NumberSystem
Question 1 
If (12x)_{3} = (123)_{x} then the value of x is
3  
3 or 4  
2  
None of these 
Question 1 Explanation:
Given, (12x)_{3} = (123)_{x}
Since LHS has 3 as the base and RHS has ‘x’ base,
1 * 3*3 + 2 * 3 + x * 1 = 1 * x*x + 2 * x + 3
9 + 6 + x = x2 + 2x + 3
x2 + x  12 = 0
x2 + 4x  3x  12 = 0
x( x + 4 )  3(x + 4) = 0
(x + 4)(x  3) = 0
x = 3 and 4
But, both the values are infeasible.
Alternative explanation :
According to the rules of number systems , the numbers present in a number system should not be greater than the base of the number system.
According to LHS , (12x)3 tells us that the value of x should be less than 3.
According to RHS , (123)x tells us that the value of x should be greater than 3 as largest digit in 123 is 3.
Therefore, any combination is not possible.
Since LHS has 3 as the base and RHS has ‘x’ base,
1 * 3*3 + 2 * 3 + x * 1 = 1 * x*x + 2 * x + 3
9 + 6 + x = x2 + 2x + 3
x2 + x  12 = 0
x2 + 4x  3x  12 = 0
x( x + 4 )  3(x + 4) = 0
(x + 4)(x  3) = 0
x = 3 and 4
But, both the values are infeasible.
Alternative explanation :
According to the rules of number systems , the numbers present in a number system should not be greater than the base of the number system.
According to LHS , (12x)3 tells us that the value of x should be less than 3.
According to RHS , (123)x tells us that the value of x should be greater than 3 as largest digit in 123 is 3.
Therefore, any combination is not possible.
Question 2 
A computer uses 8 digit mantissa and 2 digit exponent. If a = 0.052 and b = 28E + 11 then b + a – b will
result in an overflow error  
result in an underflow error  
be 0  
be 5.28 E + 11 
Question 2 Explanation:
The computer uses 8 digit mantissa and 2 digit exponent:
a = 0.052 We can represent the number in M*E So a= 0.052 = 0.52*101
mantissa = 0.52, exponent = −1.
b = 28E+11, We can represent the number in M*E So b = 28E+11= 0.28*1013
mantissa = 0.28, exponent = 13.
To add b+a, Small exponent number, a is shifted
to 13(1) = 14 places to right side
a = 0.0000000000000052E+13
From the given data computer uses only 8 digit mantissa, so digits beyond 8th position will be discarded.
So a = 0.00000000E+13 = 0.0 E+13
b + a = (0.28E + 13) + (0.0E + 13 )
= 0.28E + 13
Then b + a  b = (0.28E + 13)  (0.28E + 13)
= 0
Question 3 
Which of the following is termed as minimum error code
Binary code  
Gray code  
Excess 3 code  
Octal code 
Question 3 Explanation:
Gray codes are less errorprone for mechanical devices that involve making and breaking electrical circuits because they only change in one bit position at a time.
So, they are considered as the minimum error code.
Question 4 
The range of integers that can be represented by n bit 2’s complement number system is:
2^{n1} to (2^{n1} – 1)  
(2^{n1} – 1)to (2^{n1} – 1)  
2^{n1} to (2^{n1} )  
(2^{n1} + 1)to (2^{n1} – 1) 
Question 4 Explanation:
The range of integers that can be represented by n bit 2’s complement number system is 2^{n1} to (2^{n1} – 1).
Question 5 
A processor that has carry, overflow and sign flag bits as part of its program status word (PSW) performs addition of the following two 2’s complement numbers 01001101 and 11101001. After the execution of this addition operation, the status of the carry, overflow and sign flags, respectively will be
1, 1, 0
 
1, 0, 0  
0, 1, 0  
1, 0, 1

Question 5 Explanation:
Question 6 
The hamming distance between the octets of 0xAA and 0x55 is
7  
5  
8  
6 
Question 6 Explanation:
→ The Hamming distance between two strings of equal length is the number of positions at which the corresponding symbols are different.
→ In other words, it measures the minimum number of substitutions required to change one string into the other, or the minimum number of errors that could have transformed one string into the other.
→ Given HexaDecimal numbers are 0xAA and 0x55.
Decimal equivalent of 0xAA is 170
Binary form of 0xAA is 1010 1010
Decimal equivalent of 0x55 is 85
Binary form of 0xAA is 0101 0101
The two numbers binary length length is 8.
→ If you observe all the bits of above two binary numbers, both numbers have different bits in all positions.
→ So according to definition , the number of positions at which the corresponding symbols are different which is 8.
→ In other words, it measures the minimum number of substitutions required to change one string into the other, or the minimum number of errors that could have transformed one string into the other.
→ Given HexaDecimal numbers are 0xAA and 0x55.
Decimal equivalent of 0xAA is 170
Binary form of 0xAA is 1010 1010
Decimal equivalent of 0x55 is 85
Binary form of 0xAA is 0101 0101
The two numbers binary length length is 8.
→ If you observe all the bits of above two binary numbers, both numbers have different bits in all positions.
→ So according to definition , the number of positions at which the corresponding symbols are different which is 8.
Question 7 
The binary equivalent of the decimal number 42.75 is
101010.110  
100110.101  
101010.101  
100110.110 
Question 7 Explanation:
(42.75)10 = (?)2
There are 7 questions to complete.