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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.

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Re: Is xy>3?  [#permalink]

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fskilnik wrote:
GMATH practice exercise (Quant Class 14)

Is $$xy >3$$ ?

(1) $$7^x > 729$$
(2) $$9^y = 7$$

$$xy\,\,\mathop > \limits^? \,\,3$$

$$\left( 1 \right)\,\,{7^x} > 729\,\,\,\left\{ \matrix{ \,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {4,0} \right)\,\,\,\,\,\left[ {{7^4} = {{49}^2}} \right]\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr \,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {4,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.$$

$$\left( 2 \right)\,\,{9^y} = 7\,\,\,\, \Rightarrow \,\,\,y = {y_p}\,\,\,{\rm{unique}}\,\,{\rm{,}}\,\,\,{1 \over 2}\,\,{\rm{ < }}\,\,{y_p} < 1\,\,\,\,\left\{ \matrix{ \,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {0,{y_p}} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr \,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {6,{y_p}} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.$$

$$\left( {1 + 2} \right)\,\,\,{3^6} = 729\,\,\,\mathop < \limits^{\left( 1 \right)} \,\,\,{7^x}\,\,\mathop = \limits^{\left( 2 \right)} \,\,\,{\left( {{9^y}} \right)^x} = {3^{2xy}}\,\,\,\,\,\mathop \Rightarrow \limits^{3\,\, > \,\,1} \,\,\,2xy > 6\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle$$

The correct answer is (C).

We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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fskilnik wrote:
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.

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.

Originally posted by ocelot22 on 05 Mar 2019, 13:23.
Last edited by ocelot22 on 05 Mar 2019, 14:12, edited 3 times in total.
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Re: Is xy>3?  [#permalink]

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ocelot22 wrote:
(1) and (2) x>3 and 0<y<1. Then 0<XY<3, which gives us a definite NO

Hi ocelot22 !

Thanks for joining!

The two inequalities you mentioned (repeated above) are enough for your conclusion (in red)?

Another thing: I guarantee (1+2) is enough for a definite YES!!

My solution is short but very instructive. Think a bit more before I present it!

Regards,
Fabio.
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Re: Is xy>3?  [#permalink]

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fskilnik wrote:
ocelot22 wrote:
(1) and (2) x>3 and 0<y<1. Then 0<XY<3, which gives us a definite NO

Hi ocelot22 !

Thanks for joining!

The two inequalities you mentioned (repeated above) are enough for your conclusion (in red)?

Another thing: I guarantee (1+2) is enough for a definite YES!!

My solution is short but very instructive. Think a bit more before I present it!

Regards,
Fabio.

You are right. I use potentially inappropriate interval techniques for this problem, and have corrected my mistakes from before. Can you please however, look over my edited post, which points out some concerns I have pointed out about this problem
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Re: Is xy>3?  [#permalink]

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ocelot22 wrote:
You are right. I use potentially inappropriate interval techniques for this problem, and have corrected my mistakes from before. Can you please however, look over my edited post, which points out some concerns I have pointed out about this problem

I am glad your interest in my problem continues, ocelot22 .

ocelot22 wrote:
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 would be A NO answer to the question.

This would be true only if you had no choice but to insist on your approach... higher-level problems are harder (also) because a proper way of dealing with them are not always seen at first!

Regards,
Fabio.
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Re: Is xy>3?  [#permalink]

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ocelot22 wrote:
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.

Thanks for the comment in blue. I guess you will find the question much more interesting after analysing my solution. I will post it below right now!
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fskilnik wrote:
fskilnik wrote:
GMATH practice exercise (Quant Class 14)

Is $$xy >3$$ ?

(1) $$7^x > 729$$
(2) $$9^y = 7$$

$$xy\,\,\mathop > \limits^? \,\,3$$

$$\left( 1 \right)\,\,{7^x} > 729\,\,\,\left\{ \matrix{ \,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {4,0} \right)\,\,\,\,\,\left[ {{7^4} = {{49}^2}} \right]\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr \,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {4,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.$$

$$\left( 2 \right)\,\,{9^y} = 7\,\,\,\, \Rightarrow \,\,\,y = {y_p}\,\,\,{\rm{unique}}\,\,{\rm{,}}\,\,\,{1 \over 2}\,\,{\rm{ < }}\,\,{y_p} < 1\,\,\,\,\left\{ \matrix{ \,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {0,{y_p}} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr \,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {6,{y_p}} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.$$

$$\left( {1 + 2} \right)\,\,\,{3^6} = 729\,\,\,\mathop < \limits^{\left( 1 \right)} \,\,\,{7^x}\,\,\mathop = \limits^{\left( 2 \right)} \,\,\,{\left( {{9^y}} \right)^x} = {3^{2xy}}\,\,\,\,\,\mathop \Rightarrow \limits^{3\,\, > \,\,1} \,\,\,2xy > 6\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle$$

The correct answer is (C).

We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.

This is a very succinct solution. I Appreciate the compact notation used. This of course avoids the estimation pitfall that I ran into in my solution. I am guessing that you teach this kind of notation in your course?
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fskilnik wrote:
Is $$xy >3$$ ?

(1) $$7^x > 729$$
(2) $$9^y = 7$$

Target question: Is xy > 3 ?

Statement 1: (7^x) > 729
Since there's no information about y, we cannot answer the target question with certainty.
Statement 1 is NOT SUFFICIENT

Statement 2: (9^y) = 7
Since there's no information about x, we cannot answer the target question with certainty.
Statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that (7^x) > 729
Statement 2 tells us that (9^y) = 7

Take the inequality (7^x) > 729, and replace 7 with 9^y to get: (9^y)^x > 729
Simplify to get: 9^xy > 729
Rewrite 729 as 9^3 to get: 9^xy > 9^3
From this, we can conclude that xy > 3
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Cheers,
Brent
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Re: Is xy>3?  [#permalink]

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ocelot22 wrote:
This is a very succinct solution. I Appreciate the compact notation used. This of course avoids the estimation pitfall that I ran into in my solution. I am guessing that you teach this kind of notation in your course?
Hi, ocelot22 !

First of all, thanks for the kudos (both in the question stem and also in my solution)!

I am glad you liked our notation/solution.

YES, the precision and brevity of our exclusive notation is an IMPORTANT part of our method, especially in Data Sufficiency! I will not come into details here, due to respect for all companies, teachers, and students who have their own (probably different) opinions on the matter (or never thought about it).

In our "test drive" you will earn two credits for questions. Feel free to use one of them to ask me about this at your free trial! (*)

Regards,
Fabio.

(*) P.S.: although the number of question credits is limited per student (to avoid someone asking me, say, 10 questions a day "to avoid" thinking by himself/herself beforehand), I usually reimburse each credit used. (I am the sole creator of the method and the only person who answers questions in our preparation.)
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fskilnik wrote:
fskilnik wrote:
GMATH practice exercise (Quant Class 14)

Is $$xy >3$$ ?

(1) $$7^x > 729$$
(2) $$9^y = 7$$

$$xy\,\,\mathop > \limits^? \,\,3$$

$$\left( 1 \right)\,\,{7^x} > 729\,\,\,\left\{ \matrix{ \,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {4,0} \right)\,\,\,\,\,\left[ {{7^4} = {{49}^2}} \right]\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr \,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {4,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.$$

$$\left( 2 \right)\,\,{9^y} = 7\,\,\,\, \Rightarrow \,\,\,y = {y_p}\,\,\,{\rm{unique}}\,\,{\rm{,}}\,\,\,{1 \over 2}\,\,{\rm{ < }}\,\,{y_p} < 1\,\,\,\,\left\{ \matrix{ \,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {0,{y_p}} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr \,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {6,{y_p}} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.$$

$$\left( {1 + 2} \right)\,\,\,{3^6} = 729\,\,\,\mathop < \limits^{\left( 1 \right)} \,\,\,{7^x}\,\,\mathop = \limits^{\left( 2 \right)} \,\,\,{\left( {{9^y}} \right)^x} = {3^{2xy}}\,\,\,\,\,\mathop \Rightarrow \limits^{3\,\, > \,\,1} \,\,\,2xy > 6\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle$$

The correct answer is (C).

We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.

fskilnik,
Sir, do you want to mean 2>1 (talking about statement) in the red part?
Regards,
Attachments solution.PNG [ 57.29 KiB | Viewed 716 times ]

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Re: Is xy>3?  [#permalink]

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fskilnik,
Sir, do you want to mean 2>1 (talking about statement) in the red part?
Regards,

Sorry for the delay (very busy)!

Thank you for your REAL interest in my solution. Answering your (important) question: I said $${3^6} < {3^{2xy}}\,\,\,\mathop \Rightarrow \limits^{3\, > \,1} \,\,\,6 < 2xy$$

Meaning: from the fact that the exponential function $$y = 3^x$$ has a base that is greater than 1 (=3), it is an (strictly) increasing function:

$$x < y\,\,\,\, \Rightarrow \,\,\,{3^x} < {3^y}$$

From this fact, please note that: $$6 \ge 2xy\,\,\,\mathop \Rightarrow \limits^{3\, > \,1} \,\,\,{3^6} \ge {3^{2xy}}\,\,\,\left( {{\rm{impossible,}}\,\,{\rm{because}}\,\,{3^6} < {3^{2xy}}} \right)$$

That´s (finally!) the reason for the validity of the implication $${3^6} < {3^{2xy}}\,\,\,\mathop \Rightarrow \limits^{3\, > \,1} \,\,\,6 < 2xy$$ ...

Best Regards,
Fabio.
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Re: Is xy>3?  [#permalink]

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Thank you so much...
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Re: Is xy>3?  [#permalink]

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Thank you so much...

If you (and other readers) liked our approach and explanations, we invite you to try our test drive ("Free Trial"), so that you will have a MUCH better understanding of our METHOD, that is, our systematic (and very deep!) way of looking into the GMAT´s contents with all objectivity, subtleness and "venom" that characterizes high-level performances in the quant section of the test.

We create and post new questions there (such as this one) on an almost daily basis, by the way.

Regards and success in your studies,
Fabio.
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Re: Is xy>3?  [#permalink]

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fskilnik wrote:
Thank you so much...

If you (and other readers) liked our approach and explanations, we invite you to try our test drive ("Free Trial"), so that you will have a MUCH better understanding of our METHOD, that is, our systematic (and very deep!) way of looking into the GMAT´s contents with all objectivity, subtleness and "venom" that characterizes high-level performances in the quant section of the test.

We create and post new questions there (such as this one) on an almost daily basis, by the way.

Regards and success in your studies,
Fabio.

I'm going to try ""Free Trial"" from tomorrow. Thanks__
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Re: Is xy>3?  [#permalink]

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GMATPrepNow wrote:
fskilnik wrote:
Is $$xy >3$$ ?

(1) $$7^x > 729$$
(2) $$9^y = 7$$

Target question: Is xy > 3 ?

Statement 1: (7^x) > 729
Since there's no information about y, we cannot answer the target question with certainty.
Statement 1 is NOT SUFFICIENT

Statement 2: (9^y) = 7
Since there's no information about x, we cannot answer the target question with certainty.
Statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that (7^x) > 729
Statement 2 tells us that (9^y) = 7

Take the inequality (7^x) > 729, and replace 7 with 9^y to get: (9^y)^x > 729
Simplify to get: 9^xy > 729
Rewrite 729 as 9^3 to get: 9^xy > 9^3
From this, we can conclude that xy > 3
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Cheers,
Brent

Hi Brent,

Could you please clarify if it is always possible to replace one part of the inequality completely, just like you are doing in your solution

Quote:
Take the inequality (7^x) > 729, and replace 7 with 9^y to get: (9^y)^x > 729 Re: Is xy>3?   [#permalink] 20 May 2019, 14:01
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