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05 Mar 2019, 12:53
Kudos
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
72% (01:43) correct
28%
(01:56)
wrong
based on 173
sessions
History
Date
Time
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Not Attempted Yet
GMATH practice exercise (Quant Class 14)
Is \(xy >3\) ?
(1) \(7^x > 729\)
(2) \(9^y = 7\)
P.S.: this IS in GMAT´s quant section scope.
Is \(xy >3\) ?
(1) \(7^x > 729\)
(2) \(9^y = 7\)
P.S.: this IS in GMAT´s quant section scope.
05 Mar 2019, 14:22
Kudos
Bookmarks
fskilnik
\(xy\,\,\mathop > \limits^? \,\,3\)
\(\left( 1 \right)\,\,{7^x} > 729\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {4,0} \right)\,\,\,\,\,\left[ {{7^4} = {{49}^2}} \right]\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr \\
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {4,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\)
\(\left( 2 \right)\,\,{9^y} = 7\,\,\,\, \Rightarrow \,\,\,y = {y_p}\,\,\,{\rm{unique}}\,\,{\rm{,}}\,\,\,{1 \over 2}\,\,{\rm{ < }}\,\,{y_p} < 1\,\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {0,{y_p}} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr \\
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {6,{y_p}} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\)
\(\left( {1 + 2} \right)\,\,\,{3^6} = 729\,\,\,\mathop < \limits^{\left( 1 \right)} \,\,\,{7^x}\,\,\mathop = \limits^{\left( 2 \right)} \,\,\,{\left( {{9^y}} \right)^x} = {3^{2xy}}\,\,\,\,\,\mathop \Rightarrow \limits^{3\,\, > \,\,1} \,\,\,2xy > 6\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle\)
The correct answer is (C).
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
General Discussion
fskilnik
If YOU FIND MY SOLUTION HELPFUL, PLEASE GIVE ME KUDOS
This question is testing, among other things, our ability to estimate
(1) Notice, that powers of 7 give us 7, 49, 353, 2479 Now 729? 353, therefore 3<x<4, (although we should estimate that x is closer to 3 than 4). Lets say x = a little more than 3. Y can still = ANYTHING NS
(2) 9^Y = 7. Notice that 9^0 =1 and 9^1 =9,therefore y is a little less than 1, however X can = ANYTHING NS
(1) and (2) x is between 3 and 4, and y is a little less than 1. Assuming the numbers picked for estimation are accurate enough (y is very close to 3.4), we will get an unequivocal YES as 3.4(.9) = 3.06> 3
The problem I have with this question, is it requires us to estimate exponential relationships to a level of precision that is unrealistic on this test. For example, what if the test taker estimated x to be 3.1, and y to be .9. That would give us xy> approx 2.6, which could give the reader a potential YES and NO.
Overall this is a high quality question as far as the skills the question tests, however.
