Multiple choice

Directions: In the following question, two equations numbered (I) and (II) are given. Student should solve both the equations and mark appropriate answer. I. x3 = 729 II. 3y2 - 20y + 28 = 0

  1. If x = y or no relation can be established

  2. If x > y

  3. If x < y

  4. If x ≥ y

  5. If x ≤ y
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Solving equation (I), x^3 = 729 gives x = 9. Solving equation (II), 3y^2 - 20y + 28 = 0, we factor to (3y - 14)(y - 2) = 0, giving y = 14/3 (approx 4.67) or y = 2. Since 9 > 4.67 and 9 > 2, x > y holds.

AI explanation

From equation I, taking the cube root of 729 gives x = 9. For equation II, factoring the quadratic 3y squared minus 20y plus 28 yields (3y minus 14)(y minus 2) = 0, giving y = 14/3 or y = 2. Since 9 is greater than both 14/3 and 2, x is greater than y.

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