Question
I. 3x² - 22 x + 40 = 0  II. 4y² + 22y +
24 = 0Â Â Â In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer.Solution
I. 3x² - 22 x + 40 = 0   3x² - 12x - 10x + 40 = 0   3x (x – 4) – 10 (x – 4) (3x – 10 ) (x – 4) x = 10/3, 4 II. 4y² + 22y + 24 = 0  4y² + 16 y + 6 y + 24 = 0   4y (y + 4) + 6( y + 4) (4y + 6) (y + 4) x = - 4 , - 3/2 Hence, x > y.
625, 5, 125, 25, 25, ? , 5
11, 9, 15, 41 ,157, 789
1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 6Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ...
18, 25, 33, 42, 52, 63, ?
2187, 1458, 972, ?, 432, 288
35.25 71 146 296 624 1376
...Find the wrong number in the given series.
12, 18, 27, 42, 62, 87
4, 17, 62, 317, ?, 13267
3720 3842 ? 4092 4220 4350
64Â Â Â Â Â Â 96Â Â Â Â Â Â Â 48Â Â Â Â Â Â 72Â Â Â Â Â Â 36Â Â Â Â Â Â ?
...
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