Statistics — Medium JEE math MCQ
**[JEE Mains 2026]** If the mean and variance of observations x, y, 12, 14, 16 are 12 and 8 respectively, where x < y, then the value of x + y is:
- A. 18
- B. 20
- C. 22
- D. 16
Solution
Mean = (x + y + 12 + 14 + 16)/5 = 12, so x + y + 42 = 60, x + y = 18.
Variance = 8 means Σ(xᵢ - 12)²/5 = 8.
(x-12)² + (y-12)² + 0 + 4 + 16 = 40.
(x-12)² + (y-12)² = 20.
With x + y = 18, let x = 18 - y.
(18-y-12)² + (y-12)² = 20.
(6-y)² + (y-12)² = 20.
y² - 12y + 36 + y² - 24y + 144 = 20.
2y² - 36y + 160 = 0, y² - 18y + 80 = 0.
y = 10 or 8. So x + y = 18.