Question
13, 16, 24, 39, 64,
98 Find the odd one out in the given number seriesSolution
ATQ, 13 + 1 × 3 = 13 + 3 = 16 16 + 2 × 4 = 16 + 8 = 24 24 + 3 × 5 = 24 + 15 = 39 39 + 4 × 6 = 39 + 24 = 63 63 + 5 × 7 = 63 + 35 = 98 Therefore, 63 should be in place of 64.
I. (y – 5)2 – 9 = 0
II. x2 – 3x + 2 = 0
In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y.
I. x
I. x² - 33x + 270 = 0
II. y² - 41y + 414 = 0
Find the remainder when x⁵ − 3x⁴ + 4x³ − 6x² + 8x − 3 is divided by (x − 2).
I. x2 – 18x + 81 = 0
II. y2 – 3y - 28 = 0
I. x2 + 24x + 143 = 0
II. y2 + 12y + 35 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: 4x² - 12x + 9 = 0
Equation 2: 2y² + 8y + 6 = 0
I. 12 x ² - 3 x – 15 = 0
II. 2 y² + 12
I. 6x2 + 19x + 10 = 0
II. y2 + 10y + 25 = 0
I. 49y2 + 35y + 6 = 0
II. 12x2 + 17 x + 6 = 0