###### GATE 2001

October 7, 2023###### Nielit Scientist-D 2016 march

October 7, 2023# Digital-Logic-Design

Question 10 |

Consider three registers R1, R2 and R3 that store numbers in IEEE-754 single precision floating point format. Assume that R1 and R2 contain the values (in hexadecimal notation) 0x42200000 and 0xC1200000, respectively.

If R3 = R1/R2, what is the value stored in R3?

0x40800000 | |

0x83400000 | |

0xC8500000 | |

0xC0800000 |

Question 10 Explanation:

Given numbers are 0x42200000 and 0xC1200000 which are stored in the registers R1 and R2, respectively.

R1 = 1.0100..0 X 2

= 1.0100..0 X 2

= 101.0 X 2

= 5 X 8

= 40

R2 = (-1) x 1.0100..0 X 2

= (-1) x 1.0100..0 X 2

= (-1) x 101.0 X 2

= (-1) x5 X 2

= -10

R1 = 1.0100..0 X 2

^{132-127}= 1.0100..0 X 2

^{5}= 101.0 X 2

^{3}= 5 X 8

= 40

R2 = (-1) x 1.0100..0 X 2

^{130-127}= (-1) x 1.0100..0 X 2

^{3}= (-1) x 101.0 X 2

^{1}= (-1) x5 X 2

= -10

R3 = R1/R2

= -4

= (-1)x 1.0 x 2^{2}

Sign = 1

Mantissa = 000..0

Exponent = 2+127 = 129

R3 = 1100 0000 1000 000..0

= 0x C 0 8 0 0 0 0 0

Correct Answer: D

Question 10 Explanation:

Given numbers are 0x42200000 and 0xC1200000 which are stored in the registers R1 and R2, respectively.

R1 = 1.0100..0 X 2

= 1.0100..0 X 2

= 101.0 X 2

= 5 X 8

= 40

R2 = (-1) x 1.0100..0 X 2

= (-1) x 1.0100..0 X 2

= (-1) x 101.0 X 2

= (-1) x5 X 2

= -10

R1 = 1.0100..0 X 2

^{132-127}= 1.0100..0 X 2

^{5}= 101.0 X 2

^{3}= 5 X 8

= 40

R2 = (-1) x 1.0100..0 X 2

^{130-127}= (-1) x 1.0100..0 X 2

^{3}= (-1) x 101.0 X 2

^{1}= (-1) x5 X 2

= -10

R3 = R1/R2

= -4

= (-1)x 1.0 x 2^{2}

Sign = 1

Mantissa = 000..0

Exponent = 2+127 = 129

R3 = 1100 0000 1000 000..0

= 0x C 0 8 0 0 0 0 0

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