194  

(Let the ones digit = x. Then tens = x + 5, hundreds = x – 3. 

Solving gives digits = 194.)

Ravi = 12 years.

Sister = x, Ravi = 2x, so 3x = 18 → x = 6, Ravi = 12 years.

Fraction = 3/4. 

Let fraction = a/b.

(a+1)/b = 1 → a+1 = b → b = a+1.

a/(b+2) = ½ → a/(a+3) = ½ → 2a = a+3 → a=3, b=4.

Fraction = 3/4. 

Circumference = 2πr = 2 × 22/7 × 7 = 44 cm. 

👉 So four angles are 72° and four are 108°

Interior angles on the same side = supplementary.
So the partner angle = 180 – 72 = 108°.

The vertically opposite, corresponding, and alternate interior angles give repeats.
 

👉 Interior angles = 35°, 90°, and the one adjacent to exterior = 180 – 125 = 55°.

Exterior angle = sum of opposite interior angles.

👉 No matter what the triangle is (scalene, isosceles, right-angled), the sum of its exterior angles is always 360°.

Each exterior angle = 180° – (interior angle).
So sum of 3 exterior angles = 3 × 180° – sum of 3 interior angles.
= 540° – 180° = 360°.

• First transaction: Cost = 60. Selling Price = 60 × 1.2 = 72.

• Second transaction: Cost = 72. Loss 25% → SP = 72 × 0.75 = 54.

• Net: CP total = 60, final SP = 54 → Loss = 6.

• Loss% = 6/60 × 100 = 10% loss.

He charges for 1 kg, but gives 900 g.

Suppose CP = ₹100/kg.
He charges = 100 (for 1 kg).
CP of 900 g = 100 × (900/1000) = 90.

Profit = 100 − 90 = 10.
Profit% = 10/90 × 100 = 11.11% profit.

👉 Recall: (x + 1/x)² = x² + 1/x² + 2.
So x + 1/x = √(98+2) = √100 = 10.
Now, x³ + 1/x³ = (x + 1/x)³ − 3(x + 1/x).
= 10³ − 3×10 = 1000 − 30 = 970.

(a) Red face cards: J, Q, K of ♥ and ♦ → 6 cards. Probability = 6/52 = 3/26.

(b) Kings = 4 cards, so not a king = 52 − 4 = 48 cards. Probability = 48/52 = 12/13.

(c) Hearts = 13 cards, Black cards = 26. Overlap (black hearts) = none.

So total = 13 + 26 = 39. Probability 

D = midpoint of BC → BD = DC = 6 cm.

Triangles ABD and ADC are congruent (by SSS: AB = AC, BD = DC, AD = AD).

So AD ⟂ BC (property of median in isosceles triangle).

In ΔABD: AB = 10, BD = 6.
AD² = 10² − 6² = 100 − 36 = 64 → AD = 8 cm.

👉Angle at the center = 70°.

Angle at the circumference on the same arc = ½ × 70° = 35°.

At the circumference on the other arc = 180° − 35° = 145°.

• Let first term = a.
  
  

• For 1st AP: sum of 15 terms = (15/2)[2a+(15−1)×2].
  = (15/2)(2a+28) = 15(a+14).
  
  

• For 2nd AP: sum of 10 terms = (10/2)[2a+(10−1)×3].
  = 5(2a+27) = 10a+135.
  
  

• Equating: 15(a+14) = 10a+135.
  15a+210 = 10a+135 → 5a=−75 → a=−15.
  
  

Answer: First term = −15.

• Hypotenuse = √(15² + 20²) = 25 cm

• Formula for inradius of right triangle:
  r = (a + b − c) / 2
  where a and b = perpendicular sides, c = hypotenuse.

• Substituting:
  r = (15 + 20 − 25) / 2 = 10 / 2 = 5 cm

🎯 Answer: Radius = 5 cm

👉Let the roots be k and (k + 1).

1. Sum of roots = k + (k + 1) = 2k + 1

2. From the equation, sum of roots = −(coefficient of x) / (coefficient of x²)
   = −(−5) / 1 = 5
   So, 2k + 1 = 5 → 2k = 4 → k = 2

3. Roots are 2 and 3

4. Product of roots = 2 × 3 = 6
   But product of roots = constant term / coefficient of x² = m / 1 = m

Therefore, m = 6

🎯 Final Answer: m = 6

Using similarity of triangles (same angle of elevation of the sun):
Height of pole / Shadow of pole = Height of tree / Shadow of tree
6 / 8 = h / 20
h = (6 × 20) / 8 = 15 m

🎯 Answer: Height of tree = 15 m

In trapezium with parallel sides, triangles AOB and COD are similar.
So AO/OC = BO/OD
3/2 = BO/OD
BO:OD = 3:2

🎯 Answer: BO : OD = 3 : 2

log₂[x(x−6)] = 4 → x(x−6) = 16
x² − 6x − 16 = 0
(x−8)(x+2) = 0 → x = 8 

🎯 Answer: x = 8

• Choose positions for 2 digits = C(7,2)
  
  

• Choose digits = 10P2
  
  

• Fill remaining 5 from 30 (letters+symbols) = 30P5
  Total = C(7,2) × 10P2 × 30P5
  
  

🎯 Answer: C(7,2) × 10P2 × 30P5

Use identity: cos 2x = 1 − 2 sin^2 x.
Substitute: sin x = 1 − 2 sin^2 x.
Rearrange: 2 sin^2 x + sin x − 1 = 0.
Let s = sin x. Then 2s^2 + s − 1 = 0.
Solve quadratic: s = [−1 ± sqrt(1 + 8)] / 4 = [−1 ± 3] / 4.
So s = 1/2 or s = −1.
Solve sin x = 1/2 in [0,360): x = 30°, 150°.
Solve sin x = −1 in [0,360): x = 270°.

Final answers: x = 30°, 150°, 270°.

🔍 Check:

• Reflexive? No — a person is not usually considered their own ancestor.
  
  

• Symmetric? No — if A is ancestor of B, B is not ancestor of A.
  
  

• Transitive? Yes — if A is ancestor of B, and B is ancestor of C, then A is ancestor of C.
  
  

✅ This relation is transitive, but not reflexive or symmetric.

👉hence not equivalence

Coefficient matrix:
M =
[ 1 2 3 ]
[ 2 4 6 ]
[ 1 1 1 ]

Determinant:
det(M) = 1*(41 – 61) – 2*(21 – 61) + 3*(21 – 41)

= 1*(4 – 6) – 2*(2 – 6) + 3*(2 – 4)
= (–2) – 2*(–4) + (–6)
= –2 + 8 – 6 = 0

Since det(M) = 0, the matrix is singular.

• That means the system has no unique solution.
  
  

• Depending on the right-hand side vector b, the system may have infinitely many solutions or no solution.
  
  

• Real-life meaning: Product P2 requires resources in exactly double the proportion of P1, so the constraints are not independent.

✨ Solution (step by step)

1. Note that |sin x| means we take the positive value of sin x.
   
   

   • On [0, π], sin x ≥ 0 ⇒ |sin x| = sin x.
     
     

   • On [π, 2π], sin x ≤ 0 ⇒ |sin x| = –sin x.
     
     

2. So the required area = ∫₀^π sin x dx + ∫_π^{2π} (–sin x) dx.
   
   

3. First integral:
   ∫₀^π sin x dx = [–cos x]₀^π = (–cos π) – (–cos 0) = (–(–1)) – (–1) = 1 + 1 = 2.
   
   

4. Second integral:
   ∫_π^{2π} (–sin x) dx = [cos x]ₚᵢ^{2π} = cos(2π) – cos(π) = 1 – (–1) = 2.
   
   

5. Total area = 2 + 2 = 4.
   
   

✅ Final Answer: The total area is 4 square units.