LCM of 'x' and 'y' is 30 and their HCF is 1 such that {10 > x > y > 1}.
I. 2p²- (x +
y) p + 3y = 0
II. 2q² + (9x + 2) = (3x +
y) q

ATQ, We know that product of two numbers 'p' and 'q' = LCM (p, q) × HCF (p, q) So, x × y = 30 × 1 So, xy = 30 So, possible values of 'x' and 'y' = (1, 30) , (2, 15) , (3, 10) , (5, 6) Only two values of 'x' and 'y' satisfy the given conditions. So, x = 6 and y = 5 From I: So, 2p²- (x + y) p + 3y = 0 So, 2p²- (6 + 5) p + (3 × 5) = 0 So, 2p²- 11p + 15 = 0 Or, 2p²- 6p - 5p + 15 = 0 Or, 2p (p - 3) - 5(p - 3) = 0 Or, (2p - 5) (p - 3) = 0 So, p = 2.5 or p = 3 From II: 2q² + (9x + 2) = (3x + y) q Or, 2q² + (9 × 6 + 2) = (3 × 6 + 5) q Or, 2q² + 56 = 23q Or, 2q²- 23q + 56 = 0 Or, 2q²- 16q - 7q + 56 = 0 Or, 2q (q - 8) - 7(q - 8) = 0 Or, (2q - 7) (q - 8) = 0 So, q= 8 or q = 3.5 On comparing p and q we will get, (p < q)