# Equations and Inequalities Solved Questions:

#### 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). $$5x^2 - 4x - 12 = 0$$, (II). $$y^2 - 4y = 0$$

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: $$5x^2 - 4x - 12 = 0......(1)$$ $$5x^2 - 10x + 6x - 12 = 0$$ $$5x \ (x - 2) + 6 \ (x - 2) = 0$$ $$(x - 2) \ (5x + 6) = 0$$ $$x = 2, \ and \ x = \frac{-6}{5}$$ $$y^2 - 4y = 0......(2)$$ $$y \ (y - 4) = 0$$ $$y = 0, \ and \ y = 4$$ Hence, No relation can be established.

1. (I). $$4x + 3y = 15$$, (II). $$2x - 6y = 20$$

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: $$4x + 3y = 15......(1)$$ $$2x - 6y = 20......(2)$$ By multiplying equation (1) with 2, we get $$8x + 6y = 30......(3)$$ Now, by adding the equation (2) and (3), we get $$8x + 6y + 2x - 6y = 30 + 20$$ $$10x = 50$$ $$x = 5$$ Now, by putting the value of x in equation (1), we get $$4x + 3y = 15$$ $$4 \times 5 + 3y = 15$$ $$20 + 3y = 15$$ $$3y = 15 - 20$$ $$3y = -5$$ $$y = \frac{-5}{3}$$ $$Hence, \ x \gt y$$

1. (I). $$7x^2 + 3x - 4 = 0$$, (II). $$9y^2 - 5y - 4 = 0$$

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: $$7x^2 + 3x - 4 = 0......(1)$$ $$7x^2 + 7x - 4x - 4 = 0$$ $$7x \ (x + 1) - 4 \ (x + 1) = 0$$ $$(x + 1) \ (7x - 4) = 0$$ $$x = -1, \ and \ x = \frac{4}{7}$$ $$9y^2 - 5y - 4 = 0......(2)$$ $$9y^2 - 9y + 4y - 4 = 0$$ $$9y \ (y - 1) + 4 \ (y - 1) = 0$$ $$(y - 1) \ (9y + 4) = 0$$ $$y = 1, \ and \ y = \frac{-4}{9}$$ 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: (d) $$x \le y$$

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

1. (I). $$8x + 15y = 14$$, (II). $$2x - 3y = 17$$

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: $$8x + 15y = 14......(I)$$ $$2x - 3y = 17......(II)$$ By multiplying equation (II) with 4, we get $$8x - 12y = 68......(III)$$ Now, by subtracting equation (III) from eqaution (I), we get $$8x + 15y - 8x + 12y = 14 - 68$$ $$27y = -54$$ $$y = -2$$ By putting the value of y in equation (I), we get $$8x + 15y = 14......(I)$$ $$8x + 15 \ (-2) = 14$$ $$8x - 30 = 14$$ $$8x = 44$$ $$x = 5.5$$ $$Hence, \ x \gt y$$

1. (I). $$x^2 = 529$$, (II). $$y = \sqrt{529}$$

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: $$x^2 = 529......(I)$$ $$x = \pm 23$$ $$y = \sqrt{529}......(II)$$ $$y = 23$$ $$Hence, \ x \le y$$

1. (I). $$x^2 = 729$$, (II). $$y = \sqrt{-729}$$

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: $$x^2 = 729......(I)$$ $$x = \pm 27$$ $$y = \sqrt{-729}......(II)$$ $$y = \sqrt{-729}$$ Hence, No relation can be established.

1. (I). $$x^3 - 316 = 413$$, (II). $$y^3 - 516 = 815$$

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: $$x^3 - 316 = 413......(I)$$ $$x^3 = 413 + 316$$ $$x^3 = 729$$ $$x = \sqrt[3]{729}$$ $$x = 9$$ $$y^3 - 516 = 815......(II)$$ $$y^3 = 815 + 516$$ $$y^3 = 1331$$ $$y = \sqrt[3]{1331}$$ $$y = 11$$ $$Hence, \ x \lt y$$

1. (I). $$x - \sqrt{2401} = 0$$, (II). $$\sqrt{y} - 7 = 0$$

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: $$x - \sqrt{2401} = 0......(I)$$ $$x = \sqrt{2401}$$ $$x = 49$$ $$\sqrt{y} - 7 = 0......(II)$$ $$\sqrt{y} = 7$$ $$y = 7^2$$ $$y = 49$$ $$Hence, \ x = y$$

1. (I). $$x^2 - 36 = 0$$, (II). $$y^2 + 4y - 12 = 0$$

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: $$x^2 - 36 = 0......(I)$$ $$x^2 = 36$$ $$x = \pm 6$$ $$y^2 + 4y - 12 = 0......(II)$$ $$y^2 + 6y - 2y - 12 = 0$$ $$y \ (y + 6) - 2 \ (y + 6) = 0$$ $$(y + 6) \ (y - 2) = 0$$ $$y = -6, \ and \ y = 2$$ $$Hence, \ x \ge y$$