## Number-System

 Question 1
If (12x)3 = (123)x then the value of x is
 A 3 B 3 or 4 C 2 D None of these
Digital-Logic-Design       Number-system       ISRO CS 2008
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.
 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
 A result in an overflow error B result in an underflow error C be 0 D be 5.28 E + 11
Digital-Logic-Design       Number-system       ISRO CS 2008
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*10-1

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
 A Binary code B Gray code C Excess 3 code D Octal code
Digital-Logic-Design       Number-system       ISRO CS 2008
Question 3 Explanation:

Gray codes are less error-prone 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:
 A -2n-1 to (2n-1 – 1) B -(2n-1 – 1)to (2n-1 – 1) C -2n-1 to (2n-1 ) D -(2n-1 + 1)to (2n-1 – 1)
Digital-Logic-Design       Number-system       ISRO CS 2009
Question 4 Explanation:
The range of integers that can be represented by n bit 2’s complement number system is -2n-1 to (2n-1 – 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
 A 1, 1, 0 B 1, 0, 0 C 0, 1, 0 D 1, 0, 1
Digital-Logic-Design       Number-system       ISRO CS 2009
Question 5 Explanation:

 Question 6
The hamming distance between the octets of 0xAA and 0x55 is
 A 7 B 5 C 8 D 6
Digital-Logic-Design       Number-system       ISRO CS 2011
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.
 Question 7
The binary equivalent of the decimal number 42.75 is
 A 101010 B 100110 C 101010 D 100110
Digital-Logic-Design       Number-system       ISRO CS 2013       Video-Explanation
Question 7 Explanation:
(42.75)10 = (?)2
There are 7 questions to complete.

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