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 \(y\).

- (I). \(5x^2 - 4x - 12 = 0\), (II). \(y^2 - 4y = 0\)
- \(x \gt y\)
- \(x \lt y\)
- \(x \ge y\)
- \(x \le y\)
- \(x = y\) 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.

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.

- (I). \(4x + 3y = 15\), (II). \(2x - 6y = 20\)
- \(x \gt y\)
- \(x \lt y\)
- \(x \ge y\)
- \(x \le y\)
- \(x = y\) 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 $$ 4x + 3y = 15 $$ $$ 4 \times 5 + 3y = 15 $$ $$ 20 + 3y = 15 $$ $$ 3y = 15 - 20 $$ $$ 3y = -5 $$ $$ y = \frac{-5}{3} $$ $$ Hence, \ x \gt y $$

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 $$ 4x + 3y = 15 $$ $$ 4 \times 5 + 3y = 15 $$ $$ 20 + 3y = 15 $$ $$ 3y = 15 - 20 $$ $$ 3y = -5 $$ $$ y = \frac{-5}{3} $$ $$ Hence, \ x \gt y $$

- (I). \(7x^2 + 3x - 4 = 0\), (II). \(9y^2 - 5y - 4 = 0\)
- \(x \gt y\)
- \(x \lt y\)
- \(x \ge y\)
- \(x \le y\)
- \(x = y\) 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.

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.

- (I). \(x^2 - 5x + 6 = 0\), (II). \(y^2 - 6y + 9 = 0\)
- \(x \gt y\)
- \(x \lt y\)
- \(x \ge y\)
- \(x \le y\)
- \(x = y\) or No relation can be established between \(x\) and \(y\)

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

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

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

- (I). \(8x + 15y = 14\), (II). \(2x - 3y = 17\)
- \(x \gt y\)
- \(x \lt y\)
- \(x \ge y\)
- \(x \le y\)
- \(x = y\) 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 $$ 8x + 15y = 14......(I) $$ $$ 8x + 15 \ (-2) = 14 $$ $$ 8x - 30 = 14 $$ $$ 8x = 44 $$ $$ x = 5.5 $$ $$ Hence, \ x \gt y $$

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 $$ 8x + 15y = 14......(I) $$ $$ 8x + 15 \ (-2) = 14 $$ $$ 8x - 30 = 14 $$ $$ 8x = 44 $$ $$ x = 5.5 $$ $$ Hence, \ x \gt y $$

- (I). \(x^2 = 529\), (II). \(y = \sqrt{529}\)
- \(x \gt y\)
- \(x \lt y\)
- \(x \ge y\)
- \(x \le y\)
- \(x = y\) or No relation can be established between \(x\) and \(y\)

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

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

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

- (I). \(x^2 = 729\), (II). \(y = \sqrt{-729}\)
- \(x \gt y\)
- \(x \lt y\)
- \(x \ge y\)
- \(x \le y\)
- \(x = y\) 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.

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

- (I). \(x^3 - 316 = 413\), (II). \(y^3 - 516 = 815\)
- \(x \gt y\)
- \(x \lt y\)
- \(x \ge y\)
- \(x \le y\)
- \(x = y\) or No relation can be established between \(x\) and \(y\)

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

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

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

- (I). \(x - \sqrt{2401} = 0\), (II). \(\sqrt{y} - 7 = 0\)
- \(x \gt y\)
- \(x \lt y\)
- \(x \ge y\)
- \(x \le y\)
- \(x = y\) or No relation can be established between \(x\) and \(y\)

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

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

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

- (I). \(x^2 - 36 = 0\), (II). \(y^2 + 4y - 12 = 0\)
- \(x \gt y\)
- \(x \lt y\)
- \(x \ge y\)
- \(x \le y\)
- \(x = y\) or No relation can be established between \(x\) and \(y\)

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

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

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