We have two inequalities,

$y\ \ge\ x\ +\ 4$

$-4\ \le\ x^2\ +\ y^2\ +\ 4\left(x\ -\ y\right)\ \le\ 0$

The second inequality can be written as two separate inequalities,

$-4\ \le\ x^2\ +\ y^2\ +\ 4\left(x\ -\ y\right)$ and $\ x^2\ +\ y^2\ +\ 4\left(x\ -\ y\right)\ \le\ 0$

The first inequality can be written as,

$\ x^2\ +\ y^2\ +\ 4x\ -\ 4y\ +4\ \ge\ 0$

$\ x^2+\ 4x\ +4\ +\ y^2-\ 4y\ +4\ -4\ge\ 0$

$\left(x\ +\ 2\right)^2\ +\ \left(y\ -\ 2\right)^2\ \ge\ 4$

The second inequality can be written as,

$\ x^2\ +\ y^2\ +\ 4x\ -\ 4y\ \ \le\ \ 0$

$\ x^2+\ 4x\ +4\ +\ y^2-\ 4y\ +4\ -8\ \le\ \ 0$

$\left(x\ +\ 2\right)^2\ +\ \left(y\ -\ 2\right)^2\ \le\ \ 8$

Representing all three inequalities in the graph, we get the graph as shown above, and we must calculate the intersection of all three inequalities,

We can see that the line passes through the centre of both circles (-2, 2), and the area obtained from the second inequality is the area between the two circles. So, the area of intersection of all three graphs is half of the area between both the circles as the line divides the circle in half, and we must only consider the area above the line as per the given inequality. We know that the area of the bigger circle is $\sqrt{\ 8}$ and the area of the smaller circle is $\sqrt{\ 4}$ from the equations of the circles as we know that equation of circle as $\left(x\ -\ a\right)^2\ +\ \left(y\ -\ b\right)^2\ =\ \left(radius\right)^2$ where (a,b) is the centre of the circle.

The area of intersection 

=$ \dfrac{1}{2}\left(Area\ of\ bigger\ circle\ -\ Area\ of\ smaller\ circle\right)$

= $\dfrac{1}{2}\left(\pi\ \left(\sqrt{8}\right)^2\ -\pi\ \left(\sqrt{4}\right)^2\right)$

= $\dfrac{1}{2}\left(8\pi\ \ -4\pi\right)$

= $\dfrac{1}{2}\left(4\pi\ \right)$

= $2\pi\ $

Therefore, the correct answer is option A.

There are two critical points for the inequality to consider: x=-5 and x=3/2

Region I: x is greater than 3/2

In this scenario, both the terms would be positive; cross-multiplying, we get the relation $2x-3\le x+5$

Giving the boundary $x\le8$, hence giving us the valid range as $\frac{3}{2} \lt x \leq 8$

Region II: $-5 \lt x \lt \frac32$

In this case, the right-hand side will be a negative value, and hence, the sign would change when multiplying, giving the inequality

$2x-3\ge x+5$

Which will give x>8, which is out of bounds for this region 

Another way is to put a value in the region to check for the validity of the inequality; by putting x=0, we could see that the inequality does not hold in this region 

Region III: x less than -5

In this scenario, both the terms are negative, essentially giving us the same boundary as region 1; we take the lower bounds, giving us that x has to be less than 5

Therefore, for the given inequality to hold true x<-5 or $\frac{3}{2}$ < x $\le$ 8

Hence, Option B is the correct answer. 

We can consider the quadrants of a graph:

First quadrant: Both x and y are positive

This would change the equation to 2x+y=15 and x=20, giving a negative value of y; hence, this is not the case. 

Second quadrant: x is negative, but y is positive

This would change the equations to y=15 and x=20, giving a positive value of x, which hence can not be the case.  

Third quadrant: Both x and y are negative

This would change the equation to y=15 and x-2y=20; this gives a positive value of y and hence can not be the case. 

Fourth quadrant: x is positive, but y is negative

This would change the equations to 2x+y=15 and x-2y=20; this gives the value of x as 10 and y as -5, which would lie in the fourth quadrant. 

The value of x-y would be 10-(-5)=15

Therefore, Option B is the correct answer. 

The moduli will give out only non-negative outputs, and since we are to consider only integer values of x and y, this drastically reduces the possible cases. 

We can get 2 from either 2+0 or 1+1

We get a 2+0 form when either the first term or the second term is 0

The second term is 0; this is when x=y, in this case,|2x|=2, where x can be 1 or -1; therefore, the two cases are (1,1) and (-1,-1)

The first term is 0; this is the case when x = -y, in this case, |x- (-x)|=2, giving x=1 or -1 yet again, here the two cases are (1,-1) and (-1,1)

The other way we can get 2 is through 1+1

This is possible when one of the terms is 0; if y=0, |x|+|x|=2, where x can be 1 or -1, giving two cases (1,0) and (-1,0)

Similarly,y for x=0, we get two cases, (0,1) and (0,-1)

Therefore, there are 8 pairs of (x,y) that satisfy the given equation. 

Given,

$ x^{2} + (x - 2y - 1)^{2} = -4y(x + y) $

$ x^2 + 4xy + 4y^2 + (x-2y-1)^2 = 0 $

$ (x+2y)^2 + (x-2y-1)^2 = 0 $

For the L.H.S. of the equation to be 0, each of the square terms should be 0 (as squares cannot be negative)

$ x - 2y - 1 = 0 $ => $ x - 2y = 1 $

Let us consider 3 cases:

1) $x = 0$. This is a solution, as both L.H.S and R.H.S will be equal (0) when $x = 0$. (1 solution)

2) $x > 0$  

$\Rightarrow\ 2x\left(x^2+1\right)=5x^2$  

$\Rightarrow\ 2\left(x^2+1\right)=5x$  

$\Rightarrow\ 2x^2-5x+2=0\ \Rightarrow\ 2x^2-4x-x-2=0$  

$\Rightarrow\ 2x(x-2)-1(x-2)=0$  

$\Rightarrow\ (x-2)(2x-1)=0\ \Rightarrow\ x=2\ \text{or}\ \frac{1}{2}\ \Rightarrow$ (1 integer solution)

3) $x < 0$  

$\Rightarrow\ -2x\left(x^2+1\right)=5x^2$  

$\Rightarrow\ 2x^2+5x+2=0$  

$\Rightarrow\ 2x^2+4x+x+2=0$  

$\Rightarrow\ 2x(x+2)+1(x+2)=0$  

$\Rightarrow\ (x+2)(2x+1)=0\ \Rightarrow\ x=-2\ \text{or}\ -\frac{1}{2}\ \Rightarrow$ (1 integer solution)

So, the total number of integer solutions are $0,\,2,\,-2\ \Rightarrow\ 3$.

Given that the average marks of 4 girls and 6 boys is 24.

Let us assume 'b' is the marks scored by a boy and 'g' is the marks scored by a girl.

=> 4g + 6b = 10*24 = 240 ---(1)

Given that, $ b\le g\le2b $

We need to find the distinct possible values of 2g + 6b = 2g + 240 - 4g = 240 - 2g.

From (1)

when b = g => 10g = 240 => g = 24

when b = g/2 => 7g = 240 => g = 240/7

=> 240 - 2g varies from 240 - 2*24 to 240 - 2*240/7

=> 171.42 to 192 => Integer values of 172 to 192 => 21 values.

It is given that $ \frac{x}{y} $ < $ \frac{x+3}{y-3} $, which can be written as $ \frac{x}{y} - \frac{x+3}{y-3} $ < $ 0 $

=> $ \frac{x(y-3) - y(x+3)}{y(y-3)} $ < $ 0 $

=> $ \frac{xy - 3x - xy - 3y}{y(y-3)} $ < $ 0 $

=> $ \frac{-3(x+y)}{y(y-3)} $ < $ 0 $

=> $ \frac{3(x+y)}{y(y-3)} $ > $ 0 $

From this inequality, we can say that, when $ y $ < $ 0 \Rightarrow y(y-3) $ > $ 0 $. Now to satisfy the given equation $ \frac{3(x+y)}{y(y-3)} $ > $ 0 $,

$ (x+y) $ must be greater than zero Hence, $ x $ > $ 0 $ and $ |x| $ > $ |y| $

Therefore, the magnitude of $ x $ is greater than the magnitude of $ y $.

Hence, $ x $ > $ y $, and $ |x| $ > $ |y| \Rightarrow -x $ < $ y $ (Since the magnitude of $ x $ is greater than the magnitude of $ y $.)

The correct option is C.

It is given that if a certain amount of money is divided equally among n persons, each one receives Rs 352. Hence, the total amount of money is (352*n) = 352n ... Eq(1)

It is also known that if two persons receive Rs 506 each and the remaining amount is divided equally among the other persons, each of them receives less than or equal to Rs 330

Hence, the maximum amount of money with them = 506*2 + (n-2)*330 = 1012+330n-660 = 352+330n

Now, 352+330n ≥ 352n 

352+330n ≥ 352n 

=> 22n ≤ 352

22n ≤ 352 

=> n ≤ 16

n ≤ 16

Hence, the maximum value is 16

Given that $ 2pq - 20 = 52 - 2pq \Rightarrow 4pq = 72 \Rightarrow pq = 18 $ ---(1)

Now, $ p^2 + q^2 - 29 = 2pq - 20 \Rightarrow p^2 + q^2 - 2pq = 9 \Rightarrow (p-q)^2 = 9 \Rightarrow p - q = \pm 3 $

Also, $ p^2 + q^2 - 29 = 2pq - 20 \Rightarrow p^2 + q^2 = 2pq + 9 = 2(18) + 9 = 45 $

Now, $ p^3 - q^3 = (p-q)(p^2 + pq + q^2) = (p-q)(45 + 18) = (p-q)(63) $

=> When $ p-q = -3 $ => The value is $ 63(-3) = -189 $ and when $ p-q = 3 $ => The value is $ 63(3) = 189 $.

=> The difference = $ 189 - (-189) = 378 $.

It is given that there are exactly 41 numbers, which can be expressed as the power of two, and exist between $8^m$ and $8^n$, (where m, and n are positive integers, and m < n)

Hence, $2^{3m}$ < 41 numbers < $2^{3n}$

Since, m is a positive integer, the least value of m is 1. Therefore, $2^{3m}$ = $2^3$, hence, the 41 numbers between them are $2^4$, $2^5$, $2^6$, ..., $2^{44}$

Then the lowest possible value of $8^n$ is $2^{45}$. Hence, the smallest value of n is $2^{45}$ = $8^n$ => $2^{3n}$ = $2^{45}$ => n = 15

Hence, the smallest value of m+n is (15+1) = 16

It is given that $5^{n−1}$ < $3^{n+1}$, where n is a natural number. By inspection, we can say that the inequality holds when n = 1, 2, 3, 4, and 5.

Now, we need to find the least integer value of m that satisfies $3^{n+1}$ < $2^{n+m}$

For, n =1, the least integer value of m is 3.

For, n = 2, the least integer value of m is 3

For, n = 3, the least integer value of m is 4.

For, n = 4, the least integer value of m is 4.

For, n = 5, the least integer value of m is 5.

Hence, the least integer value of m such that for all the values of n, the equation holds is 5.3

It is given that the population of the town in 2020 was 100000. The population decreased by y% from the year 2020 to 2021 and increased by x% from the year 2021 to 2022, where x and y are two natural numbers.

Hence, the population in 2021 is $ 100000 \left(\frac{100-y}{100}\right) $.

The population in 2022 is $ 100000 \left(\frac{100-y}{100}\right)\left(\frac{100+x}{100}\right) $

It is also given that the population in 2022 was greater than the population in 2020, and the difference between x and y is 10.

Hence,

$ 100000 \left(\frac{100-y}{100}\right)\left(\frac{100+x}{100}\right) \gt 100000 $, and (x-y) = 10

=> $ 100000 \left(\frac{100-y}{100}\right)\left(\frac{110+y}{100}\right) \gt 100000 $

=> $ \frac{100-y}{100} \left(\frac{110+y}{100}\right) \gt 1 $

To get the minimum possible value of 2021, we need to increase the value of y as much as possible.

Hence, $ (100-y)\{(100+y) + 10\} \gt 10000 $

=> $ 10000 - y^2 + 1000 - 10y \gt 10000 $

=> $ y^2 + 10y \lt 1000 $

=> $ y^2 + 10y + 25 \lt 1025 $

=> $ (y + 5)^2 = 1024 \lt 1025 $

=> $ (y + 5)^2 = 32^2 $

=> $ y = 27 $

Hence, the population in 2021 is 100000*(100-27) = 73000

Let $\log_2 n = y$

$\frac{4 - y}{3 - \frac{y}{2}}$ < $0$

$(4 - y)\left(3 - \frac{y}{2}\right)$ < $0$

$(4 - y)(6 - y)$ < $0$

$(y - 4)(y - 6)$ < $0$

$4$ < $y$ < $6$

$4$ < $\log_2 n$ < $6$

$2^4$ < $n$ < $2^6$

$16$ < $n$ < $64$

n can take values from 17 to 63 (inclusive).

Case 1: When $x^2$ − 3x − 10 = 0 and $x^2$ − 10 ≠ 0

$x^2$ − 3x − 10 = 0 or, (x−5) (x+2) = 0

or, x= 5 or -2

Case 2: $x^2$ − 10 = 1

$x^2$ − 11 = 0

No integer solutions

Case 3: $x^2$ − 10 = −1 and $x^2$ − 3x − 10 is even

$x^2$ − 9 = 0

or, (x+3)(x-3) = 0

or, x = -3 and 3

for x = -3 and +3$x^2$ − 3x − 10 is even

In total 4 values of x satisfy the equations