# 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). $$\sqrt{x} - \frac{\sqrt{2}}{\sqrt{x}} = 0$$, (II). $$y^3 = 2^{\frac{3}{2}}$$

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: $$\sqrt{x} - \frac{\sqrt{2}}{\sqrt{x}} = 0....(1)$$ $$\frac{\sqrt{x} \times \sqrt{x} - \sqrt{2}}{\sqrt{x}} = 0$$ $$x - \sqrt{2} = 0$$ $$x = \sqrt{2}$$ $$y^3 = 2^{\frac{3}{2}}....(2)$$ $$y^3 = 2^{\frac{1}{2} \times 3}$$ $$y^3 = (\sqrt{2})^3$$ $$y = \sqrt{2}$$ Hence, $$x = y$$.

1. (I). $$9x - 49 = 41 - 6x$$, (II). $$\sqrt{y + 220} = \sqrt{49} + \sqrt{64}$$

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

Solution: $$9x - 49 = 41 - 6x......(1)$$ $$9x + 6x = 41 + 49$$ $$15x = 90$$ $$x = 6$$ $$\sqrt{y + 220} = \sqrt{49} + \sqrt{64}......(2)$$ $$\sqrt{y + 220} = \pm \ (7 + 8)$$ $$\sqrt{y + 220} = \pm \ 15$$ $$y + 220 = (\pm \ 15)^2$$ $$y + 220 = 225$$ $$y = 5$$ Hence, $$x \gt y$$.

1. (I). $$81 \ x^2 - 10 = 6$$, (II). $$9y + \sqrt{4} = \sqrt{36}$$

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: $$81 \ x^2 - 10 = 6...(1)$$ $$81 \ x^2 = 10 + 6$$ $$81 \ x^2 = 16$$ $$x^2 = \frac{16}{81}$$ $$x = \sqrt{\frac{16}{81}}$$ $$x = \pm \ \frac{4}{9}$$ $$9y + \sqrt{4} = \sqrt{36}...(2)$$ $$9y = \sqrt{36} - \sqrt{4}$$ $$9y = \pm \ (6 - 2)$$ $$9y = \pm \ 4$$ $$y = \pm \ \frac{4}{9}$$ Hence, $$x = y$$.

1. (I). $$\frac{\sqrt{x}}{4} + \frac{2 \sqrt{x}}{8} = \frac{1}{\sqrt{x}}$$, (II). $$\frac{12}{\sqrt{y}} - \frac{4}{\sqrt{y}} = 2 \sqrt{y}$$

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: $$\frac{\sqrt{x}}{4} + \frac{2 \sqrt{x}}{8} = \frac{1}{\sqrt{x}}...(1)$$ $$\frac{2 \sqrt{x} + 2 \sqrt{x}}{8} = \frac{1}{\sqrt{x}}$$ $$\frac{4 \sqrt{x}}{8} = \frac{1}{\sqrt{x}}$$ $$4x = 8$$ $$x = 2$$ $$\frac{12}{\sqrt{y}} - \frac{4}{\sqrt{y}} = 2 \sqrt{y}...(2)$$ $$\frac{12 - 4}{\sqrt{y}} = 2 \sqrt{y}$$ $$2y = 8$$ $$y = 4$$ Hence, $$x \lt y$$.

1. (I). $$x^2 - 26 = 199$$, (II). $$\sqrt{y} - \sqrt{196} = - \sqrt{144}$$

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: $$x^2 - 26 = 199...(1)$$ $$x^2 = 199 + 26$$ $$x^2 = 225$$ $$x = \sqrt{225}$$ $$x = \pm \ 15$$ $$\sqrt{y} - \sqrt{196} = - \sqrt{144}...(2)$$ $$\sqrt{y} = \sqrt{196} - \sqrt{144}$$ $$\sqrt{y} = \pm \ (14 - 12)$$ $$\sqrt{y} = \pm \ 2$$ $$y = (\pm \ 2)^2$$ $$y = 4$$ Hence, No relation can be established.

1. (I). $$x - \sqrt{144} = 0$$, (II). $$y^2 - 144 = 0$$

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) $$x = y$$

Solution: $$x - \sqrt{144} = 0...(1)$$ $$x = \sqrt{144}$$ $$x = \pm \ 12$$ $$y^2 - 144 = 0.....(2)$$ $$y^2 = 144$$ $$y = \sqrt{144}$$ $$y = \pm \ 12$$ $$Hence, \ x = y$$

1. (I). $$x^2 - 355 = 374$$, (II). $$y - \sqrt{441} = \sqrt{36}$$

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: $$x^2 - 355 = 374...(1)$$ $$x^2 = 355 + 374$$ $$x^2 = 729$$ $$x = \sqrt{729}$$ $$x = \pm \ 27$$ $$y - \sqrt{441} = \sqrt{36}...(2)$$ $$y = \sqrt{441} + \sqrt{36}$$ $$y = \pm \ (21 + 6)$$ $$y = \pm \ 27$$ $$Hence, \ x = y$$

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

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

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

1. (I). $$\sqrt{900} \ x + \sqrt{3600} = 0$$, (II). $$(216)^{\frac{1}{3}} \ y - (144)^{\frac{1}{2}} = 0$$

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: $$\sqrt{900} \ x + \sqrt{3600} = 0....(1)$$ $$\sqrt{900} \ x = - \sqrt{3600}$$ $$30x = 60$$ $$x = - 2$$ $$(216)^{\frac{1}{3}} \ y - (144)^{\frac{1}{2}} = 0....(2)$$ $$(216)^{\frac{1}{3}} \ y = (144)^{\frac{1}{2}}$$ $$(6)^{3 \times \frac{1}{3}} \ y = (12)^{2 \times \frac{1}{2}}$$ $$6y = 12$$ $$y = 2$$ $$Hence, \ x \lt y$$

1. (I). $$\frac{(2)^4 + (12)^2}{4} = x^3$$, (II). $$3 \ y^3 = - (145 \div 5) + 4 \ y^3$$

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: $$\frac{(2)^4 + (12)^2}{4} = x^3...(1)$$ $$\frac{16 + 144}{4} = x^3$$ $$\frac{160}{4} = x^3$$ $$x^3 = 40$$ $$3 \ y^3 = - (145 \div 5) + 4 \ y^3....(2)$$ $$3 \ y^3 - 4 \ y^3 = - \frac{145}{5}$$ $$- y^3 = - 29$$ $$y^3 = 29$$ Hence, $$x \gt y$$.