Multiple choice

In the following question two equation numbered I and II are given .You have to solve both the equations and give answer if - I. (x^2 - 22x + 120 = 0) II. (y^2 - 26y + 168 = 0)

  1. $\(x > y\)$
  2. $\(x \ge y\)$
  3. $\(x < y\)$
  4. $\(x \le y\)$

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

Equation I: x²-22x+120=(x-10)(x-12) gives x=10,12. Equation II: y²-26y+168=(y-12)(y-14) gives y=12,14. Since x can be 12 (equal to y's 12) but never exceeds 14, and x=10<12,14, we have x≤y.

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