Basic learning:
$1 + 0.5 = 1.5$
This can be written as a fraction:
$1.5 = 1 \frac{1}{2}$
This can also be written as:
$1 \frac{1}{2} = 1 + \frac{1}{2}$

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Hence, question states:
$999\frac{1}{7} + 999\frac{2}{7} + 999\frac{3}{7} + 999\frac{4}{7} + 999\frac{5}{7} + 999\frac{6}{7}$
This can be written as:
$999 + \frac{1}{7} + 999 + \frac{2}{7} + 999 + \frac{3}{7} + 999 + \frac{4}{7} + 999 + \frac{5}{7} + 999 + \frac{6}{7}$
We can add the fractions separately. Also, $999$ appears six times:
$\Rightarrow 6 \times 999 + \left( \frac{1}{7} + \frac{2}{7} + \frac{3}{7} + \frac{4}{7} + \frac{5}{7} + \frac{6}{7} \right)$
$\Rightarrow 5994 + \left( \frac{1}{7} + \frac{2}{7} + \frac{3}{7} + \frac{4}{7} + \frac{5}{7} + \frac{6}{7} \right)$
$\Rightarrow 5994 + \frac{21}{7}$
$\Rightarrow 5994 + 3 = \boxed{5997}$

There will be two cases:

1. A speaks the truth and B lies
   
2. A lies and B speaks the truth.
   

Hence,
$\Rightarrow [75\% \times (100-80)\%]+[(100-75)\%\times80\%]$
$\Rightarrow \dfrac{75\times20+25\times80}{10000}=\dfrac{3500}{1000}=\dfrac{35}{100}=35\%$

For equation: $ax^2+bx+c=0$
Sum of roots $=-\frac{b}{a}$
Product of roots $= \frac{c}{a}$
Let's assume the equation here to be $x^2+bx+c=0$, we need not worry ourself with the value of $a$ as it doesn't matter here!
Amit got the constant term wrong, but he has gotten the coefficient of $x$ right.
So from his equation, we can derive $-b=4+3=7 \Rightarrow \boxed{-b=7}$

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Similarly, Alok has gotten the coefficient of $x$ wrong, but he has gotten the constant term right.
So from his equation, we can derive $c=3\times2=6 \Rightarrow \boxed{c=6}$

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Now we know sum of roots is $7$ and product of roots is $6$.
We can go through the options to see what matches these (add both the numbers for sum of roots; multiply both the numbers for product).
Roots as per Option 2: $(6,1)$
Sum $= 6+1 = 7$
Product $= 6 \times 1 = 6$

Note: Repetition is not allowed
For a three-digit even, the unit place has two possibilities.
Case 1: The unit place is 2 i.e. _ _ 2
Now, Hundreds place can be filled in 4 ways (1,3,4,5) and Tens place can be filled in 3 ways.
Case 1 $=4 \times 3 = 12$ ways
Case 2: The unit place is 4 i.e. _ _ 4
Now, Hundreds place can be filled in 4 ways (1,2,3,5) and Tens place can be filled in 3 ways.
Case 2 $=4 \times 3 = 12$ ways
Total cases $= 12 + 12 = 24$
Hence, the correct option is Option 3.

Let's simplify : $2^{122} + 4^{62} + 8^{42} + 4^{64} + 2^{130}$
$= 2^{122} + 2^{2 \times{62}} + 2^{3 \times{42}} + 2^{2 \times {64}} + 2^{130}$
$= 2^{122} + 2^{124} + 2^{126} + 2^{128} + 2^{130}$
$= 2^{122} + 2^{122} \times 2^2 + 2^{122} \times 2^4 + 2^{122} \times 2^6 +2^{122} \times 2^8$
$= 2^{122} (1 + 4 + 16 + 64 + 256)$
$= 2^{122} (341)$
$341$ is divisible by $11$ , hence, entire expression will be divisible by 11.
Hence, the correct answer is Option 4.

The cost of fencing an equilateral triangular park and a square park is the same, implying the perimeter of both parts is equal.
Let $a$ be the side of the equilateral triangular park and $s$ be the side of the square park.
Thus, $3a = 4s \rightarrow s = \dfrac{3}{4}a$
Now, area of triangular park : $\dfrac{\sqrt3}{4} a^2 = 16 \sqrt3$
Hence, $a^2 = 64$
$a = 8$ units (since side of an equilateral triangle can't be negative)
Thus, $s = \dfrac{3}{4} \times 8 = 6$ units.
Now, length of diagonal $= \sqrt 2 \times s = \sqrt2 \times 6 = 6\sqrt2$ units
Hence, the correct option is Option 4.

Let the sum of money be $x$.
The rate is $10 \%$ per annum, compounded semi-annually, so the rate per half-year is $5\%$,
The number of periods in $1.5$ years is $3$.
Using the compound interest formula:
$CI =$ $x\left(1+\dfrac{5}{100}\right)^{3} -x$
$CI =x\left(\dfrac{21}{20}\right)^{3}-x$
$CI=x\left(\dfrac{441 \times 21}{8000}\right)-x$
$CI=x\left(\dfrac{9261}{8000}\right)-x$
$CI=x\left(\dfrac{1261}{8000}\right)$
Now, simple interest:
$SI = \dfrac{x \cdot 3 \cdot 5}{100}=\dfrac{15 x}{100}=\dfrac{3 x}{20}$.
The difference between Cl and SI is given as ₹183, so:
$\Rightarrow \dfrac{1261 x}{8000}-\dfrac{3 x}{20}=183$
$\Rightarrow \dfrac{1261 x-1200 x}{8000}=183$
$\Rightarrow \dfrac{61 x}{8000}=183$
$\Rightarrow x=3 \times 8000=24,000$
Hence, the correct option is Option 2.

It is best to first check the (B) statement from the given options.
B. $(243)^{0.16} \times (9)^{0.1} = 0.3$
LHS = $(3)^{5 \times{0.16}} \times (3)^{2 \times {0.1}} = 3^{(0.80)} \times 3^{(0.20)} = 3^1 = 3$
Thus, B is an incorrect statement.
Hence, the only possible option is Option 4.

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A. If $2^x = 3^y = 6^{-z}$, then $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$
Given: $2^{x}=3^{y}=6^{-z}$
Let: $2^{x}=3^{y}=6^{-z}=k \quad \text { (say) }$
Now take logs:

• From $2^{x}=k$, we get $x=\log _{2} k$
  • From $3^{y}=k$, we get $y=\log _{3} k$
  • From $6^{-z}=k$, take log: $-z=\log _{6} k \Rightarrow z=-\log _{6} k$

Now evaluate:
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{\log _{2} k}+\frac{1}{\log _{3} k}+\frac{1}{-\log _{6} k}$
Use change of base: $\frac{1}{\log _{b} k}=\log _{k} b$
So: $=\log _{k} 2+\log _{k} 3-\log _{k} 6=\log _{k}\left(\frac{2 \times 3}{6}\right)=\log _{k}(1)=0$

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C. $3.105 \times 10^{P}=0.00239+0.000715=0.003105$
Thus, $3.105 \times 10^{P}=0.003105 \Rightarrow 10^{P}=\frac{0.003105}{3.105}=10^{-3} \Rightarrow P=-3$

Let $x, y$ and $z$ be the no. of litres purchased in three consecutive years, respectively.

Years	1	2	3
No. of litres	x	y	z
Rate	17	19	20
Total Expense	6460	6460	6460

From the above table,
$x = \dfrac{6460}{17} = 380$
$y = \dfrac{6460}{19} = 340$
$z= \dfrac{6460}{20} = 323$
Thus, the average cost per litre $= \dfrac{\text{Total expense}}{\text{Total No. of litres}} = \dfrac{6460 + 6460 + 6460}{380 + 340 +323}$
$= \dfrac{3 \times 6460}{1043} = \dfrac{19380}{1043} = \dfrac{19.38}{1.043}$
From here, we understand that the average has to be lower than $₹ 19.38$. However, we still have two options, $₹18.49$ or $₹18.58$
To get a unique answer, let's assume the average is $₹18.5$
Thus, total expense $= 18.5 \times 1043 = ₹ 19295.5$ which is lower than actual expense of $₹ 19380$
Hence, the average is greater than $18.5$, which gives us $18.58$ as the unique solution.
Hence, the correct option is Option 2.