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, \(x \gt y\).