Multiple choice

In the following questions, two equations I and II are given. You have to solve both the equations and give answer. (I: x^2 + (59319)^{1/3} = (529 \div 23)\times8 - 24 ) (II: y^2 - 26y + 169 = 0)

  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
C Correct answer
Explanation

Equation I: x² + 59319^(1/3) = (529 ÷ 23) × 8 - 24. 59319^(1/3) = 39. 529 ÷ 23 = 23. So x² + 39 = 23 × 8 - 24 = 184 - 24 = 160. Therefore x² = 121, so x = ±11. Equation II: y² - 26y + 169 = 0 gives (y - 13)² = 0, so y = 13. Comparing x = -11 or 11 with y = 13, we get X < Y.

Keep practicing — related questions