JEE Main Maths: Ratio and Proportion questions with solutions

Q1. If a/b = c/d = e/f, find the value of (2*a⁴*b² + 3*a²*c² - 5*e⁴*f)/(2*b⁶ + 3*b²*d² - 5*f⁵), expressed in terms of a and b.

1. a⁴/b⁴
2. a²/b²
3. a/b
4. a³/b³

Answer: a⁴/b⁴

Let a/b = c/d = e/f = k. Then a = bk, c = dk, e = fk. Numerator: 2(bk)⁴ b² + 3(bk)²(dk)² - 5(fk)⁴ f = 2b⁶ k⁴ + 3 b² d² k⁴ - 5 f⁵ k⁴ = k⁴ (2b⁶ + 3 b² d² - 5 f⁵). Dividing by the denominator (2b⁶ + 3 b² d² - 5 f⁵) gives k⁴ = (a/b)⁴ = a⁴/b⁴.

Q2. If x: a = y: b = z: c, is the identity (a² + b² + c²)(x² + y² + z²) = (ax + by + cz)² true?

1. True
2. False
3. Only if a = b = c
4. Only if x = y = z

Answer: True

Let x = ka, y = kb, z = kc. LHS = (a² + b² + c²)(k² a² + k² b² + k² c²) = k² (a² + b² + c²)². RHS = (a*ka + b*kb + c*kc)² = (k(a² + b² + c²))² = k² (a² + b² + c²)². LHS = RHS, so the identity is true (it is the equality case of Cauchy-Schwarz).