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

→ The modality of a relationship is 0 if there is no explicit need for the relationship to occur or the relationship is optional.
→ The modality is 1 if an occurrence of the relationship is mandatory

→ Imagine that we want to design a code with m message bits and r check bits that will allow all single errors to be corrected.
→ Each of the 2​ m​ legal messages has n illegal codewords at a distance of 11 from it.
→ These are formed by systematically inverting each of the n bits in the n-bit codeword formed from it. Thus, each of the 2​ ^m​ legal messages requires n+1 bit patterns dedicated to it.
→ Since the total number of bit patterns is 2​ ⁿ​ ,​ We must have (n+1)2​ ^m​ ≤ 2^{​ n}​ .
→ Using n=m+r, this requirement becomes
= (m+r+1) ≤ 2​ ^r
= 40+6+1 ≤ 2⁶
= 47 ≤ 2
r=6
Message size will be 6+40=46

● The term "bit stuffing" broadly refers to a technique whereby extra bits are added to a data stream, which do not themselves carry any information, but either assist in management of the communications or deal with other issues.
● The receiver knows how to detect and remove or disregard the stuffed bits.

→ In hardwired control, we saw how all the control signals required inside the CPU can be generated using a state counter and a PLA circuit.
→ There is an alternative approach by which the control signals required inside the CPU can be generated . This alternative approach is known as microprogrammed control unit.
→ In microprogrammed control unit, the logic of the control unit is specified by a microprogram. A microprogram consists of a sequence of instructions in a microprogramming language. These are instructions that specify microoperations.
→ A microprogrammed control unit is a relatively simple logic circuit that is capable of
(1) sequencing through microinstructions.
(2) generating control signals to execute each microinstruction.
→ The concept of microprogram is similar to computer program. In computer program the complete instructions of the program is stored in main memory and during execution it fetches the instructions from main memory one after another.
→ The sequence of instruction fetch is controlled by program counter (PC)

→ Shannon Capacity (Noisy Channel) In presence of Gaussian band-limited white noise, Shannon-Hartley theorem gives the maximum data rate capacity.
C=B log​ ₂​ (1+S/N).
where S and N are the signal and noise power, respectively, at the output of the channel.
→ This theorem gives an upper bound of the data rate which can be reliably transmitted over a thermal-noise limited channel. =6 * log​ ₂​ (1+16)
=6 * 4.08746284125034
=24.74 kbps

→ Power in sidebands may be calculated as modulation index.
→ Each sideband is 25%. Total modulation index=1.

→ Shannon Capacity (Noisy Channel) In presence of Gaussian band-limited white noise, Shannon-Hartley theorem gives the maximum data rate capacity.
C=W log​ ₂​ (1+S/N).
where S and N are the signal and noise power, respectively, at the output of the channel.
→ This theorem gives an upper bound of the data rate which can be reliably transmitted over a thermal-noise limited channel.

The Floyd Warshall Algorithm is for solving the All Pairs Shortest Path problem. The problem is to find shortest distances between every pair of vertices in edge weighted directed Graph.

In this problem, we have to check all possibilities.


So, we will prefer max(f(n),g(n)).

→ Primitive recursive functions are a class of functions that are defined using composition and primitive recursion
→ They are a strict subset of those μ-recursive functions (also called partial recursive functions) which are also total functions.
→ Primitive recursive functions form an important building block on the way to a full formalization of computability.
→ A total recursive function is also a partial recursive function

● Shared locks exist when two transactions are granted read access.
● One transaction gets the shared lock on data and when the second transaction requests the same data it is also given a shared lock.
● Both transactions are in a read-only mode, updating the data is not allowed until the shared lock is released.
● There is no conflict with the shared lock because nothing is being updated.
● Shared locks last as long as they need to last; it depends on the level of the transaction that holds the lock.

Total seek time= 10+12+2+18+38+34+32
= 146*6 ms
= 876 ms

This query counts the number of employees who get more than the minimum salary. From the 10 employees, you need to exclude all those employees who are getting the minimum salary. Since the SALARY column is UNIQUE, only one employee will be getting the minimum salary.

RAID is a technology that is used to increase the performance and/or reliability of data storage.
Following are the various RAID levels:
RAID 0: Stripping
RAID 1: Mirroring
RAID 2: Hamming code for error detection
RAID 3: Byte-level striping with a dedicated parity disk
RAID 4: Block-level striping with block-level striping with two parity blocks parity
RAID 5: Block-level striping with distributed parity
RAID 6: Block-level striping with double distributed parity.

→ The instruction opcode is the portion of a machine language instruction that specifies the operation to be performed. Beside the opcode itself, most instructions also specify the data they will process, in the form of operands. In addition to opcodes used in the instruction set architectures of various CPUs, which are hardware devices, they can also be used in abstract computing machines as part of their byte code specifications.

Step-1: putchar(105) will print the ASCII equivalent of 105 is ' i '.
Step-2: The printf statement prints the current value of i=5 because we are given post decrement.
Step-3: The same process will perform until it fail the condition.

The above recurrence is in the form of masters theorem.
Case-1: a>b^k
Here, a=8,b=2,k=1,p=0
O(n(log​ ₂​ 8))=O(n(log​ ₂​ 2​ ³​ ))=O(n​ ³​ (log_{​ 2}​ 2)) [ log_{​ 2}​ 2=1]
=O(n​ ³​ )

The convergence of the bisection method is very slow. Although the error, in general, does not decrease monotonically, the average rate of convergence is 1/2 and so, slightly changing the definition of order of convergence, it is possible to say that the method converges linearly with rate 1/2.
Note:For the bisection you simply have that ε​ _i+1​ /ε_{​ i}​ =1/2, so, by definition the order of convergence is 1 (linearly).

Wait-die scheme​ :
It is a non-preemptive technique for deadlock prevention. When transaction T​ _i​ requests a data item currently held by T​ _j​ , T​ _i​ is allowed to wait only if it has a timestamp smaller than that of T​ _j​ (That is T​ _i​ is older than T_{​ j}​ ), otherwise T_{​ i}​ is rolled back (dies).
→ In this scheme, if a transaction requests to lock a resource (data item), which is already held with a conflicting lock by another transaction, then one of the two possibilities may occur
1. If TS(T​ _i​ ) < TS(T_{​ j}​ ) − that is T_{​ i}​ , which is requesting a conflicting lock, is older than T​ _j −then T_i is allowed to wait until the data-item is available.
2. If TS(T​ _i​ ) > TS(T_{​ j}​ ) − that is T_{​ i}​ is younger than T_{​ j}​ − then T​ _i​ dies. T​ _i​ is restarted later with a random delay but with the same timestamp.
→ This scheme allows the older transaction to wait but kills the younger one.
Example:
Suppose that transaction T22, T23, T24 have time-stamps 5, 10 and 15 respectively. If T22 requests a data item held by T23 then T22 will wait. If T24 requests a data item held by T23, then T24 will be rolled back.

Step-1: initial a=3 but we are given storage class is static.
Step-2: In printf( ) we are given post decrement. So, it will print 3 then decrements.
Step-3: if(2) then we are calling main() function, a=2 because of static otherwise it will be 3 only.
Step-4: It will print 2 and next iteration will print 1.

Web links are stored within the page itself and when you wish to 'jump' to the page that is linked, we select the hotspot or anchor. This technique is called hypertext (or) Hypermedia

The final value theorem is used to find the steady state value of the system output

Let x[n] be causal signal given by x[n] = a​ ⁿ​ u[n]
→ The Z-Transform of x[n] is given by

The ROC is defined by | az ⁻¹ | < 1 or | z| > | a| .
Region of Convergence(RoC)
Region of Convergence for a discrete time signal x[n] is dened as a continuous region in z plane where the Z-Transform converges. In order to determine RoC, it is convenient to represent the Z-Transform as a is rational X(z)=P(z)/Q(z)
1. The roots of the equation P(z) = 0 correspond to the ’zeros’ of X(z)
2. The roots of the equation Q(z) = 0 correspond to the ’poles’ of X(z)
3. The RoC of the Z-transform depends on the convergence of the polynomials P(z) and Q(z).

Given, the signal
V (t) = 3 0 sin 100t + 10 cos 300t + 6 sin (500t + π/4)
So, we have
ω ₁ = 100 rads
ω ₂ = 300 rads
ω ₃ = 500 rads
∴ The respective time periods are
T ₁ = 2π/ω₁ = 2π/100 sec
T ₂ = 2π/ω₂ = 2π/300 sec
T ₃ = 2π/ω₃ = 2π/500 sec
So, the fundamental time period of the signal is
LCM (T ₁ , T ₂ , T ₃ ) =LCM (2π,2π,2π)/HCF (100,300,500)
as T ₀ =2π/100
∴ The fundamental frequency, ω₀ =2π/T ₀ = 100 rad/s

(S/N _q)_dB
= (1.76 + 6 n) _dB
(SQNR) ₁ = 1 .76 + 6 n
(SQNR) ₂ = 1 .76 + 6 n(n + 1 ) = 1 .76 + 6 n + 6
(SQNR) ₂ − (SQNR) ₁ = 1 .76 + 6 n + 6 − 1 .76 − 6 n = 6 dB
So for every one bit increase in bits per sample will result is 6 dB improvement in signal to quantization ratio.

Here, PI=Present Input
PS=Present state
NS=Next State
Q​ _n+1​ =D=f(PI,PS)=f(x,y,Q_{​ n}​ )
D=X’Z+YZ

→ In synchronous counter time delay is constant while in Ripple it is additive.
→ In Ripple counter (or) Asynchronous counter each flip flop waits for its previous flip flops output.
R= bit size*propagation delay
= 4*10ns
= 40ns
→ In Synchronous counter all flip flops are triggered by same clock. It will gives output of all four flip flops at the same time.
S=10ns

In MOSFET fabrication channel length is defined during Poly-Silicon gate patterning process

Given data,
The open loop transfer function of a feedback control system is G(s).H(s)=1/(S+1)​ ³
Gain margin=?
→ Here, the formula is -3tan​ ^-1​ (W_{​ pc}​ ) = -180^{​ 0}
tan​ ^-1​ (W_{​ pc}​ ) = -180^{​ 0}​ /3
= -60​ ⁰
W​ _pc​ =√3 rad/sec
G(s).H(s)=1/(S+1)​ ³
|G(s).H(s)|=X= 1/(1+W​ _pc​ ²​ )
=1⁄8
G(s).H(s)=1/X
= 8

An ideal op-amp lead us to an op-amp with input resistance is infinite, so no current flows into either input terminal (the “current rule”) and that the differential input offset voltage is zero (the “voltage rule”).
Ideal operational amplification will have
➝ High input impedance ( R _in = ∞ ) and low output impedance ( R _out = 0 )

➝ R _in = ∞ , so zero input current
➝ Infinite voltage gain
➝ So, it is a voltage controlled voltage source device.

A ripple counter is an asynchronous counter where only the first flip-flop is clocked by an external clock.
→ All subsequent flip-flops are clocked by the output of the preceding flip-flop. Asynchronous counters are also called ripple-counters because of the way the clock pulse ripples it way through the flip-flops.

So, answer is clock input of one flip-flop.

Number of entries for virtual page number = 2​ ^{virtual address​} / Pages =2^{​ 32}​ /2KB= 2^{​ 32}​ /2^{​ 11}​ =2​ ²¹
Number of entries for physical page number = 2^{​ physical address​} / Pages =2​ ²⁸​ /2KB= 2^{​ 28}​ /2​ ¹¹​ =2^{​ 17}
The number bits for virtual page number are 21.
The number bits for physical page number are 21.

An open interval does not include its endpoints, and is indicated with parentheses. For example, (0,1) means greater than 0 and less than 1.
A closed interval is an interval which includes all its limit points, and is denoted with square brackets. For example, [0,1] means greater than or equal to 0 and less than or equal to 1.
A half-open interval includes only one of its endpoints, and is denoted by mixing the notations for open and closed intervals. (0,1] means greater than 0 and less than or equal to 1, while [0,1) means greater than or equal to 0 and less than 1.
Given function f(x)=2x​ ²​ -2x +6 and the given interval is [0,2]
According to the given interval , we need to check the function at the values 0,1,2.
f(0)=2x0-2x0+6=6
f(1)=2x1​ ²​ -2x1+6=2-2+6=6
f(2)=2x2​ ²​ -2x2+6=8-4+6=10

The integral value of x​ ²​ sin(x)dx is [−x​ ²​ cos(x)+2xsin(x)+2cos(x)+C]​ with the interval [π/2,0]
=[−(​ π/2)​ ²​ cos(​ π/2​ )+2​ π/2​ sin(​ π/2​ )+2cos(​ π/2​ )+C]​ - ​ [−(​ 0)​ ²​ cos(​ π/2​ )+2(0)sin(​ π/2​ )+2cos(​ 0 ​ )+C]
=[0+​ π+0+C-2-C]
=π-2

→ A communication protocol is a system of rules that allow two or more entities of a communications system to transmit information via any kind of variation of a physical quantity. The protocol defines the rules, syntax, semantics and synchronization of communication and possible error recovery methods.
→ There are many properties of a transmission that a protocol can define. Common ones include: packet size, transmission speed, error correction types, handshaking and synchronization techniques, address mapping, acknowledgement processes, flow control, packet sequence controls, routing, address formatting

The language L={ x​ ⁿ​ y^{​ n}​ , n>=1 }
For n=1, xy
n=2,x​ ²​ y^{​ 2}
n=3,x​ ³​ y^{​ 3}
In the above language , there are equal number of x’s followed by y’s
Definition-1​ :E→ xEy |xy → xEy → xxEyy → xxxEyyy → xxxxyyyy
Which is nothing but equal number of x’s followed by equal number of y’s
X​
+​
means one or more x’s
Y​
+
​ means one or more y’s
Definition -2 : The expression xy|x​ ⁺​ x yy^{​ +} ​ gives xy, xxyy , xxxxyy , xxyyyyy and so on
The definition-2 won’t give equal number of x’s followed by y’s
It generates any number of x’s followed by any number of y’s.
Definition -3 :
The expression x​ ⁺​ y^{​ +}​ gives xy, xxy,xxxy,xyy,xyyy and so on strings.
The definition-3 won’t give equal number of x’s followed by y’s
It generates any number of x’s followed by any number of y’s.

● Highly coupled have program units dependent on each other.
● Loosely coupled are made up of units that are independent or almost independent.
● The desirable property is High cohesion and low coupling
● From the above options , the suitable option is independency

● Given the recurrence relation a​ _r​ =a​ _r-1​ +2a​ _r-2 and a​ ₀​ =2,a_{​ 1}​ =7.
● For r =2, a​ ₂=​ a​ _2-1​ +2a_{​ 2-2}​ =a​ ₁​ +2a_{​ 0}​ =7+2*2=7+4=11 ● For r=3,a​ 3=​ a​ 3-1​ +2a​ 3-2​ =a​ 2​ +2a​ 1​ =11+2*7=11+14=25
● From the above options ,Substitute the r values 0,1,2,3 then option -D gives the solution to recurrence relation.

● From the matrix property : If each element of a row (or a column) of a determinant is multiplied by a constant k, then its value gets multiplied by k.
● Consider matrix order is 3, there are 3 rows and each row is multiplied by 2 means we need to multiply 8 to the existing determinant.
● The given existing determinant is 5 and each row multiplied by 2 means 8 *5 =40.

This above code is similar to binary search
First iteration:
I=0,j=9
K=(0+9)/2=4
Elements with duplicate values of “2” the condition if(Y[k] Second iteration :
I=0, j=k=4
K= (0+4)/2=2.
Third iteration:
I=0, j=k=2
K= (0+2)/2=1
Fourth iteration:
I=0, j=k=1
K= (0+1)/2=0
The while condition is Y[k] != x && i < j is true
→ The above program doesn’t work for the cases where element to be searched is the last element of Y[ ] or greater than the last element (or maximum element) in Y[ ].
→ For such cases, program goes in an infinite loop because i is assigned value as k in all iterations, and i never becomes equal to or greater than j.
So, while condition never becomes false.

Memory page numbers will be calculated by using corresponding to given address as follows
The page number is 1 from the record numbers 1-100,
The page number is 2 from the record numbers 101 -200
The page number is 3 from the record numbers 201 -300
The page number is 4 from the record numbers 301 -400
The page number is 5 from the record numbers 401 -500
The page number is 6 from the record numbers 501 -600
The page number is 7 from the record numbers 601 -700
For a given address sequence 0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240,0260, 0320, 0370
The page number numbers sequence is 1,2,5,5,6,6,6,2,3,3,3,4,4
The number of available frame is 1, So we need to use only one frame for executing demand page process.

So total number of page faults are 7.

The condition for a grammar to be LL(1)
Theorem: A context free grammar G=(V​ _T​ , V_{​ N}​ , S,P) is LL(1) if and if only if for every nonterminal A and every strings of symbols ⍺,β such that ⍺≠β and A → ⍺,β we have
1. First(⍺) ∩ First (β) ∩ Follow(A)=Θ.
2. If ⍺ *⇒ ε then First(β) ∩ Follow(A)= Θ.
If grammar is epsilon free then condition 2 is not required.
Now as per condition 1, for every non-terminal A, if we have A → u and A→v
And First(u) First(v) = φ and CFG is epsilon free then it must be LL(1) and if epsilon free CFG is LL(1) then it must satisfy the condition 1.

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.

Here in this question 5000g means 5kg only
So from that question Aamir (A) Birju (B) Charles (C)
Let Aamir, Birju and Charles alone can complete the work in A, B and C minutes respectively, then
1 min. work of (A+B)=1/20, (B+C)=1/40 and (C+A)=1/30
A + B + C = 13/240
1 min. work of C = (A+B+C) - (A+B) = 13/240 - 1/20 = 1/240
So C alone can complete the work in 240 min.

Amount of copper in first alloy is 5x/8 and zinc is 3x/8.
Amount of copper in second alloy is 8x/13 and tin is 5x/13.
So after mixing tin is available in 2 alloy only that means in 3rd alloy 5x/13/2x
Tin present in 3rd alloy is 5/26.

A number that is divisible only by itself and 1 is called prime number.

Given data,
AB=24cm
CB=12cm
θC=5cm(Assume)
θB=5​ ²​ +12​ ²​ =169
θB=13cm
Note: They forgot to mention the value 5. If we are considering value=5 then answer should be A.

This option is both near the center of town and in a location where children and their parents are sure to be around. This is the only option that meets both of Monisha’s requirements.

Option-A FALSE: The author never says that reading is dull.
Option-B,C FALSE: Not related to given paragraph.
Option-D TRUE: The author stresses the need to read critically by performing thoughtful and careful operations on the text.

In between r and T there is letter is s
Likewise m and O letter is n
Same h is the next letter to fit in that sequence

Qmelaqali→ fruitcake
We can divide artificial language into parts.
Qmela→ fruit
qali→ cake
Qalitiimmeo→ cakewalk
We can divide artificial language into parts.
Qali→ cake
tiimmeo→ walk
Useguamao → buttercup
We can divide artificial language into parts.
Usegu→ butter
amao→ cup
cupcake→ ?
amao→ cup
qali→ cake
So, option-D is correct answer.

In the word DROVE we have two vowels
Those are O and E
After consonants are arranged in alphabetically we got the word like DRV
But we have two place vowels in between two pairs of consonants we got finally DEROV
So right from fourth letter is E

From that question we can decide Pali Quereshi sant and timber but Rohan is not mentioned there. So Data inadequate is answer.

Q $ R means R’s father is Q
R @ T means T’s brother is R
T * M means M’s Daughter T
Finally M’s Daughter is T and T’s Brother is R and R’s father is Q so Q’s wife is M.

For this type of questions there is one formula that is
[ n (n+1)(2n+1) ] / 6
Here n is 4 because it’s 4 x 4
[ 4 (4+1)(8+1) ] / 6
30

Explanation: Let CP = 100%
Sold on 12% profit means sold on 112%
now 18% would have been gained means 118% if Rs 18 was more it means 118% - 112% = 18 Rs
so 6%=18Rs
so 100% = (18/6)×100 = 3× 100 = 300 Rs
CP = 300

7 people, the average age is found to be 17 years means 17 * 7 = 119
Two more joined with an average age 19 years means 19 * 2 = 38 So total is 157
But one person left from that group whose age is 25 years 157 - 25 = 132
Remaining average of 8 people is 132 / 8
16.5 Year

Speed = Distance/time = 300/18 = 50/3 m/sec
Let the length of the platform be x meters
Distance = Speed * Time
x+300 = (50/3) * 39
x = 350

From that question 60 * n = 50 (n + 1)
n = 5
so sum of money is 60 * 5 = 300

A tank can be filled in 10 mins
another one is in 30 mins and those taps are opened for 5 mins
so in one min one tap is 1/10 another one is 1/30
both can tank open for 5 mins so (1/10 + 1/30 ) * 5
= (2/15) * 5
= 10/15
= 2⁄3
Remaining work is 1- (2⁄3)
= 1⁄3
The left work has to be done by first Tap because first is shut off.
Work will be completed by Second Tap in= (1⁄3) * 30 = 10

Wealth is opposite to Poverty and Prodigal opposite word is chary.
Wealth: an abundance of valuable possessions or money
Poverty: The state of being extremely poor.
Prodigal: spending money or using resources freely and recklessly
chary: Cautiously or suspiciously reluctant to do something.

Misfortune is related to Catastrophe and knowledge is related to learning.
Misfortune: An unfortunate condition or event.
Catastrophe: An event causing great and usually sudden damage or suffering.

Molecule is a group of atoms bonded together.
Tissue is any of the distinct types of material of which animals or plants are made, consisting of specialized cells and their products.

Limp is related to walk and shutter is related to talk.
Limp: Walk with difficulty, typically because of a damaged or stiff leg or foot.

Riddle is related to Solve and tangle is related to unravel
Riddle: A question or statement intentionally phrased so as to require ingenuity in ascertaining its answer or meaning.
Tangle: Twist together into a confused mass.
unravel: undo (twisted, knitted, or woven threads).

According to venn diagram, (i) is FALSE because male birds never lay eggs. (ii) and (iii) are TRUE

Only son of her grandfather -- Her father;
Man's brother's father -- man's father.
So, man's father is her father i.e., She is the man's sister.

Since the person who is telling has no brother or sister, so his father son is he himself.
So the man in the photograph is his son.

→ The word "quadrilateral" is derived from the Latin words quadri, a variant of four, and latus, meaning "side".
→ A quadrilateral is a polygon with four edges (or sides) and four vertices or corners.
→ According to figure, we can label first and get total quadrilaterals.

Quadrilaterals are 1234, 1245, 1246, 1248, 3481, 3451, 3461, 4517, 4561, 6178 and 1748.

A convex polygon is a simple polygon (not self-intersecting) in which no line segment between two points on the boundary ever goes outside the polygon.

The pentagons are 12468, 34682, 56824, 78246, 13467, 35681, 57823, 71245, 24578, 46712, 68134 and 82356.

Total cost of the items he purchased = Rs.25
Given that out of this Rs.25, 30 Paise is given as tax => Total tax incurred = 30 Paise = Rs.30/100
Let the cost of the tax free items = x
Given that tax rate = 6%
( 25 - 30/100 - x ) * (6/100) = 30 / 100
6(25−0.3−x)=30
x=25−0.3−5=19.7

Sum = (1560 * 100 ) / ( 3 * 6 + 8 * 5 + 2 * 10 ) = 2000

Let B joins for X months.
Then,
(Profit of A)/(Profit of B) = 3/1
(85000 * 12)/(42500 *X) = 3/1
85000 * 12 = 42500 * 3 * X
X = 8.
B Joined for 8 months

Speed= 50 miles/ hour Time=2.5 hour
speed=70 miles / hour Time=1.5 hour
Speed= Distance/ time
Distance= speed x Time
=50×2.5+70×1.5
=125+105
=230 miles

SP of 45 limes is Rs.40
loss = 20%
So CP of 45 limes would be 40*100/80= Rs. 50
To gain 20% She should sell these 45 limes for =50*20/100=Rs. 60
Now If she sells for Rs. 60 = 45 limes
So for Rs. 24 he would sell = (24/60)*45 = 18 limes

Record holds only specific and one type of information.

‘.JPG’ extension used for image files.