GATE CSE 2022 | Question: 31

Question

Consider three floating point numbers $\textit{A, B}$ and $\textit{C}$ stored in registers $\text{R}_{\text{A}}, \text{R}_{\text{B}}$ and $\text{R}_{\text{C}},$ respectively as per $\textsf{IEEE-754}$ single precision floating point format. The $\text{32-bit}$ content stored in these registers $\text{(in hexadecimal form)}$ are as follows.

$\begin{array} {|c|c|c|} \hline \text{R}_{\text{A}} = \text{0xC1400000} & \text{R}_{\text{B}} = \text{0x42100000} & \text{R}_{\text{C}} = \text{0x41400000} \\\hline \end{array}$

Which one of the following is $\text{FALSE}?$

• A. $\textit{A + C} = 0$
• B. $\textit{C = A + B}$
• C. $\textit{B =3C}$
• D. $\textit{(B – C)} > 0$

Comments

i missed the word "false" while reading this question.

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wow

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Similar questions:
Digital Logic: GATE CSE 2020 | Question: 29
Digital Logic: GATE CSE 2023 | Question: 35
Digital Logic: GATE CSE 2025 | Set 2 | Question: 39

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A=-12, B=36, C=+12

Option A is False Only

So, Option A is correct

Answer 1

Here, option B is the right answer. See the image attched below:

Comments

thanks

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@Arjun Sir... Please make my answer as the best answer since it received highest number of upvotes

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This is wrong ans =c

Answer 2

Given that Numbers are in IEEE-754 single precision.

Representation :: $\text{1 sign bit, 8 exponent bits and 23 Mantissa Bits}$

$\text{Decimal value} = (-1)^{\text{Sign bit}}\times(1.Mantissa)\times(2)^{exponent-127}$

 

$R_A = 0xC1400000 = 1100 \;0001\;0100 \; 0000\; 0000\; 0000\; 0000\; 0000$

$R_A = 0xC1400000 = \underbrace{1}\underbrace{100 \;0001\;0}\underbrace{100 \; 0000\; 0000\; 0000\; 0000\; 0000}$

$\text{Decimal value of R}_A = (-1)^{\text{1}}\times(1.1000000000..)\times(2)^{130-127}=-(1100)_2 = -12$

 

$R_B = 0x42100000 = 0100 \;0010\;0001 \; 0000\; 0000\; 0000\; 0000\; 0000$

$R_B = 0x42100000 = \underbrace{0}\underbrace{100 \;0010\;0}\underbrace{001 \; 0000\; 0000\; 0000\; 0000\; 0000}$

$\text{Decimal value of R}_B = (-1)^{\text{0}}\times(1.001000000000..)\times(2)^{132-127}=(100100)_2 = 36$

 

$R_C = 0x41400000 = 0100 \;0001\;0100 \; 0000\; 0000\; 0000\; 0000\; 0000$

$R_C = 0x41400000 = \underbrace{0}\underbrace{100 \;0001\;0}\underbrace{100 \; 0000\; 0000\; 0000\; 0000\; 0000}$

$\text{Decimal value of R}_C= (-1)^{\text{0}}\times(1.1000000000..)\times(2)^{130-127}=(1100)_2 = 12$

 

$\therefore C=A+B\;\; is\; false.$

 

Alternate :

$R_A = 0xC1400000, R_B = 0x42100000 \;\;and\;\; R_C = 0x41400000$

1. By observing $R_A$ and $R_C$ only MSB bit is different ===> $R_C$ = -$R_A$ ===> $R_C$ + $R_A$ = 0
2. By observing $R_B$ and $R_C$, $R_B$ exponent part is higher than $R_C$ exponent part ===> $R_B$ > $R_C$ ===> $R_B$ - $R_C$ > 0

 

Now Verify other two options.

Comments

in R(B) it should be +(100100).

you had given correct sign in the begining but in last part there is typo mistake

Answer 3

For conceptual & practice purpose, it is good to solve in procedural way. But in exam this question was not meant to be solved in procedural way.

If we observe carefully, A & C are only differ in sign.

So, let A = -x & C = +x.

This makes option A true & B false. Solution ends here.

Comments

Wrong. What if value of B is 2x. Think about it

Answer 4

ans =c

Comments

@Sonu123x

Bro, you did everything correct. 

But near the options, you should tick option B instead of option A. 😅

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🤣🤣...........hope you understand why people get negative marks

Answer 5

Here is the Details Explanation answer is option B ✅