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\frac{\frac{[square_root]10[/square_root]}{[square_root]8[/s
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15 Jul 2019, 02:59
15 Jul 2019, 02:59
Expert Reply
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Question Stats:
82% (01:31) correct
17% (01:44) wrong based on 79 sessions
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Quantity A |
Quantity B |
\(\dfrac{ \frac{\sqrt{10}}{ \sqrt{8}}}{\frac{\sqrt{9}}{\sqrt{10}}}\) |
\(\dfrac{ \frac{\sqrt{11}}{\sqrt{9}}}{\frac{\sqrt{10}}{\sqrt{11}}}\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Re: \frac{\frac{[square_root]10[/square_root]}{[square_root]8[/s
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15 Jul 2019, 07:49
15 Jul 2019, 07:49
how
Re: \frac{\frac{[square_root]10[/square_root]}{[square_root]8[/s
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15 Jul 2019, 14:16
15 Jul 2019, 14:16
Expert Reply
A)
\(\frac{\sqrt{10}}{\sqrt{8}} \times \frac{\sqrt{10}}{\sqrt{9}}\)
\(= \frac{10}{\sqrt{72}}\) \(= \frac{10}{6 \sqrt{2} }\) = \(\frac{5}{3 \sqrt{2} }\)
B)
\(\frac{\sqrt{11}}{\sqrt{9}} \times \frac{\sqrt{11}}{\sqrt{10}}\)
\(= \frac{11}{3 \sqrt{10}}\)
A > B clearly because you divide a greater quantity
Similar questions to practice
https://gre.myprepclub.com/forum/search.ph ... mit=Search
\(\frac{\sqrt{10}}{\sqrt{8}} \times \frac{\sqrt{10}}{\sqrt{9}}\)
\(= \frac{10}{\sqrt{72}}\) \(= \frac{10}{6 \sqrt{2} }\) = \(\frac{5}{3 \sqrt{2} }\)
B)
\(\frac{\sqrt{11}}{\sqrt{9}} \times \frac{\sqrt{11}}{\sqrt{10}}\)
\(= \frac{11}{3 \sqrt{10}}\)
A > B clearly because you divide a greater quantity
Similar questions to practice
https://gre.myprepclub.com/forum/search.ph ... mit=Search
