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 \(y\).
(I). \(\frac{6}{\sqrt{x}} + \frac{4}{\sqrt{x}} = \sqrt{x}\), (II). \(y^2 - \frac{10^\frac{5}{2}}{\sqrt{y}} = 0\)
\(x \gt y\)
\(x \lt y\)
\(x \ge y\)
\(x \le y\)
\(x = y\) or No relation can be established between \(x\) and \(y\)
Answer & Solution
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, \(x = y\).
(I). \(\sqrt{49} x + \sqrt{441} = 0\), (II). \(8y - \sqrt{256} = 0\)
\(x \gt y\)
\(x \lt y\)
\(x \ge y\)
\(x \le y\)
\(x = y\) or No relation can be established between \(x\) and \(y\)
Answer & Solution
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, \(x \lt y\).
(I). \(\frac{12}{\sqrt{x}} - \frac{2}{\sqrt{x}} = 5 \sqrt{x}\), (II). \(\frac{\sqrt{y}}{3} + \frac{3 \sqrt{y}}{18} = \frac{1}{\sqrt{y}}\)
\(x \gt y\)
\(x \lt y\)
\(x \ge y\)
\(x \le y\)
\(x = y\) or No relation can be established between \(x\) and \(y\)
Answer & Solution
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, \(x = y\).
(I). \(x^2 - 256 = 144\), (II). \(y^2 - 35y + 306 = 0\)
\(x \gt y\)
\(x \lt y\)
\(x \ge y\)
\(x \le y\)
\(x = y\) or No relation can be established between \(x\) and \(y\)
Answer & Solution
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.
(I). \(2 \sqrt{x} - \frac{1}{3 \sqrt{x}} = \sqrt{x}\), (II). \(\frac{1}{\sqrt{y}} + \frac{2}{\sqrt{y}} = 12 \sqrt{y}\)
\(x \gt y\)
\(x \lt y\)
\(x \ge y\)
\(x \le y\)
\(x = y\) or No relation can be established between \(x\) and \(y\)
Answer & Solution
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, \(x \gt y\).
(I). \(\frac{12}{x^2} + \frac{5}{x^2} - \frac{7}{x^2} = \frac{20}{x}\), (II). \(5.25 + 2.50 = y + 7\)
\(x \gt y\)
\(x \lt y\)
\(x \ge y\)
\(x \le y\)
\(x = y\) or No relation can be established between \(x\) and \(y\)
Answer & Solution
Answer: (b) \(x \lt y\)Solution: $$ \frac{12}{x^2} + \frac{5}{x^2} - \frac{7}{x^2} = \frac{20}{x}...(1) $$ $$ \frac{12 + 5 + 7}{x^2} = \frac{20}{x} $$ $$ \frac{10}{x^2} = \frac{20}{x} $$ $$ 20x^2 = 10x $$ $$ x = \frac{1}{2} = 0.5 $$ $$ 5.25 + 2.50 = y + 7.....(2) $$ $$ 7.75 = y + 7 $$ $$ 7.75 - 7 = y $$ $$ y = 0.75 $$ $$ Hence, \ x \lt y $$
(I). \(\frac{x^{\frac{1}{2}}}{64} = \frac{81}{x^{\frac{3}{2}}}\), (II). \(y^{\frac{1}{3}} \times y^{\frac{2}{3}} \times 576 = 8 \times y^2\)
\(x \gt y\)
\(x \lt y\)
\(x \ge y\)
\(x \le y\)
\(x = y\) or No relation can be established between \(x\) and \(y\)
Answer & Solution
Answer: (d) \(x \le y\)Solution: $$ \frac{x^{\frac{1}{2}}}{64} = \frac{81}{x^{\frac{3}{2}}}...(1) $$ $$ x^{\frac{1}{2}} \times x^{\frac{3}{2}} = 81 \times 64 $$ $$ x^2 = 81 \times 64 $$ $$ x = \sqrt{81 \times 64} $$ $$ x = \pm \ 9 \times 8 $$ $$ x = \pm \ 72 $$ $$ y^{\frac{1}{3}} \times y^{\frac{2}{3}} \times 576 = 8 \times y^2...(2) $$ $$ y \times 576 = 8 \times y^2 $$ $$ 576 = 8y $$ $$ y = 72 $$ $$ Hence, \ x \le y $$
(I). \(\frac{2}{\sqrt{x}} + \frac{3}{\sqrt{x}} = \sqrt{x}\), (II). \(y^2 - \frac{5^\frac{5}{2}}{\sqrt{y}} = 0\)
\(x \gt y\)
\(x \lt y\)
\(x \ge y\)
\(x \le y\)
\(x = y\) or No relation can be established between \(x\) and \(y\)
Answer & Solution
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, \(x = y\).
(I). \(5x - 2x = 22.50 + 4.50\), (II). \(\sqrt{y + 40} = 4 + 3\)
\(x \gt y\)
\(x \lt y\)
\(x \ge y\)
\(x \le y\)
\(x = y\) or No relation can be established between \(x\) and \(y\)
Answer & Solution
Answer: (e) \(x = y\)Solution: $$ 5x - 2x = 22.50 + 4.50....(1) $$ $$ 3x = 27 $$ $$ x = 9 $$ $$ \sqrt{y + 40} = 4 + 3....(2) $$ $$ \sqrt{y + 40} = 7 $$ $$ y + 40 = 7^2 $$ $$ y + 40 = 49 $$ $$ y = 9 $$ $$ Hence, \ x = y $$
(I). \(\sqrt{x + 12} = \sqrt{100} - \sqrt{36}\), (II). \(y^2 = 235 - 210\)
\(x \gt y\)
\(x \lt y\)
\(x \ge y\)
\(x \le y\)
\(x = y\) or No relation can be established between \(x\) and \(y\)
Answer & Solution
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.