Multiple choice

(\text{In the following questions, two equations x and y are given. You have to solve both the equations and give answer.}\\text{I.} \frac{(40)^3}{80 \times 16} \div 25 = 32x\\text{II.} \frac{y}{1089} = \frac{64}{y})

  1. x > y

  2. x ≥ y

  3. x < y

  4. x ≤ y

  5. x = y or the relation cannot be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solving Equation I: (40)³ = 64000, 80 × 16 = 1280, so 64000/1280 = 50, then 50 ÷ 25 = 2. Therefore 32x = 2, giving x = 2/32 = 1/16. Solving Equation II: y/1089 = 64/y, cross-multiplying gives y² = 64 × 1089 = 69696, so y = ±264. Since y can be either 264 or -264, the relation between x and y is ambiguous - if y = 264 then x < y, but if y = -264 then x > y. Therefore the relation cannot be established.

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