# 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). $$\frac{6}{\sqrt{x}} + \frac{4}{\sqrt{x}} = \sqrt{x}$$, (II). $$y^2 - \frac{10^\frac{5}{2}}{\sqrt{y}} = 0$$

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: $$\frac{6}{\sqrt{x}} + \frac{4}{\sqrt{x}} = \sqrt{x}....(1)$$ $$\frac{6 + 4}{\sqrt{x}} = \sqrt{x}$$ $$6 + 4 = \sqrt{x} \times \sqrt{x}$$ $$x = 10$$ $$y^2 - \frac{10^\frac{5}{2}}{\sqrt{y}} = 0....(2)$$ $$\frac{y^{2 + \frac{1}{2}} - 10^\frac{5}{2}}{\sqrt{y}} = 0$$ $$y^{\frac{5}{2}} - 10^\frac{5}{2} = 0$$ $$y^{\frac{5}{2}} = 10^\frac{5}{2}$$ $$y = 10$$ Hence, $$x = y$$.

1. (I). $$\sqrt{49} x + \sqrt{441} = 0$$, (II). $$8y - \sqrt{256} = 0$$

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: $$\sqrt{49} x + \sqrt{441} = 0......(1)$$ $$7x + 21 = 0$$ $$7x = -21$$ $$x = - \frac{21}{7}$$ $$x = -3$$ $$8y - \sqrt{256} = 0......(2)$$ $$8y - 16 = 0$$ $$8y = 16$$ $$y = 2$$ Hence, $$x \lt y$$.

1. (I). $$\frac{12}{\sqrt{x}} - \frac{2}{\sqrt{x}} = 5 \sqrt{x}$$, (II). $$\frac{\sqrt{y}}{3} + \frac{3 \sqrt{y}}{18} = \frac{1}{\sqrt{y}}$$

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: (e) $$x = y$$

Solution: $$\frac{12}{\sqrt{x}} - \frac{2}{\sqrt{x}} = 5 \sqrt{x}...(1)$$ $$\frac{12 - 2}{\sqrt{x}} = 5 \sqrt{x}$$ $$10 = 5x$$ $$x = 2$$ $$\frac{\sqrt{y}}{3} + \frac{3 \sqrt{y}}{18} = \frac{1}{\sqrt{y}}...(2)$$ $$\frac{6 \sqrt{y} + 3 \sqrt{y}}{18} = \frac{1}{\sqrt{y}}$$ $$\frac{9 \sqrt{y}}{18} = \frac{1}{\sqrt{y}}$$ $$9y = 18$$ $$y = 2$$ Hence, $$x = y$$.

1. (I). $$x^2 - 256 = 144$$, (II). $$y^2 - 35y + 306 = 0$$

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: $$x^2 - 256 = 144......(1)$$ $$x^2 = 144 + 256$$ $$x^2 = 400$$ $$x = \sqrt{400}$$ $$x = \pm 20$$ $$y^2 - 35y + 306 = 0......(2)$$ $$y^2 - 17y - 18y + 306 = 0$$ $$y \ (y - 17) - 18 \ (y - 17) = 0$$ $$(y - 17) \ (y - 18) = 0$$ $$y = 17 \ and \ y = 18$$ Hence, No relation can be established.

1. (I). $$2 \sqrt{x} - \frac{1}{3 \sqrt{x}} = \sqrt{x}$$, (II). $$\frac{1}{\sqrt{y}} + \frac{2}{\sqrt{y}} = 12 \sqrt{y}$$

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

Solution: $$2 \sqrt{x} - \frac{1}{3 \sqrt{x}} = \sqrt{x}...(1)$$ $$\frac{6x - 1}{3 \sqrt{x}} = \sqrt{x}$$ $$6x - 1 = 3x$$ $$3x = 1$$ $$x = \frac{1}{3}$$ $$\frac{1}{\sqrt{y}} + \frac{2}{\sqrt{y}} = 12 \sqrt{y}...(2)$$ $$\frac{1 + 2}{\sqrt{y}} = 12 \sqrt{y}$$ $$12y = 3$$ $$y = \frac{1}{4}$$ Hence, $$x \gt y$$.

1. (I). $$\frac{12}{x^2} + \frac{5}{x^2} - \frac{7}{x^2} = \frac{20}{x}$$, (II). $$5.25 + 2.50 = y + 7$$

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: $$\frac{12}{x^2} + \frac{5}{x^2} - \frac{7}{x^2} = \frac{20}{x}...(1)$$ $$\frac{12 + 5 + 7}{x^2} = \frac{20}{x}$$ $$\frac{10}{x^2} = \frac{20}{x}$$ $$20x^2 = 10x$$ $$x = \frac{1}{2} = 0.5$$ $$5.25 + 2.50 = y + 7.....(2)$$ $$7.75 = y + 7$$ $$7.75 - 7 = y$$ $$y = 0.75$$ $$Hence, \ x \lt y$$

1. (I). $$\frac{x^{\frac{1}{2}}}{64} = \frac{81}{x^{\frac{3}{2}}}$$, (II). $$y^{\frac{1}{3}} \times y^{\frac{2}{3}} \times 576 = 8 \times y^2$$

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

Solution: $$\frac{x^{\frac{1}{2}}}{64} = \frac{81}{x^{\frac{3}{2}}}...(1)$$ $$x^{\frac{1}{2}} \times x^{\frac{3}{2}} = 81 \times 64$$ $$x^2 = 81 \times 64$$ $$x = \sqrt{81 \times 64}$$ $$x = \pm \ 9 \times 8$$ $$x = \pm \ 72$$ $$y^{\frac{1}{3}} \times y^{\frac{2}{3}} \times 576 = 8 \times y^2...(2)$$ $$y \times 576 = 8 \times y^2$$ $$576 = 8y$$ $$y = 72$$ $$Hence, \ x \le y$$

1. (I). $$\frac{2}{\sqrt{x}} + \frac{3}{\sqrt{x}} = \sqrt{x}$$, (II). $$y^2 - \frac{5^\frac{5}{2}}{\sqrt{y}} = 0$$

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: $$\frac{2}{\sqrt{x}} + \frac{3}{\sqrt{x}} = \sqrt{x}....(1)$$ $$\frac{2 + 3}{\sqrt{x}} = \sqrt{x}$$ $$2 + 3 = \sqrt{x} \times \sqrt{x}$$ $$x = 5$$ $$y^2 - \frac{5^\frac{5}{2}}{\sqrt{y}} = 0....(2)$$ $$\frac{y^{2 + \frac{1}{2}} - 5^\frac{5}{2}}{\sqrt{y}} = 0$$ $$y^{\frac{5}{2}} - 5^\frac{5}{2} = 0$$ $$y^{\frac{5}{2}} = 5^\frac{5}{2}$$ $$y = 5$$ Hence, $$x = y$$.

1. (I). $$5x - 2x = 22.50 + 4.50$$, (II). $$\sqrt{y + 40} = 4 + 3$$

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: (e) $$x = y$$

Solution: $$5x - 2x = 22.50 + 4.50....(1)$$ $$3x = 27$$ $$x = 9$$ $$\sqrt{y + 40} = 4 + 3....(2)$$ $$\sqrt{y + 40} = 7$$ $$y + 40 = 7^2$$ $$y + 40 = 49$$ $$y = 9$$ $$Hence, \ x = y$$

1. (I). $$\sqrt{x + 12} = \sqrt{100} - \sqrt{36}$$, (II). $$y^2 = 235 - 210$$

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: $$\sqrt{x + 12} = \sqrt{100} - \sqrt{36}...(1)$$ $$\sqrt{x + 12} = \pm \ (10 - 6)$$ $$\sqrt{x + 12} = \pm \ 4$$ $$x + 12 = (\pm \ 4)^2$$ $$x + 12 = 16$$ $$x = 4$$ $$y^2 = 235 - 210....(2)$$ $$y^2 = 25$$ $$y = \sqrt{25}$$ $$y = \pm \ 5$$ Hence, No relation can be established.