Consider three digits(1,2,3) of decimal numbers.Maximum number, we can generate by that three digits are 10^{​ 3}​ -1 which is 999.
Then, Decimal number has 25 digits, so maximum number is 10^{​ 25}​ -1
Similarly, in the binary representation with “n” bits the maximum number is 2​ ²⁵​ -1
So we can write 10​ ²⁵​ –1 = 2^{​ n}​ – 1 → 10^{​ 25}​ = 2^{​ n}
After taking log​ ₂​ on both sides
log​ _2​ 2​ ⁿ​ =log _{​ 2}​ 10​ ²⁵
n log​ ₂​ 2=25 log ​ ₂​ 10
n = 25 log ​ ₂​ 10
n = 25 x 3.3 [ log​ ₂​ 2=1 & log​ ₂​ 10 =3.322]
n = 82.5
Note: Original question paper given option D is 60. But actual answer is 82.5.

→ "Gray code" as an alternative name is "reflected binary code". one of those also lists "minimum error code" and "cyclic permutation code" among the names.
→ Gray codes are widely used to facilitate error correction in digital communications such as digital terrestrial television and some cable TV systems.

3 address instruction:
Two operand locations and a result location are explicitly contained in the instruction word.
e.g., Y = A − B
2 address instruction:
One of the addresses is used to specify both an operand and the result location.
e.g., Y = Y + X
1 address instruction:
Two addresses are implied in the instruction and accumulator based operations.
e.g., A CC = A CC + X
0 address instructions:
They are applicable to a special memory organization called a stack. It interact with a stack using push and pop operations. All addresses are implied as in register based operations.
T = Tap(T − 1 )

Preorder traversal yields (Root, left, right)
Option D:
Let draw binary search tree for the given sequence,

After traversing through this tree we will get same sequence.

Sum of degree of vertices = 2*no. of edges
d*n = 2*|E|
∴ |E| = (d*n)/2

Then the time complexity is Best Case : Ω(1) & Worst Case: O(√n)

A language L for which there exists a TM,, 'T', that accepts every word in L and either rejects or loops for every word that is not in L, is said to be recursively enumerable

● A ​ Synthesized attribute ​ is an attribute of the nonterminal on the left-hand side of a production.
● Synthesized attributes represent information that is being passed up the parse tree.
● LR-attributed grammars allow the attributes to be evaluated on LR parsing. As a result, attribute evaluation in LR-attributed grammars can be incorporated conveniently in bottom-up parsing.

→ The pigeonhole principle is nothing more than the obvious remark: if you have fewer pigeon holes than pigeons and you put every pigeon in a pigeon hole, then there must result at least one pigeon hole with more than one pigeon. It is surprising how useful this can be as a proof strategy.
→ In the theory of formal languages in computability theory, a pumping lemma or pumping argument states that, for a particular language to be a member of a language class, any sufficiently long string in the language contains a section, or sections, that can be removed, or repeated any number of times, with the resulting string remaining in that language. The proofs of these lemmas typically require counting arguments such as the pigeonhole principle.

r has no duplicate, if r can have duplicates it can be remove in the final state. s in non-empty if s is empty then r*s becomes empty.

→ A weak entity is an entity that cannot be uniquely identified by its attributes alone.
→ It must use a foreign key in conjunction with its attributes to create a primary key.
→ The foreign key is typically a primary key of an entity it is related to.

● Tools are also in the open domain which can be downloaded and used.
● They do not usually have very good user interfaces.

Risk Exposure(RE) is determined using the following relationship RE = P*L
where P is the probability of occurrence for a risk, and L is the cost to the project should the risk occur.

A diagram will make it easy to comprehend.

It is easy to find that the can labeled 25 Paise must have 5 paise coins in it. So, the can labeled 10 Paise, must have 25 Paise coins in it.

→ Like above other input combinations also repeat after these inputs.
→ We have 4 input pairs and produces output either 0 or 1.
→ Total functions possible=2*2*2*2 (or) 2​ ⁴ =16

Let principal be ‘x’
S.I = P T R / 100
S.I = X * 3 * 10 / 100
C.I = [ P (1 + (R / 100)^T) - P]
C.I = [ X (1+ (10 / 100)³) - X]
C.I = 331X / 1000
Difference between Compound interest and Simple interest is 77.5
( 331X / 1000 ) - ( 3X / 10) = 77.5
31x / 1000 = 77.5
x = 2500.