Question
I. 20y² - 13y + 2 = 0 II. 6x² - 25x +
14 = 0 In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer.Solution
I. 20y² - 13y + 2 = 0 20y² - 8y - 5y + 2 = 0 4y (5y – 2) – 1(5y – 2) = 0 (4y – 1) (5y – 2) = 0 y =2/5, 1/4 II. 6x² - 25x + 14 = 0 6x² - 21x - 4x + 14 = 0 3x (2x – 7) - 2 (2x – 7) = 0 (3x – 2) (2x – 7) = 0 x = 7/2 , 2/3 Hence, x>y
(i) 2x² – 12x + 16 = 0
(ii) 2y² – 20y + 48 = 0
I. x2 – 13x + 36 = 0
II. 3y2 – 29y + 18 = 0
Solve: 3x² − 7x − 6 = 0
I. 5x + 2y = 31
II. 3x + 7y = 36
What is the average salary of 10 officers , if a clerk’s salary is Rs. 15,000?
Statement I: The total salary of 10 officers and 5 clerk is Rs....
For what values of k does the equation x² – (k+1)x + k = 0 have two distinct real roots, both greater than 1?
I. x2 + x – 42 = 0
II. y2 + 6y – 27 = 0
I. 25p + 2(2p2 – 1) = 8(p + 5)
II. 8q2 + 35q – 78 = 0
I. x2 + 24x + 143 = 0
II. y2 + 12y + 35 = 0
I. 2x² - 12x + 16 = 0  Â
II. 4y² - 8y - 12 = 0  Â
Relevant for Exams:
