square-root-and-cube-root aptitude question and answers -Page 11

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Square Root and Cube Root problems

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3. If x�y = x + y + xy , the value of 6x24 is:
Options
A. 41
B. 42
C. 43
D. 44

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4. What should come in place of x in the following equation? x 128 = 162 x
Options
A. 13
B. 12
C. 17
D. 16

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Correct Answer: 2

Explanation:

$\sqrt{(\frac{(0.75)^{3}}{1 - 0.75} + [0.75 + (0.75)^{2} + 1])} = \sqrt{\frac{(0.75)^{3} + (1 - 0.75) [(1)^{2} + (0.75)^{2} + 1 � 0.75]}{1 - 0.75}}$

= $\sqrt{\frac{(0.75)^{3} + [(1)^{3} - (0.75)^{3}]}{1 - 0.75}} = \sqrt{\frac{1}{0.25}} = \sqrt{4} = 2$

7. If a = 5 + 15 - 1 b = 5 - 15 + 1 then find the value of a2+ab+b2a2-ab+b2 is ::
Options
A. 3/4
B. 4/3
C. 3/5
D. 5/3

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Correct Answer: 5sqrt{3}

Explanation:

Given Exp = $\frac{3 \sqrt{2}}{\sqrt{6} - \sqrt{3}} � \frac{\sqrt{6} + \sqrt{3}}{\sqrt{6} + \sqrt{3}} - \frac{4 \sqrt{3}}{\sqrt{6} - \sqrt{2}} � \frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} + \sqrt{2}} - \frac{6}{2 (\sqrt{2} - \sqrt{3})}$

= $\frac{3 \sqrt{2} (\sqrt{6} + \sqrt{3})}{6 - 3} - \frac{4 \sqrt{3} (\sqrt{6} + \sqrt{2})}{6 - 2} + \frac{3}{(\sqrt{3} - \sqrt{2})} � \frac{(\sqrt{3} + \sqrt{2})}{(\sqrt{3} + \sqrt{2})}$

= $\sqrt{2} (\sqrt{6} + \sqrt{3}) - \sqrt{3} (\sqrt{6} + \sqrt{2}) + 3 (\sqrt{3} + \sqrt{2})$

= $\sqrt{12} + \sqrt{6} - \sqrt{18} - \sqrt{6} + 3 \sqrt{3} + 3 \sqrt{2}$

= $5 \sqrt{3}$

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