Question
A gets 36% of the full marks in an exam and fails by 90
marks. B gets 39% of the full marks in the same exam and he also fails by 75 marks. What is the minimum percentage of marks a student should obtain to pass the exam?Solution
Let maximum marks = 100x ∴ 36x + 90 = 39x + 75 => 3x = 15 => x = 5 ∴ Max marks = 100x = 500 ∴ Minimum percentage marks to be pass = {[(36×5) + 90]/500} × 100 = 54%
I. 2p² - 11p + 12 = 0
II. 2q² - 17q + 36 = 0
- Determine the remainder when equation 4p³- 5p² + 2p + 1 is divided by (4p - 3).
I. 99x² + 161 x + 26 = 0
II. 26 y² + 161 y + 99 = 0
I. x2 - 4x – 21 = 0
II. y2 + 12y + 20 = 0
I. 3q² -29q +18 = 0
II. 9p² - 4 = 0
I. 8x2 - 2x – 15 = 0
II. 12y2 - 17y – 40 = 0
In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between 'p' and 'q' and choose...
Solve the quadratic equations and determine the relation between x and y:
Equation 1: x² - 29x + 210 = 0
Equation 2: y² - 27y + 182 = 0
I. 15y2 + 4y – 4 = 0
II. 15x2 + x – 6 = 0
I. p2 +7p + 10 = 0 II. q2 - q – 6 = 0
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