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Linear Inequalities and Linear Programming

88522 If $\frac{2 x + 3}{5} < \frac{4 x - 1}{2}$ , then $x$ lies $x$ in the interval

1

[0, \frac{11}{16})

2

[\frac{11}{16}, \infty)

3

(0, \frac{11}{16})

4

(- \infty, \frac{11}{16})

5

(\frac{11}{16}, \infty)

Explanation:

(E) : We have,
$\frac{2 x + 3}{5} < \frac{4 x - 1}{2}$
$\frac{4 x - 1}{2} - \frac{2 x + 3}{5} > 0$
$\frac{20 x - 5 - (4 x + 6)}{10} > 0$
$\frac{16 x - 11}{10} > 0$
$16 x - 11 > 0$
$x > \frac{11}{16}$

Kerala CEE-2012

Linear Inequalities and Linear Programming

88523 The solution set of $\frac{x + 3}{x - 2} \leq 2$ is

1

(- \infty, \infty)

2

(- \infty, 2] \cup [7, \infty)

3

(- \infty, 2) \cup [7, \infty)

4

[7, \infty)

5

(- \infty, 2)

Explanation:

(C) : Given,
$\frac{x + 3}{x - 2} \leq 2$
$\frac{x + 3}{x - 2} - 2 \leq 0$
$\frac{x + 3 - 2 x + 4}{x - 2} \leq 0$
$\frac{- x + 7}{x - 2} \leq 0$
$\frac{x - 7}{x - 2} \geq 0$
$\frac{(x - 7) (x - 2)}{(x - 2)^{2}} \geq 0$
$\frac{+)^{2} - (+}{7} + \infty$
$x \in (- \infty, 2) \cup [7, \infty)$

Kerala CEE-2011

Linear Inequalities and Linear Programming

88524 If $| 2 x - 3 | < | x + 5 |$ , then $x$ lies in the interval

1

(- 3, 5)

2

(5, 9)

3

(- \frac{2}{3}, 8)

4

(- 8, \frac{2}{3})

5

(- 5, \frac{2}{3})

Explanation:

(C) : Given,
$| 2 x - 3 | < | x + 5 |$
$(2 x - 3)^{2} < (x + 5)^{2}$
$3 x^{2} - 22 x - 16 < 0$
$(3 x + 2) (x - 8) < 0$
$x \in (- \frac{2}{3}, 8)$

Kerala CEE-2011

Linear Inequalities and Linear Programming

88525 If $3 \leq 3 t - 18 \leq 18$ , then which one of the following is true?

1

15 \leq 2 t + 1 \leq 20

2

8 \leq t < 12

3

8 \leq t + 1 \leq 13

4

21 \leq 3 t \leq 24

5

t \leq 7

or

t \geq 12

Explanation:

(C) : Given,
$3 \leq 3 t - 18 \leq 18$
$3 + 18 \leq 3 t \leq 18 + 18$
$21 \leq 3 t \leq 36$
$7 \leq t \leq 12$
$8 \leq t + 1 \leq 13$

Kerala CEE-2010

Linear Inequalities and Linear Programming

88522 If $\frac{2 x + 3}{5} < \frac{4 x - 1}{2}$ , then $x$ lies $x$ in the interval

1

[0, \frac{11}{16})

2

[\frac{11}{16}, \infty)

3

(0, \frac{11}{16})

4

(- \infty, \frac{11}{16})

5

(\frac{11}{16}, \infty)

Explanation:

(E) : We have,
$\frac{2 x + 3}{5} < \frac{4 x - 1}{2}$
$\frac{4 x - 1}{2} - \frac{2 x + 3}{5} > 0$
$\frac{20 x - 5 - (4 x + 6)}{10} > 0$
$\frac{16 x - 11}{10} > 0$
$16 x - 11 > 0$
$x > \frac{11}{16}$

Kerala CEE-2012

Linear Inequalities and Linear Programming

88523 The solution set of $\frac{x + 3}{x - 2} \leq 2$ is

1

(- \infty, \infty)

2

(- \infty, 2] \cup [7, \infty)

3

(- \infty, 2) \cup [7, \infty)

4

[7, \infty)

5

(- \infty, 2)

Explanation:

(C) : Given,
$\frac{x + 3}{x - 2} \leq 2$
$\frac{x + 3}{x - 2} - 2 \leq 0$
$\frac{x + 3 - 2 x + 4}{x - 2} \leq 0$
$\frac{- x + 7}{x - 2} \leq 0$
$\frac{x - 7}{x - 2} \geq 0$
$\frac{(x - 7) (x - 2)}{(x - 2)^{2}} \geq 0$
$\frac{+)^{2} - (+}{7} + \infty$
$x \in (- \infty, 2) \cup [7, \infty)$

Kerala CEE-2011

Linear Inequalities and Linear Programming

88524 If $| 2 x - 3 | < | x + 5 |$ , then $x$ lies in the interval

1

(- 3, 5)

2

(5, 9)

3

(- \frac{2}{3}, 8)

4

(- 8, \frac{2}{3})

5

(- 5, \frac{2}{3})

Explanation:

(C) : Given,
$| 2 x - 3 | < | x + 5 |$
$(2 x - 3)^{2} < (x + 5)^{2}$
$3 x^{2} - 22 x - 16 < 0$
$(3 x + 2) (x - 8) < 0$
$x \in (- \frac{2}{3}, 8)$

Kerala CEE-2011

Linear Inequalities and Linear Programming

88525 If $3 \leq 3 t - 18 \leq 18$ , then which one of the following is true?

1

15 \leq 2 t + 1 \leq 20

2

8 \leq t < 12

3

8 \leq t + 1 \leq 13

4

21 \leq 3 t \leq 24

5

t \leq 7

or

t \geq 12

Explanation:

(C) : Given,
$3 \leq 3 t - 18 \leq 18$
$3 + 18 \leq 3 t \leq 18 + 18$
$21 \leq 3 t \leq 36$
$7 \leq t \leq 12$
$8 \leq t + 1 \leq 13$

Kerala CEE-2010

Linear Inequalities and Linear Programming

88522 If $\frac{2 x + 3}{5} < \frac{4 x - 1}{2}$ , then $x$ lies $x$ in the interval

1

[0, \frac{11}{16})

2

[\frac{11}{16}, \infty)

3

(0, \frac{11}{16})

4

(- \infty, \frac{11}{16})

5

(\frac{11}{16}, \infty)

Explanation:

(E) : We have,
$\frac{2 x + 3}{5} < \frac{4 x - 1}{2}$
$\frac{4 x - 1}{2} - \frac{2 x + 3}{5} > 0$
$\frac{20 x - 5 - (4 x + 6)}{10} > 0$
$\frac{16 x - 11}{10} > 0$
$16 x - 11 > 0$
$x > \frac{11}{16}$

Kerala CEE-2012

Linear Inequalities and Linear Programming

88523 The solution set of $\frac{x + 3}{x - 2} \leq 2$ is

1

(- \infty, \infty)

2

(- \infty, 2] \cup [7, \infty)

3

(- \infty, 2) \cup [7, \infty)

4

[7, \infty)

5

(- \infty, 2)

Explanation:

(C) : Given,
$\frac{x + 3}{x - 2} \leq 2$
$\frac{x + 3}{x - 2} - 2 \leq 0$
$\frac{x + 3 - 2 x + 4}{x - 2} \leq 0$
$\frac{- x + 7}{x - 2} \leq 0$
$\frac{x - 7}{x - 2} \geq 0$
$\frac{(x - 7) (x - 2)}{(x - 2)^{2}} \geq 0$
$\frac{+)^{2} - (+}{7} + \infty$
$x \in (- \infty, 2) \cup [7, \infty)$

Kerala CEE-2011

Linear Inequalities and Linear Programming

88524 If $| 2 x - 3 | < | x + 5 |$ , then $x$ lies in the interval

1

(- 3, 5)

2

(5, 9)

3

(- \frac{2}{3}, 8)

4

(- 8, \frac{2}{3})

5

(- 5, \frac{2}{3})

Explanation:

(C) : Given,
$| 2 x - 3 | < | x + 5 |$
$(2 x - 3)^{2} < (x + 5)^{2}$
$3 x^{2} - 22 x - 16 < 0$
$(3 x + 2) (x - 8) < 0$
$x \in (- \frac{2}{3}, 8)$

Kerala CEE-2011

Linear Inequalities and Linear Programming

88525 If $3 \leq 3 t - 18 \leq 18$ , then which one of the following is true?

1

15 \leq 2 t + 1 \leq 20

2

8 \leq t < 12

3

8 \leq t + 1 \leq 13

4

21 \leq 3 t \leq 24

5

t \leq 7

or

t \geq 12

Explanation:

(C) : Given,
$3 \leq 3 t - 18 \leq 18$
$3 + 18 \leq 3 t \leq 18 + 18$
$21 \leq 3 t \leq 36$
$7 \leq t \leq 12$
$8 \leq t + 1 \leq 13$

Kerala CEE-2010

Linear Inequalities and Linear Programming

88522 If $\frac{2 x + 3}{5} < \frac{4 x - 1}{2}$ , then $x$ lies $x$ in the interval

1

[0, \frac{11}{16})

2

[\frac{11}{16}, \infty)

3

(0, \frac{11}{16})

4

(- \infty, \frac{11}{16})

5

(\frac{11}{16}, \infty)

Explanation:

(E) : We have,
$\frac{2 x + 3}{5} < \frac{4 x - 1}{2}$
$\frac{4 x - 1}{2} - \frac{2 x + 3}{5} > 0$
$\frac{20 x - 5 - (4 x + 6)}{10} > 0$
$\frac{16 x - 11}{10} > 0$
$16 x - 11 > 0$
$x > \frac{11}{16}$

Kerala CEE-2012

Linear Inequalities and Linear Programming

88523 The solution set of $\frac{x + 3}{x - 2} \leq 2$ is

1

(- \infty, \infty)

2

(- \infty, 2] \cup [7, \infty)

3

(- \infty, 2) \cup [7, \infty)

4

[7, \infty)

5

(- \infty, 2)

Explanation:

(C) : Given,
$\frac{x + 3}{x - 2} \leq 2$
$\frac{x + 3}{x - 2} - 2 \leq 0$
$\frac{x + 3 - 2 x + 4}{x - 2} \leq 0$
$\frac{- x + 7}{x - 2} \leq 0$
$\frac{x - 7}{x - 2} \geq 0$
$\frac{(x - 7) (x - 2)}{(x - 2)^{2}} \geq 0$
$\frac{+)^{2} - (+}{7} + \infty$
$x \in (- \infty, 2) \cup [7, \infty)$

Kerala CEE-2011

Linear Inequalities and Linear Programming

88524 If $| 2 x - 3 | < | x + 5 |$ , then $x$ lies in the interval

1

(- 3, 5)

2

(5, 9)

3

(- \frac{2}{3}, 8)

4

(- 8, \frac{2}{3})

5

(- 5, \frac{2}{3})

Explanation:

(C) : Given,
$| 2 x - 3 | < | x + 5 |$
$(2 x - 3)^{2} < (x + 5)^{2}$
$3 x^{2} - 22 x - 16 < 0$
$(3 x + 2) (x - 8) < 0$
$x \in (- \frac{2}{3}, 8)$

Kerala CEE-2011

Linear Inequalities and Linear Programming

88525 If $3 \leq 3 t - 18 \leq 18$ , then which one of the following is true?

1

15 \leq 2 t + 1 \leq 20

2

8 \leq t < 12

3

8 \leq t + 1 \leq 13

4

21 \leq 3 t \leq 24

5

t \leq 7

or

t \geq 12

Explanation:

(C) : Given,
$3 \leq 3 t - 18 \leq 18$
$3 + 18 \leq 3 t \leq 18 + 18$
$21 \leq 3 t \leq 36$
$7 \leq t \leq 12$
$8 \leq t + 1 \leq 13$

Kerala CEE-2010