# Equations and Inequalities Questions and Answers:

#### Overview:

 Questions and Answers Type: MCQ (Multiple Choice Questions). Main Topic: Quantitative Aptitude. Quantitative Aptitude Sub-topic: Equations and Inequalities Aptitude Questions and Answers. Number of Questions: 10 Questions with Solutions.

Directions: Two equations (I) and (II) are given in each question. On the basis of these equations, you have to decide the relation between $$x$$ and $$y$$.

1. (I). $$3x^2 + 14x + 15 = 0$$, (II). $$2y^2 + 18y + 36 = 0$$

1. $$x \gt y$$
2. $$x \lt y$$
3. $$x \ge y$$
4. $$x \le y$$
5. $$x = y$$ or No relation can be established between $$x$$ and $$y$$

Answer: (c) $$x \ge y$$

Solution: $$3x^2 + 14x + 15 = 0......(1)$$ $$3x^2 + 9x + 5x + 15 = 0$$ $$3x \ (x + 3) + 5 \ (x + 3) = 0$$ $$(x + 3) \ (3x + 5) = 0$$ $$x = -3, \ and \ x = \frac{-5}{3}$$ $$2y^2 + 18y + 36 = 0......(2)$$ $$2y^2 + 12y + 6y + 36 = 0$$ $$2y \ (y + 6) + 6 \ (y + 6) = 0$$ $$(y + 6) \ (2y + 6) = 0$$ $$y = -6, \ and \ y = -3$$ $$Hence, \ x \ge y$$

1. (I). $$8x^2 - 10x - 3 = 0$$, (II). $$7y^2 - 12y - 4 = 0$$

1. $$x \gt y$$
2. $$x \lt y$$
3. $$x \ge y$$
4. $$x \le y$$
5. $$x = y$$ or No relation can be established between $$x$$ and $$y$$

Answer: (e) No relation can be established between $$x$$ and $$y$$

Solution: $$8x^2 - 10x - 3 = 0......(1)$$ $$8x^2 - 12x + 2x - 3 = 0$$ $$4x \ (2x - 3) + 1 \ (2x - 3) = 0$$ $$(2x - 3) \ (4x + 1) = 0$$ $$x = \frac{3}{2}, \ and \ x = \frac{-1}{4}$$ $$7y^2 - 12y - 4 = 0......(2)$$ $$7y^2 - 14y + 2y - 4 = 0$$ $$7y \ (y - 2) + 2 \ (y - 2) = 0$$ $$(y - 2) \ (7y + 2) = 0$$ $$y = 2, \ and \ y = \frac{-2}{7}$$ Hence, No relation can be established.

1. (I). $$x^2 + 5x - 6 = 0$$, (II). $$y^2 - 6y + 9 = 0$$

1. $$x \gt y$$
2. $$x \lt y$$
3. $$x \ge y$$
4. $$x \le y$$
5. $$x = y$$ or No relation can be established between $$x$$ and $$y$$

Answer: (b) $$x \lt y$$

Solution: $$x^2 + 5x - 6 = 0......(1)$$ $$x^2 + 6x - x - 6 = 0$$ $$x \ (x + 6) - 1 \ (x + 6) = 0$$ $$(x + 6) \ (x - 1) = 0$$ $$x = -6, \ and \ x = 1$$ $$y^2 - 6y + 9 = 0......(2)$$ $$y^2 - 3y - 3y + 9 = 0$$ $$y \ (y - 3) - 3 \ (y - 3) = 0$$ $$y = 3, \ and \ y = 3$$ $$Hence, \ x \lt y$$

1. (I). $$x^2 = 64$$, (II). $$y = \sqrt{64}$$

1. $$x \gt y$$
2. $$x \lt y$$
3. $$x \ge y$$
4. $$x \le y$$
5. $$x = y$$ or No relation can be established between $$x$$ and $$y$$

Answer: (d) $$x \le y$$

Solution: $$x^2 = 64......(1)$$ $$x = \pm 8$$ $$y = \sqrt{64}......(2)$$ $$y = 8$$ $$Hence, \ x \le y$$

1. (I). $$15x - 12y = 18$$, (II). $$5x + 6y = 8$$

1. $$x \gt y$$
2. $$x \lt y$$
3. $$x \ge y$$
4. $$x \le y$$
5. $$x = y$$ or No relation can be established between $$x$$ and $$y$$

Answer: (b) $$x \lt y$$

Solution: $$15x - 12y = 18......(1)$$ $$5x + 6y = 8......(2)$$ By multiplying equation (2) with 3, we get $$15x + 18y = 24......(3)$$ Now, by adding equations (1) and (3), we get $$15x - 12y + 15x + 18y = 18 + 24$$ $$6y = 42$$ $$y = 7$$ Now, by putting the value of y in equation, we get $$15x - 12y = 18......(1)$$ $$15x - 12 \times 7 = 18$$ $$15x - 84 = 18$$ $$15x = 102$$ $$x = 6.8$$ $$Hence, \ x \lt y$$

1. (I). $$6x^2 - 9x - 6 = 0$$, (II). $$4y^2 - 4y - 3 = 0$$

1. $$x \gt y$$
2. $$x \lt y$$
3. $$x \ge y$$
4. $$x \le y$$
5. $$x = y$$ or No relation can be established between $$x$$ and $$y$$

Answer: (c) $$x \ge y$$

Solution: $$6x^2 - 9x - 6 = 0......(1)$$ $$6x^2 - 12x + 3x - 6 = 0$$ $$6x \ (x - 2) + 3 \ (x - 2) = 0$$ $$(x - 2) \ (6x + 3) = 0$$ $$x = 2, \ and \ x = \frac{-1}{2}$$ $$4y^2 - 4y - 3 = 0......(2)$$ $$4y^2 - 6y + 2y - 3 = 0$$ $$2y \ (2y - 3) + 1 \ (2y - 3) = 0$$ $$(2y - 3) \ (2y + 1) = 0$$ $$y = \frac{3}{2}, \ and \ y = \frac{-1}{2}$$ $$Hence, \ x \ge y$$

1. (I). $$x = \sqrt{1296}$$, (II). $$\sqrt{y} = 6$$

1. $$x \gt y$$
2. $$x \lt y$$
3. $$x \ge y$$
4. $$x \le y$$
5. $$x = y$$ or No relation can be established between $$x$$ and $$y$$

Answer: (e) $$x = y$$

Solution: $$x = \sqrt{1296}......(1)$$ $$x = 36$$ $$\sqrt{y} = 6......(2)$$ $$y = 6^2$$ $$y = 36$$ $$Hence, \ x = y$$

1. (I). $$2x + 2y = 4$$, (II). $$x - 2y = 5$$

1. $$x \gt y$$
2. $$x \lt y$$
3. $$x \ge y$$
4. $$x \le y$$
5. $$x = y$$ or No relation can be established between $$x$$ and $$y$$

Answer: (a) $$x \gt y$$

Solution: $$2x + 2y = 4......(1)$$ $$x - 2y = 5......(2)$$ By adding equations (1) and (2), we get $$2x + 2y + x - 2y = 4 + 5$$ $$3x = 9$$ $$x = 3$$ Now, by putting the value of x in equation (1), we get $$2x + 2y = 4......(1)$$ $$2 \times 3 + 2y = 4$$ $$6 + 2y = 4$$ $$2y = -2$$ $$y = -1$$ $$Hence, \ x \gt y$$

1. (I). $$9x + 4y = 18$$, (II). $$5x + 4y = 6$$

1. $$x \gt y$$
2. $$x \lt y$$
3. $$x \ge y$$
4. $$x \le y$$
5. $$x = y$$ or No relation can be established between $$x$$ and $$y$$

Answer: (a) $$x \gt y$$

Solution: $$9x + 4y = 18......(1)$$ $$5x + 4y = 6......(2)$$ By subtracting equation (2) with equation (1), we get $$9x + 4y - (5x + 4y) = 18 - 6$$ $$9x + 4y - 5x - 4y = 12$$ $$4x = 12$$ $$x = 3$$ Now, by putting the value of x in equation (1), we get $$9x + 4y = 18......(1)$$ $$9 \times 3 + 4y = 18$$ $$27 + 4y = 18$$ $$4y = -9$$ $$y = \frac{-9}{4}$$ $$y = -2.25$$ $$Hence, \ x \gt y$$

1. (I). $$3x^2 - 5x - 12 = 0$$, (II). $$5y^2 + 9y + 4 = 0$$

1. $$x \gt y$$
2. $$x \lt y$$
3. $$x \ge y$$
4. $$x \le y$$
5. $$x = y$$ or No relation can be established between $$x$$ and $$y$$

Answer: (e) No relation can be established between $$x$$ and $$y$$

Solution: $$3x^2 - 5x - 12 = 0......(1)$$ $$3x^2 - 9x + 4x - 12 = 0$$ $$3x \ (x - 3) + 4 \ (x - 3) = 0$$ $$(x - 3) \ (3x + 4) = 0$$ $$x = 3, \ and \ x = \frac{-4}{3}$$ $$5y^2 + 9y + 4 = 0......(2)$$ $$5y^2 + 5y + 4y + 4 = 0$$ $$5y \ (y + 1) + 4 \ (y + 1) = 0$$ $$(y + 1) \ (5y + 4) = 0$$ $$y = -1, \ and \ y = \frac{-4}{5}$$ Hence, No relation can be established.