Sol: (1/2 × 1/5)⁻¹ ÷ 1/4 = (1/10)⁻¹ × 4 = 10 × 4 = 40.

Sol: Between n² and (n+1)², there are 2n numbers. Here n=25, so 2×25 = 50.

Sol: Number ends in 6, so cube root ends in 6 (since 6³=216).

Sol: Inverse proportion. 15 × 48 = x × 30 ⇒ x = 720/30 = 24.

Sol: Discount = 15% of 1200 = 180. SP = 1200 - 180 = 1020.

Sol: Rate becomes R/2 and time becomes 2n.

Sol: Identity (a-b)² = a² - 2ab + b². Here p² - 2(p)(5) + 5².

Sol: Highest power of the variable x is 3.

Sol: 5x - 3x = -5 + 3 ⇒ 2x = -2 ⇒ x = -1.

Sol: 4x + 5x = 180° ⇒ 9x = 180° ⇒ x = 20°. Angles are 80° and 100°.

Sol: n = 360°/Exterior Angle = 360°/45° = 8.

Sol: Point on y-axis has x-coordinate 0.

Sol: (2a)³ = 8a³.

Sol: Even prime number is only {2}. P(E) = 1/6.

Sol: (x - 1/x)² = x² + 1/x² - 2 ⇒ 5² = A - 2 ⇒ A = 25 + 2 = 27.

Sol: Sum = (n-2)×180° = (5-2)×180° = 3×180° = 540°.

Sol: Range = Max(30) - Min(5) = 25.

(3/5)⁻³ × (3/5)⁻⁶ = (3/5)²ˣ⁻¹
Using aᵐ × aⁿ = aᵐ⁺ⁿ:
-3 + (-6) = 2x - 1
-9 = 2x - 1
2x = -8 ⇒ x = -4

Let Marked Price be x.
SP = MP - Discount% of MP
1120 = x - 20% of x
1120 = 0.8x
x = 1120 / 0.8 = ₹1400

LCM is 6. Multiply entire equation by 6:
6m - 3(m - 1) = 6 - 2(m - 2)
6m - 3m + 3 = 6 - 2m + 4
3m + 3 = 10 - 2m
5m = 7 ⇒ m = 7/5 or 1.4

(i) x = 100° (Opposite angle to B)
(ii) ∠DAB + ∠B = 180° (Adjacent angles) ⇒ ∠DAB = 80°.
∠DAC + ∠CAB = 80° ⇒ 40° + z = 80° ⇒ z = 40°
(iii) y = 40° (Alternate interior angle to z)

LCM of 4, 9, 10 is 180.
Prime factors of 180 = 2 × 2 × 3 × 3 × 5 = 2² × 3² × 5.
To make it a perfect square, multiply by 5.
Required Number = 180 × 5 = 900.

P = 10800, n = 3, R = 25/2 %.
A = P(1 + R/100)ⁿ = 10800 (1 + 1/8)³ = 10800 (9/8)³
A = 10800 × (729/512) ≈ 15377.34
CI = A - P = 15377.34 - 10800 = ₹4577.34

(i) x² + 6x + 8 = x² + 4x + 2x + 8 = x(x+4) + 2(x+4) = (x+2)(x+4)
(ii) q² - 10q + 21 = q² - 7q - 3q + 21 = q(q-7) - 3(q-7) = (q-3)(q-7)

Case 1 (Truck): Direct Proportion
14 km in 25 min. In 5 hrs (300 min):
14/25 = x/300 ⇒ x = (14 × 300)/25 = 14 × 12 = 168 km.
OR Case 2 (Pipes): Inverse Proportion
6 pipes take 80 mins. 5 pipes take y mins.
6 × 80 = 5 × y ⇒ y = 480/5 = 96 mins (1 hr 36 min).

Circumference = 2πr = 110 ⇒ 2 × 22/7 × r = 110 ⇒ r = 17.5 cm.
CSA = 2πrh = 4400 ⇒ 110 × h = 4400 ⇒ h = 40 cm.
Volume = πr²h = (22/7) × 17.5 × 17.5 × 40 = 38500 cm³.

Dividend = 44x²(x² - 5x - 24). Factorize quadratic: (x-8)(x+3).
Expression = 44x²(x-8)(x+3).
Divide by 11x(x-8):
Result = (44x²(x-8)(x+3)) / (11x(x-8)) = 4x(x+3).

Diagonals of a rectangle are equal and bisect each other.
Therefore, half diagonals OR and OT are equal.
2x + 4 = 3x + 1
4 - 1 = 3x - 2x ⇒ x = 3.

Let digits be x and y. Number = 10x + y.
(10x + y) + (10y + x) = 143 ⇒ 11(x+y) = 143 ⇒ x + y = 13.
Given difference x - y = 3 (or y - x = 3).
Solving x+y=13 and x-y=3 gives x=8, y=5. Number: 85.
Solving x+y=13 and y-x=3 gives y=8, x=5. Number: 58.

OR (Ages):
Let Shriya = S, Arjun = 2S.
5 years ago: (2S - 5) = 3(S - 5) ⇒ 2S - 5 = 3S - 15 ⇒ S = 10.
Present Ages: Shriya = 10 years, Arjun = 20 years.

(i) 2001 (Principal): 54000 = P(1 + 5/100)² ⇒ P = 54000 / 1.1025 ≈ 48,980.
(ii) 2005 (Amount): A = 54000(1 + 5/100)² = 54000 × 1.1025 = 59,535.
(iii) Depreciation is reduction in value over time (e.g., car value decreasing). Formula: A = P(1 - R/100)ⁿ.

(i) (1000 - 2)² = 1000000 - 4000 + 4 = 996004.
(ii) (100 + 3)(100 + 4) = 10000 + (3+4)100 + 12 = 10712.
(iii) (a+b)(a-b) = a²-b². Then (a²-b²)(a²+b²) = a⁴-b⁴. Finally (a⁴-b⁴)(a⁴+b⁴) = a⁸ - b⁸.

Diameter = 84cm, r = 0.42m. Length h = 1m.
CSA = 2πrh = 2 × 22/7 × 0.42 × 1 = 2.64 m².
Total Area = 750 × 2.64 = 1980 m².

(i) LSA = 2h(l+b) = 2(7)(25) = 350 m².
(ii) Total = LSA + Ceiling(l×b) = 350 + 150 = 500 m².
(iii) Cans = 500 / 100 = 5 cans.

(i) 10% of Total = 20 ⇒ Total = 200 people.
(ii) Light Music (40% is maximum slice).
(iii) 30% of 1000 = 300 CDs.

(i) 4 units = 1 hour (or based on grid visibility).
(ii) 8:00 am to 11:30 am = 3.5 hours.
(iii) Yes, from 10:00 am to 10:30 am (horizontal line indicates no distance change).