Question

It is given that (6a² + 7ab + b²) ∶ (a² + 3ab + b²) ∶∶ 4 ∶ 1 and (11a² + 2ac + c²) ∶ (7a² + 3ac + c²) ∷ 7 ∶ 5. If a + b + c = 66, then what is the value of a? (a, b, c > 0)
1. 12
2. 18
3. 21
4. 36
5. 40

Answer 1

Correct Answer - Option 4 : 36

Let a/b = x and a/c = y

⇒ (6a² + 7ab + b²)/(a² + 3ab + b²) = 4 / 1

Dividing numerator and denominator by b²

⇒ {6(a/b)² + 7(a/b) + 1}/{(a/b)² + 3(a/b) + 1} = 4/1

⇒ {6x² + 7x + 1}/{x² + 3x + 1} = 4/1

⇒ 6x² + 7x + 1 = 4x² + 12x + 4

⇒ 2x² – 5x – 3 = 0

⇒ 2x² – 6x + x – 3 = 0

⇒ 2x (x – 3) + 1(x – 3) = 0

⇒ (x – 3) (2x + 1) = 0

⇒ x = 3 (∵ a and b both are positive)

∴ b = a/3

⇒ (11a² + 2ac + c²)/(7a² + 3ac + c²) = 7/ 5

Dividing numerator and denominator by c²

⇒ {11(a/c)² + 2(a/c) + 1}/{7(a/c)² + 3(a/c) + 1} = 7/5

⇒ {11y² + 2y + 1}/{7y² + 3y + 1} = 7/5

⇒ 55y² + 10y + 5 = 49y² + 21y + 7

⇒ 6y² – 11y – 2 = 0

⇒ 6y² – 12y + y – 2 = 0

⇒ 6y (y – 2) + 1(y – 2) = 0

⇒ (y – 2) (6y + 1) = 0

∴ y = 2

⇒ c = a/2

⇒ a + b + c = 66

⇒ a + a/3 + a/2 = 66

⇒ (6a + 2a + 3a) /6 = 66

⇒ 11a/6 = 66

⇒ a = 36

∴ The value of a is 36.