The mode of the observation $2 x + 3$ , $3 x - 2$ , $4 x + 3$ , $x - 1$ , $3 x - 1$ , $5 x + 2$ (x is a positive integer) can be

3

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5

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7

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9

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Solution

The correct option is C $7$
Since, smallest positive integer is 1.
given, x is a positive integer.
$∴ x = 1$
Given obsevations are $2 x + 3, 3 x - 2, 4 x + 3, x - 1, 3 x - 1, 5 x + 2$
put $x = 1$ in given obsevations. we get,
$5, 1, 7, 0, 2, 7$
7 appears twice, all the rest appear only once, so Mode is 7.
Option C is correct.

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$(i) \frac{1}{3} x - 6 = \frac{5}{2} (i i) \frac{2 x}{3} - \frac{3 x}{8} = \frac{7}{12} (i i i) (x + 2) (x + 3) + (x - 3) (x - 2) - 2 x (x + 1) = 0 (i v) \frac{1}{10} - \frac{7}{x} = 35 (v) 13 (x - 4) - 3 (x - 9) - 4 (x + 4) = 0 (v i) x + 7 - \frac{8 x}{3} = \frac{17 x}{6} - \frac{5 x}{8} (v i i) \frac{3 x - 2}{4} - \frac{2 x + 3}{3} = \frac{2}{3} - x (v i i i) \frac{x + 2}{6} - (\frac{11 - x}{3} - \frac{1}{4}) = \frac{3 x - 4}{12} (i x) \frac{2}{5 x} - \frac{5}{3 x} = \frac{1}{15} (x) \frac{x + 2}{3} - \frac{x + 1}{5} = \frac{x - 3}{4} - 1 (x i) \frac{3 x - 2}{3} + \frac{2 x + 3}{2} = x + \frac{7}{6} (x i i) x - \frac{x - 1}{2} = 1 - \frac{x - 2}{3} (x i i i) \frac{9 x + 7}{2} - (x - \frac{x - 2}{7}) = 36 (x i v) \frac{6 x + 1}{2} + 1 = \frac{7 x - 3}{3}$