y ≥ x + 4  
represents the portion above the line y = x + 4,  
whose intercepts are (-4, 0) and (0, 4).  

-4 ≤ x^2 + y^2 + 4(x − y) ≤ 0  
-4 ≤ x^2 + y^2 + 4x − 4y ≤ 0  
-4 ≤ x^2 + 4x + y^2 − 4y ≤ 0  
4 ≤ x^2 + 4x + 4 + y^2 − 4y + 4 ≤ 8  
4 ≤ (x + 2)^2 + (y − 2)^2 ≤ 8  

This is the region between two concentric circles with center (-2, 2)  
and radii 2√2 and 2.  

Thus, the common region is the portion above the line and within the circles.  

Area of the shaded region:  
= 1/2 [π(2√2)^2 − π(2)^2]  
= 1/2 [8π − 4π]  
= 1/2 [4π]  
= 2π

In right-angled triangle ADE:  
AD^2 + DE^2 = AE^2  

AD = 45, DE = 28  

Calculating hypotenuse AE:  
AE^2 = 45^2 + 28^2  
AE^2 = 2025 + 784  
AE^2 = 2809  
AE = √2809 = 53  

In-radius of a right-angled triangle = (a + b − h)/2  
where a and b are the lengths of the legs, and h is the hypotenuse.  

In-radius of △ADE:  
r = (45 + 28 − 53)/2  
r = 20/2  
r = 10 cm

When given n distinct numbers, and asked to find the sum of the possible n-digit numbers that can be formed,  
we use the formula:  

(Inserting n = 4)  

We are told that this value equals:  

Since n is a single-digit natural number, the total value cannot be that much greater, and 6666 should perfectly divide a49, a50.  

Formula:  
(10^(n-1) + 10^(n-2) + 10^(n-3) + ...) × ((n-1)!) × (Sum of numbers)  

For n = 4:  
(1000 + 100 + 10 + 1) * (3!) * (a + b + c + d)  
= (6666) * (a + b + c + d)  

Given value: 153310 + n  

Upon dividing 153310 by 6666, we get quotient ≈ 22.99  
Nearest value being 23, we take 6666 * 23 = 153318  

Hence:  
n = 8  
a + b + c + d = 23  

Therefore:  
a + b + c + d + n = 23 + 8 = 31

We are told that, the incomes of Kamal, Amal and Vimal are in the ratio 8 ∶ 6 ∶ 5
Lets assume them to be 80X, 60X, 50X respectively.
Money that each one of them pays towards rent,
Kamal 15% which comes to 12X
Amal 12% which comes to 7.2X
Vimal 18% which comes to 9X
Total Rental expenditure will be, 28.2X
Their incomes increase by 10%, 12% and 15% respectively,
Kamal 10% which comes to 88X
Amal 12% which comes to 67.2X
Vimal 15% which comes to 57.5X
Total income will be 212.7X
We are told the rent is the same, so it will be 28.2X
Percentage of income going towards rent in October will be,
Hence, the answer is 13.26%

Largest integer not exceeding √n for a1, a2, a3 = 1  
Largest integer not exceeding √n for a4, a5, a6, a7, a8 = 2  
Largest integer not exceeding √n for a49 = 7  

Calculating sums for ranges:  
a1 to a3 = 3 * 1 = 3  
a4 to a8 = 5 * 2 = 10  
a9 to a15 = 7 * 3 = 21  
a16 to a24 = 9 * 4 = 36  
a25 to a35 = 11 * 5 = 55  
a36 to a48 = 13 * 6 = 78  
a49 = 7  
a50 = 7  

Total sum:  
a1 + a2 + ⋯ + a50 = 3 + 10 + 21 + 36 + 55 + 78 + 7 + 7 = 217

We need remainder of 10^100 ÷ 7.
Check pattern of remainders:
10^1 ≡ 3 (mod 7)
10^2 ≡ 2 (mod 7)
10^3 ≡ 6 (mod 7)
10^4 ≡ 4 (mod 7)
10^5 ≡ 5 (mod 7)
10^6 ≡ 1 (mod 7)

Thus, the cycle length is 6.

Now, 100 ÷ 6 gives remainder 4.
So 10^100 ≡ 10^4 (mod 7) ≡ 4.

Let the amount of work done by Renu in one hour be R.
Let the amount of work done by Seema in one hour be S.

Renu: 15 days × 4 hours/day ⇒ total work by Renu = 15 × 4 × R = 60R.
Seema: 8 days × 5 hours/day ⇒ total work by Seema = 8 × 5 × S = 40S.

Since both complete the same job individually,
60R = 40S  ⇒  S = 1.5R.

Now suppose Seema works for X days and Renu works for Y hours per day in the new arrangement.
Seema works 2Y hours per day and Renu works for 2X days.

Work done by Seema = X × (2Y) × S = 2XY S.
Work done by Renu = (2X) × Y × R = 2XY R.

Given Y = 2, substitute Y = 2 and S = 1.5R:

Work by Seema = 2X(2)×1.5R = 6XR.
Work by Renu = 2X×2×R = 4XR.

Total work = 6XR + 4XR = 10XR, which must equal the whole job (same as 60R or 40S).

Solving gives X = 6.

Thus, the number of days Seema will work is 6.

xn = a + (n − 1) d
2x6 + 2x9 = x11 + x13 x100
2 (a + 5d) + 2 (a + 8d) = a + 10d + a + 12d
4a + 26d = 2a + 22d
2a = −4d
a = −2d
x5 = −4
a + 4d = −4
2d = −4
d = −2
a = 4
x100 = a + 99d
x100 = 4 − 198 = −194

When given more than one equations, stating the fact that there is a common root,
we need to equate the two equations to get discernible values for x.

Here, we are given three equations with the values of m, n:

x^2 + m x + 9 = x^2 + (m + n) x + 35
m x + 9 = m x + n x + 35
n x = -26

Similarly, we can do it for the other equation as well,

x^2 + n x + 17 = x^2 + (m + n) x + 35
m x = -18

Substituting the value of either m x or n x in the original equations, we get:

x^2 - 18 + 9 = 0
x^2 = 9
x = ±3

Since we are given that the root is negative, x = -3.

n = -26 / -3
n = 26/3

m = -18 / -3
m = 6

3n = 26
2m = 12
2m + 3n = 38