GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 19 Oct 2019, 03:06 GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  If ab >0, is (ab)^2 < (ab)^(1/2) ?

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

Hide Tags

Senior Manager  V
Joined: 25 Sep 2018
Posts: 426
Location: United States (CA)
Concentration: Finance, Strategy
GMAT 1: 640 Q47 V30 GPA: 3.97
WE: Investment Banking (Investment Banking)
If ab >0, is (ab)^2 < (ab)^(1/2) ?  [#permalink]

Show Tags

4 00:00

Difficulty:

(N/A)

Question Stats: 26% (02:09) correct 74% (02:08) wrong based on 84 sessions

HideShow timer Statistics

If $$ab >0$$, is $$(ab) ^2<\sqrt{ab}$$

(1) $$\frac{4}{a}>7b$$

(2) $$a-16 > -16$$

_________________
Why do we fall?...So we can learn to pick ourselves up again

Originally posted by Abhi077 on 15 Feb 2019, 10:39.
Last edited by Bunuel on 16 Feb 2019, 03:18, edited 2 times in total.
Renamed the topic and edited the question.
GMAT Tutor G
Joined: 24 Jun 2008
Posts: 1806
Re: If ab >0, is (ab)^2 < (ab)^(1/2) ?  [#permalink]

Show Tags

Abhi077 wrote:
If $$ab >0$$, is $$(ab) ^2$$ < $$\sqrt{ab}$$

1.$$\frac{4}{a}$$ > $$7ab$$
2. $$a-16 > -16$$

edit: the OP fixed a typo in the original post above, so this solution is to the question I have quoted above (with the typo in it) and not to the now-edited question at the top of the page (which I solve in a separate post below)

First, if we replace "ab" with "x", the question is asking:

Is x^2 <√x ?

and x^2 < √x is only true if 0 < x < 1, so the question is asking "Is ab < 1?"

Statement 1 tells us:

4/a > 7ab

Before determining if this is sufficient, notice that we know ab > 0. So the right side of this inequality is positive, which means 4/a is greater than some positive number, and therefore a itself is positive. So Statement 1 guarantees that a is positive. But that's the only information you get from Statement 2. So you don't learn anything new by combining the two statements that you don't know from Statement 1 on its own, and the OA provided cannot be correct -- the answer can only be A or E, since Statement 2 alone is clearly not sufficient.

Using Statement 1 alone (which we can rewrite, since a is positive, as 4 > 7a^2 b or as 4/7 > a^2 b if we want to), it's easy to see that ab < 1 is possible, using the values a = 0.1 and b = 1, say. But it's also easy enough to find values where ab > 1. We can take, say, a = 0.1 and b = 11.

So we can't say if ab < 1 is true, and the answer is E unless there's a typo in the question.
_________________
GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Originally posted by IanStewart on 15 Feb 2019, 14:19.
Last edited by IanStewart on 15 Feb 2019, 22:06, edited 1 time in total.
Senior Manager  V
Joined: 25 Sep 2018
Posts: 426
Location: United States (CA)
Concentration: Finance, Strategy
GMAT 1: 640 Q47 V30 GPA: 3.97
WE: Investment Banking (Investment Banking)
If ab >0, is (ab)^2 < (ab)^(1/2) ?  [#permalink]

Show Tags

IanStewart wrote:
Abhi077 wrote:
If $$ab >0$$, is $$(ab) ^2$$ < $$\sqrt{ab}$$

1.$$\frac{4}{a}$$ > $$7ab$$
2. $$a-16 > -16$$

First, if we replace "ab" with "x", the question is asking:

Is x^2 <√x ?

and x^2 < √x is only true if 0 < x < 1, so the question is asking "Is ab < 1?"

Statement 1 tells us:

4/a > 7ab

Before determining if this is sufficient, notice that we know ab > 0. So the right side of this inequality is positive, which means 4/a is greater than some positive number, and therefore a itself is positive. So Statement 1 guarantees that a is positive. But that's the only information you get from Statement 2. So you don't learn anything new by combining the two statements that you don't know from Statement 1 on its own, and the OA provided cannot be correct -- the answer can only be A or E, since Statement 2 alone is clearly not sufficient.

Using Statement 1 alone (which we can rewrite, since a is positive, as 4 > 7a^2 b or as 4/7 > a^2 b if we want to), it's easy to see that ab < 1 is possible, using the values a = 0.1 and b = 1, say. But it's also easy enough to find values where ab > 1. We can take, say, a = 0.1 and b = 11.

So we can't say if ab < 1 is true, and the answer is E unless there's a typo in the question.

Yes there was a typo. I'm so sorry. just corrected it
_________________
Why do we fall?...So we can learn to pick ourselves up again

Originally posted by Abhi077 on 15 Feb 2019, 21:17.
Last edited by Abhi077 on 18 Feb 2019, 03:38, edited 1 time in total.
GMAT Tutor G
Joined: 24 Jun 2008
Posts: 1806
Re: If ab >0, is (ab)^2 < (ab)^(1/2) ?  [#permalink]

Show Tags

Abhi077 wrote:
If $$ab >0$$, is $$(ab) ^2$$ < $$\sqrt{ab}$$

1.$$\frac{4}{a}$$ > $$7b$$
2. $$a-16 > -16$$

With the correction, it's much easier to analyze the question algebraically. As I explained above, the only way (ab)^2 < √(ab) is true is if 0 < ab < 1, so that's what we want to know: is ab < 1?

Using only Statement 1, we have two cases:

- if a > 0, then we can multiply both sides of the inequality by a without reversing the inequality. So we then have

4/a > 7b
4 > 7ab
4/7 > ab

So in this case, it is indeed true that ab < 1.

- if a < 0, then we can again multiply both sides of the inequality by a, but because we're multiplying by a negative number, we must reverse the inequality:

4/a > 7b
4 < 7ab
4/7 < ab

So when a (and also b of course, since ab > 0) is negative, then the value of ab can be anything greater than 4/7, and can indeed be larger than 1. It is easy enough to confirm this just by choosing numbers: we might have, say, that a = -4 and b = -100, for example, in which case 4/a > 7b is clearly true, since -1 > -700, and in this case ab is much larger than 1.

Statement 2 alone is clearly not sufficient, since it only tells us a > 0. But once we know that a > 0, that rules out the second case above in our analysis of Statement 1. So we know only the first case is possible, and that ab < 4/7, so the two statements together are sufficient, and the answer is C.
_________________
GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com
Director  G
Joined: 09 Mar 2018
Posts: 994
Location: India
Re: If ab >0, is (ab)^2 < (ab)^(1/2) ?  [#permalink]

Show Tags

Abhi077 wrote:
If $$ab >0$$, is $$(ab) ^2$$ < $$\sqrt{ab}$$

1.$$\frac{4}{a}$$ > $$7b$$
2. $$a-16 > -16$$

So i would say, nice , Lets see, given is important here ab> 0

Lets manipulate $$(ab)^4$$ < ab

take common and it becomes => (ab) {$$(ab)^3$$ -1 } < 0,For representation purpose only (x) (y) < 0

Now for the above to be true, it can be that x > 0 & y< 0 or x < 0 & y > 0, So from given, only bold is valid

which means we have to only focus on (ab)^3 < 1 part

from 1) $$\frac{4}{a}$$ > $$7b$$

Can be true when a=-1, b=-1, this makes the question as No
Can be true when a=1, b=-1, this makes the question as Yes

From 2) a-16 > -16, this can only be true when a> 0, dont know anything about B

When we combine we can directly pinpoint on the condition that

a > 0 and b < 0

Sufficient to answer the question

C
_________________
If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.
Director  P
Joined: 24 Nov 2016
Posts: 611
Location: United States
Re: If ab >0, is (ab)^2 < (ab)^(1/2) ?  [#permalink]

Show Tags

Abhi077 wrote:
If $$ab >0$$, is $$(ab) ^2<\sqrt{ab}$$

(1) $$\frac{4}{a}>7b$$
(2) $$a-16 > -16$$

$$ab>0…{{a,b}}=same.sign≠0$$
$$(ab) ^2<\sqrt{ab}…ab=x…x^2<\sqrt{x}$$?
rearrange: $$x^2<\sqrt{x}…x^4<x…x(x^3-1)<0$$?
if $$x<0$$ is $$(x^3-1)>0…x^3>1…x>1…ab>1$$; invalid, since $$x<0$$;
if $$x>0$$ is $$(x^3-1)<0…x^3<1…x<1…ab<1$$? find this.

(1) $$\frac{4}{a}>7b$$:
if $${{a,b}}=positive…\frac{4}{a}>7b…4/7>ab…ab<1$$
if $${{a,b}}=negative…\frac{4}{-a}>7•(-b)…4/7<ab…ab>1$$

(2) $$a-16 > -16$$: $$a>0…b>0$$ but we don't know if $$ab<1$$, insufi.

(1&2): $$a>0$$ then $${{a,b}}=positive…\frac{4}{a}>7b…4/7>ab…ab<1$$; sufic. Re: If ab >0, is (ab)^2 < (ab)^(1/2) ?   [#permalink] 31 Aug 2019, 09:32
Display posts from previous: Sort by

If ab >0, is (ab)^2 < (ab)^(1/2) ?

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  