Question
A 5-digit number is created using the digits 2, 3, 4, 5,
and 6 without repetition. What is the probability that the number is divisible by 5?Solution
ATQ,
Total numbers formed = 5! = 120 For divisibility by 5, the number must end in either 0 or 5. But 0 is not among the digits, so only 5 can be at the unit place. Fix 5 at the end → Remaining 4 digits can be arranged in 4! = 24 ways Total such numbers formed = 24 × 1 = 24 Required probability = 24 / 120 = 1 / 5
I. Â 3y2Â + 13y - 16 = 0
II. 3x2 – 13x + 14 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: 31x² - 170x + 216 = 0
Equation 2: 22y² - 132y + ...
I. 20y² - 13y + 2 = 0
II. 6x² - 25x + 14 = 0
I. 2y2Â + 11y + 15 = 0
II. 3x2Â + 4x - 4= 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...
I. 5x² = 19x – 12
II. 5y² + 11y = 12
I. x2 – 13x + 40 = 0
II. 2y2 – 15y + 13 = 0Â
I. x² + 3x – 154 = 0
II. y² + 5y – 126 = 0
I. x² + 11x + 24 = 0
II. y² + 17y + 72 = 0
I. 4p² + 17p + 15 = 0
II. 3q² + 19q + 28 = 0
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