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01 Nov 2010, 05:41
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D
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Question Stats:
57% (01:23) correct
43%
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wrong
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Is the length of the diagonal of the rectangle greater than \(\sqrt{6}\) centimeters ?
(1) The shorter side of the rectangle measures 2 centimeters.
(2) The longer side of the rectangle measures 3 centimeters.
(1) The shorter side of the rectangle measures 2 centimeters.
(2) The longer side of the rectangle measures 3 centimeters.
Hi,
In Geomtry I, question 3 the DS question is:
Question: The OA assumes sides as integers, otherwise it should be another answer. Why is assuming integers and not decimals?
Thank you,
In Geomtry I, question 3 the DS question is:
Question: The OA assumes sides as integers, otherwise it should be another answer. Why is assuming integers and not decimals?
Thank you,
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Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
01 Nov 2010, 06:02
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You don't need to assume sides as integers.
Statement (1) The shorter side of the rectangle is 2 which means the greater side will be more than 2. Let us say it is just a little more than 2, say 2.01.
The diagonal will be \(\sqrt{{2^2 + (2.01)^2}} = \sqrt{8.04}.\) This is definitely greater than \(\sqrt{6}\). So it doesn't matter what the greater side is, the diagonal will be greater than root 6. Statement (1) is sufficient.
Statement (2) The greater side of the rectangle is 3. The smaller side can be as small as possible, let's say a little more than 0. Still, the diagonal will be \(\sqrt{{3^2 + (0.01)^2}} = \sqrt{9.0001}.\) This is definitely greater than \(\sqrt{6}\). Therefore, statement (2) is sufficient.
Answer (D).
Statement (1) The shorter side of the rectangle is 2 which means the greater side will be more than 2. Let us say it is just a little more than 2, say 2.01.
The diagonal will be \(\sqrt{{2^2 + (2.01)^2}} = \sqrt{8.04}.\) This is definitely greater than \(\sqrt{6}\). So it doesn't matter what the greater side is, the diagonal will be greater than root 6. Statement (1) is sufficient.
Statement (2) The greater side of the rectangle is 3. The smaller side can be as small as possible, let's say a little more than 0. Still, the diagonal will be \(\sqrt{{3^2 + (0.01)^2}} = \sqrt{9.0001}.\) This is definitely greater than \(\sqrt{6}\). Therefore, statement (2) is sufficient.
Answer (D).
02 Nov 2018, 18:03
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vivaslluis
\(a \ge b > 0\,\,\,\,\,\left[ {{\rm{rectangle}}\,\,{\rm{dimensions}}} \right]\)
\({a^2} + {b^2}\,\,\mathop > \limits^? \,\,6\)
\(\left( 1 \right)\,\,a > b = 2\,\,\,\, \Rightarrow \,\,\,{a^2} + {b^2} > {2^2} + {2^2} = 8\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle\)
\(\left( 2 \right)\,\,3 = a > b > 0\,\,\,\, \Rightarrow \,\,\,{a^2} + {b^2} > {3^2} = 9\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle\)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
