

Similarly 

      RY = 0×40C00000 = 0100 0000 1100 0000 0000 0000 0000 0000

                           Sign = 0 ( Positive )

                          Biased  Exponent (E) = (10000001)₂ = 129

                           Actual  Exponent (e) = 129 - 127 =  2 

                           Mantissa (M) = 10000000000000000000000 

                        

                           Actual Number = (-1)⁰ X 1.10000000000000000000000  X 2² = 6

RY= 6

 

Finally 

            RZ= 0×41400000 = 0100 0001 0100 0000 0000 0000 0000 0000

                            Sign = 0 ( Positive )

                           Biased  Exponent(E) = (10000010)₂ = 130

                           Actual  Exponent (e) = 130 - 127 =  3

                          Mantissa (M) = 10000000000000000000000

 

                          Actual  Number = (-1)⁰ X 1.10000000000000000000000  X 2³ = 12

RZ = 12

              

 

  Now verify all options 

                         A)  4 (X + Y ) + Z =0  Substitute  X = −9 , Y = 6 ,  and  Z = 12:

                                                4( −9 + 6 ) + 12 = 4 ( −3 ) + 12 = −12 + 12 = 0 

 

                          B) 2 Y − Z = 0  Substitute  Y = 6  and  Z = 12:

                                                2( 6 ) − 12 = 12 − 12 = 0

 

                          C)  4 X + 3 Z = 0 Substitute X=−9 and Z=12:

                                                4( −9 ) + 3 ( 12 ) = − 36 + 36 = 0

 

                          D)  X + Y + Z = 0  Substitute  X = −9 , Y = 6 ,  and  Z = 12:

                                                −9 + 6 + 12 = −9 + 18 = 9

 

A, B, C are correct

Comments

great!!

Answer 2

 IEEE 754 Single-Precision Format Recap

A single precision floating-point number uses 32 bits, divided as follows:

• Bit 31: sign ($s$) : 0 for positive, 1 for negative.  
• Bits 30–23: exponent ($E$) : an 8-bit unsigned integer stored with a bias of 127.  
• Bits 22–0: mantissa ($M$) : a 23-bit fraction interpreted as  

$$
M = \sum_{i=1}^{23} b_i \cdot 2^{-i}.
$$

The real value represented is  
$$
(-1)^s \cdot (1 + M) \cdot 2^{E - 127}.
$$

We decode the three given hexadecimal words using this rule.

Bit Field Extraction

$$
\begin{array}{|c|c|c|c|c|}
\hline
\text{Register} & \text{Hexadecimal} & \text{Sign (1 bit)} & \text{Exponent (8 bits)} & \text{Mantissa (23 bits)} \\
\hline
R_X & \text{C1100000} & 1 & 10000010 & 00100000000000000000000 \\
\hline
R_Y & \text{40C00000} & 0 & 10000001 & 10000000000000000000000 \\
\hline
R_Z & \text{41400000} & 0 & 10000010 & 10000000000000000000000 \\
\hline
\end{array}
$$

Decimal Value Computation

$$
\begin{array}{|c|c|c|c|c|l|}
\hline
\text{Register} & s & E_{\text{(bin)}} & E_{\text{(dec)}} & M_{\text{(dec)}} & \text{Value} \\
\hline
R_X & 1 & 10000010 & 130 & 2^{-3} = 0.125 & (-1)^1 (1 + 0.125) \cdot 2^{130 - 127} = -1.125 \cdot 2^{3} = -9.0 \\
\hline
R_Y & 0 & 10000001 & 129 & 2^{-1} = 0.5 & (+1) (1 + 0.5) \cdot 2^{129 - 127} = 1.5 \cdot 2^{2} = 6.0 \\
\hline
R_Z & 0 & 10000010 & 130 & 2^{-1} = 0.5 & (+1) (1 + 0.5) \cdot 2^{130 - 127} = 1.5 \cdot 2^{3} = 12.0 \\
\hline
\end{array}
$$

Thus,  
$$
X = -9.0, \quad Y = 6.0, \quad Z = 12.0.
$$

Evaluation of Statements

$$
\begin{array}{|c|c|c|c|}
\hline
\text{Option} & \text{Expression} & \text{Substitution} & \text{Result} \\
\hline
\text{A} & 4(X + Y) + Z & 4(-9.0 + 6.0) + 12.0 & 4(-3.0) + 12.0 = -12.0 + 12.0 = 0 \\
\hline
\text{B} & 2Y - Z & 2(6.0) - 12.0 & 12.0 - 12.0 = 0 \\
\hline
\text{C} & 4X + 3Z & 4(-9.0) + 3(12.0) & -36.0 + 36.0 = 0 \\
\hline
\text{D} & X + Y + Z & -9.0 + 6.0 + 12.0 & 9.0 \neq 0 \\
\hline
\end{array}
$$

 

$$
\color{gold} \boxed{\text{Correct options: A, B, C}}
$$

Answer 3

Given values: \[ Rx = C1100000 = -9, \quad Ry = 40C00000 = 6, \quad Rz = 41400000 = 12 \] This implies: \[ X = -9, \quad Y = 6, \quad Z = 12 \] A) \( 4(X + Y) + Z = 0 \) Substitute \( X = -9 \), \( Y = 6 \), and \( Z = 12 \): \[ 4(-9 + 6) + 12 = 4(-3) + 12 = -12 + 12 = 0 \] B) \( 2Y - Z = 0 \) Substitute \( Y = 6 \) and \( Z = 12 \): \[ 2(6) - 12 = 12 - 12 = 0 \] C) \( 4X + 3Z = 0 \) Substitute \( X = -9 \) and \( Z = 12 \): \[ 4(-9) + 3(12) = -36 + 36 = 0 \] D) \( X + Y + Z = 0 \) Substitute \( X = -9 \), \( Y = 6 \), and \( Z = 12 \): \[ -9 + 6 + 12 = -9 + 18 = 9 \]

A, B, C are correct

Answer 4

Sorry For Bad Handwriting , hope this helps

Answer 5

Option A,B,C are correct

Answer in detaill is given g drive below

https://drive.google.com/file/d/1cDfD1l07m6QK6ms0cks5YUovpmQJJAqN/view?usp=drivesdk

Answer 6

X=-9

Y=6

Z=12