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NIMCET Previous Year Questions (PYQs)

NIMCET Number System PYQ


NIMCET PYQ 2021
The number (2217)8 is equivalent to





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NIMCET Previous Year PYQNIMCET NIMCET 2021 PYQ

Solution

(2217)8 =(010010001111)8
Pair of 4 Bits 
1111= F
1000=8
0100=4
(010010001111)=(48F)16

NIMCET PYQ 2021
The binary multiplication 00*11 will give





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NIMCET Previous Year PYQNIMCET NIMCET 2021 PYQ

Solution

0*3=0

NIMCET PYQ 2020
The binary equivalent of (234.125)10?





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NIMCET Previous Year PYQNIMCET NIMCET 2020 PYQ

Solution


NIMCET PYQ 2020
Determine the octal equivalent of (432267)10?





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NIMCET Previous Year PYQNIMCET NIMCET 2020 PYQ

Solution


NIMCET PYQ 2020
One Exabyte is equal to …





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NIMCET Previous Year PYQNIMCET NIMCET 2020 PYQ

Solution


NIMCET PYQ 2020
The logic XOR operation of (4AC0)16 and (B53F)16 results





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NIMCET Previous Year PYQNIMCET NIMCET 2020 PYQ

Solution


NIMCET PYQ 2023
Equivalent of the decimal number (25.375)10 in  binary form





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NIMCET Previous Year PYQNIMCET NIMCET 2023 PYQ

Solution


NIMCET PYQ 2022
The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively is





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NIMCET Previous Year PYQNIMCET NIMCET 2022 PYQ

Solution

The first number = (1657 - 6) = 1651

Second number = (2037 - 5) = 2032

Taking the HCF of two numbers 1651, 2032 we get 127.

So, if we divide 1657 and 2037 by 127 we will get remainders 6 and 5 respectively.

NIMCET PYQ 2022
Suppose the largest n bit number requires ‘d’ digits in decimal representation. Which of the following relations between ‘n’ and ‘d’ is approximately correct 





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NIMCET Previous Year PYQNIMCET NIMCET 2022 PYQ

Solution

n bits binary number required d -decimal digits.

So, 10d>2n

Take on both side 

log10(10d)>log10(2n)

d>nlog10(2)

NIMCET PYQ 2022
‘Floating point representation' is used to represent





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NIMCET Previous Year PYQNIMCET NIMCET 2022 PYQ

Solution


NIMCET PYQ 2019
Each of A, B and C is a different digit among 1 to 9. How many different values of the sum of A, B and C are possible, if  ABA X AA=ACCA ?





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NIMCET Previous Year PYQNIMCET NIMCET 2019 PYQ

Solution



NIMCET PYQ 2021
Choose the correct option for the remainder when X = 1! + 2! + 3! + ...........+ 100! is divided by 24





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NIMCET Previous Year PYQNIMCET NIMCET 2021 PYQ

Solution


NIMCET PYQ 2021
Insert the missing number: 16, 33, 65, 131, 261, ?





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NIMCET Previous Year PYQNIMCET NIMCET 2021 PYQ

Solution



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