Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 23 May 2017, 23:12

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If AB/7 > 1/14 and A = B, which of the following must be gre

Author Message
TAGS:

### Hide Tags

Intern
Joined: 17 Jan 2012
Posts: 29
Location: India
GMAT 1: 650 Q48 V31
WE: Information Technology (Telecommunications)
Followers: 0

Kudos [?]: 16 [1] , given: 13

If AB/7 > 1/14 and A = B, which of the following must be gre [#permalink]

### Show Tags

19 Feb 2014, 02:10
1
KUDOS
3
This post was
BOOKMARKED
00:00

Difficulty:

35% (medium)

Question Stats:

69% (02:06) correct 31% (01:15) wrong based on 184 sessions

### HideShow timer Statistics

If AB/7 > 1/14 and A = B, which of the following must be greater than 1?

A. A+B
B. 1-A
C. 2A^2
D. A^2 - 1/2
E. A

[Reveal] Spoiler:
OA is C but It seems A also satisfies this condition...

I calculated it as below;
AB/7 > 1/14
multiplying both sides by 14
2AB > 1
as A=B
2A^2 >1 and 2 B^2 > 1
Option (C) is true as per this logic
While analyzing other options this is what I got...
2A^2>1
A^2>1/2
A>1/sqrt(2)
as A=B, B >1/sqrt(2)

so A+B > 1/sqrt(2) + 1/sqrt(2)
A+B > 2/sqrt(2)
A+B > sqrt(2)
This is also going to be greater than 1
[Reveal] Spoiler: OA

Last edited by Bunuel on 19 Feb 2014, 03:52, edited 1 time in total.
Renamed the topic and edited the question.
Math Expert
Joined: 02 Sep 2009
Posts: 38846
Followers: 7720

Kudos [?]: 105948 [1] , given: 11602

### Show Tags

19 Feb 2014, 03:51
1
KUDOS
Expert's post
2
This post was
BOOKMARKED
This problem is from MGMAT Algebra Strategy Guide:

If AB/7 > 1/14 and A=B While of the following must be greater than 1 ?
a) A+B
b) 1-A
c) 2A^2
d) A^2-1/2
e) A

OA is C but It seems A also satisfies this condition...

I calculated it as below;
AB/7 > 1/14
multiplying both sides by 14
2AB > 1
as A=B
2A^2 >1 and 2 B^2 > 1
Option (C) is true as per this logic
While analyzing other options this is what I got...
2A^2>1
A^2>1/2
A>1/sqrt(2)
as A=B, B >1/sqrt(2)

so A+B > 1/sqrt(2) + 1/sqrt(2)
A+B > 2/sqrt(2)
A+B > sqrt(2)
This is also going to be greater than 1

If AB/7 > 1/14 and A = B, which of the following must be greater than 1?

A. A+B
B. 1-A
C. 2A^2
D. A^2 - 1/2
E. A

First of all notice that the question asks which of the following MUST be true, not COULD be true.

$$\frac{AB}{7} > \frac{1}{14}$$ --> $$2AB>1$$. Since A = B, then $$2A^2>1$$.

As for option A: notice that from $$A^2>\frac{1}{2}$$ it follows that A, and therefore B, since A = B, can be negative numbers, for example, -1, and in this case A + B = -2 < 1. The problem with your solution is that $$A^2>\frac{1}{2}$$ means that $$A>\frac{1}{\sqrt{2}}$$ OR $$A<-\frac{1}{\sqrt{2}}$$, not only $$A>\frac{1}{\sqrt{2}}$$.

Hope it's clear.

_________________
Intern
Joined: 17 Jan 2012
Posts: 29
Location: India
GMAT 1: 650 Q48 V31
WE: Information Technology (Telecommunications)
Followers: 0

Kudos [?]: 16 [0], given: 13

### Show Tags

19 Feb 2014, 05:01
Thanks Bunuel, Your explanation is awesome as usual.

But I am a bit confused about below part.
Bunuel wrote:
The problem with your solution is that $$A^2>\frac{1}{2}$$ means that $$A>\frac{1}{\sqrt{2}}$$ OR $$A<-\frac{1}{\sqrt{2}}$$, not only $$A>\frac{1}{\sqrt{2}}$$.
.

As per my understanding
$$A^2>\frac{1}{2}$$ means that $$A>+\frac{1}{\sqrt{2}}$$ OR $$A>-\frac{1}{\sqrt{2}}$$. Could you please elaborate why $$A<-\frac{1}{\sqrt{2}}$$ ?
Math Expert
Joined: 02 Sep 2009
Posts: 38846
Followers: 7720

Kudos [?]: 105948 [0], given: 11602

### Show Tags

19 Feb 2014, 05:06
Thanks Bunuel, Your explanation is awesome as usual.

But I am a bit confused about below part.
Bunuel wrote:
The problem with your solution is that $$A^2>\frac{1}{2}$$ means that $$A>\frac{1}{\sqrt{2}}$$ OR $$A<-\frac{1}{\sqrt{2}}$$, not only $$A>\frac{1}{\sqrt{2}}$$.
.

As per my understanding
$$A^2>\frac{1}{2}$$ means that $$A>+\frac{1}{\sqrt{2}}$$ OR $$A>-\frac{1}{\sqrt{2}}$$. Could you please elaborate why $$A<-\frac{1}{\sqrt{2}}$$ ?

Let me ask you a question: does $$x^2>4$$ mean $$x>2$$ or $$x>-2$$? What does $$x>2$$ or $$x>-2$$ even mean?

$$x^2>4$$ --> $$|x|>2$$ --> $$x>2$$ or $$x<-2$$.

Go through the links below to brush up fundamentals on inequalities:

Hope this helps.
_________________
Intern
Joined: 17 Jan 2012
Posts: 29
Location: India
GMAT 1: 650 Q48 V31
WE: Information Technology (Telecommunications)
Followers: 0

Kudos [?]: 16 [0], given: 13

### Show Tags

19 Feb 2014, 07:56
Bunuel wrote:

Let me ask you a question: does $$x^2>4$$ mean $$x>2$$ or $$x>-2$$? What does $$x>2$$ or $$x>-2$$ even mean?

$$x^2>4$$ --> $$|x|>2$$ --> $$x>2$$ or $$x<-2$$.

This conclusion is based on below logic.
$$x^2=25$$ has TWO solutions, +5 and -5 so I thought $$x^2>4$$ would also have two solutions +2 & -2, this is how I got $$x>2$$ or $$x>-2$$
Could you please let me know why this is not valid ?

I am asking very basic question but I know only after getting my basics clear, I will be able to get good score.... Thanks for bearing with me and my stupid questions .
Math Expert
Joined: 02 Sep 2009
Posts: 38846
Followers: 7720

Kudos [?]: 105948 [0], given: 11602

### Show Tags

19 Feb 2014, 08:03
Bunuel wrote:

Let me ask you a question: does $$x^2>4$$ mean $$x>2$$ or $$x>-2$$? What does $$x>2$$ or $$x>-2$$ even mean?

$$x^2>4$$ --> $$|x|>2$$ --> $$x>2$$ or $$x<-2$$.

This conclusion is based on below logic.
$$x^2=25$$ has TWO solutions, +5 and -5 so I thought $$x^2>4$$ would also have two solutions +2 & -2, this is how I got $$x>2$$ or $$x>-2$$
Could you please let me know why this is not valid ?

I am asking very basic question but I know only after getting my basics clear, I will be able to get good score.... Thanks for bearing with me and my stupid questions .

Again, what does x is more than 2 or x is more than -2 mean? What are the possible values of x in this case? For example, can x be 1, since it's more than -2?

$$x^2>4$$ indeed has two ranges, which are $$x>2$$ or $$x<-2$$. Please follow the links in my previous post for more.
_________________
Intern
Joined: 05 Nov 2014
Posts: 2
Followers: 0

Kudos [?]: 0 [0], given: 0

If AB/7 > 1/14 and A = B, which of the following must be gre [#permalink]

### Show Tags

15 Nov 2014, 10:06
I don't understand, can someone please explain what I've done wrong?

$$\frac{AB}{7} > \frac{1}{14}$$

$$AB > \frac{7}{14}$$

$$AB > \frac{1}{2}$$

$$A^2 > \frac{1}{2}$$

$$\sqrt{A^2} > \sqrt{\frac{1}{2}}$$

$$|A| > \sqrt{\frac{1}{2}}$$

so it's either

$$A > \sqrt{\frac{1}{2}}$$

or

$$A < -(\sqrt{\frac{1}{2}})$$

following this logic i find that answer C does not necessarily produce a number greater than 1, for example:

A=-1/2

then:

2*($$\frac{-1}{2}^2$$) = 2*($$\frac{1}{4}$$) = $$\frac{1}{2}$$

Current Student
Joined: 14 Nov 2014
Posts: 108
Location: United States
GMAT 1: 740 Q49 V41
GPA: 3.34
WE: General Management (Aerospace and Defense)
Followers: 1

Kudos [?]: 30 [0], given: 0

If AB/7 > 1/14 and A = B, which of the following must be gre [#permalink]

### Show Tags

15 Nov 2014, 10:49
pietropietro wrote:
I don't understand, can someone please explain what I've done wrong?

$$\frac{AB}{7} > \frac{1}{14}$$

$$AB > \frac{7}{14}$$

$$AB > \frac{1}{2}$$

$$A^2 > \frac{1}{2}$$

$$\sqrt{A^2} > \sqrt{\frac{1}{2}}$$

$$|A| > \sqrt{\frac{1}{2}}$$

so it's either

$$A > \sqrt{\frac{1}{2}}$$

or

$$A < -(\sqrt{\frac{1}{2}})$$

following this logic i find that answer C does not necessarily produce a number greater than 1, for example:

A=-1/2

then:

2*($$\frac{-1}{2}^2$$) = 2*($$\frac{1}{4}$$) = $$\frac{1}{2}$$

You picked A = -1/2. Is -1/2 less than -sqrt(1/2)? In addition, does A = -1/2 satisfy the original condition?
SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1857
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Followers: 51

Kudos [?]: 2164 [0], given: 193

Re: If AB/7 > 1/14 and A = B, which of the following must be gre [#permalink]

### Show Tags

20 Nov 2014, 02:08
pietropietro wrote:
I don't understand, can someone please explain what I've done wrong?

$$\frac{AB}{7} > \frac{1}{14}$$

$$AB > \frac{7}{14}$$

$$AB > \frac{1}{2}$$

$$A^2 > \frac{1}{2}$$

$$\sqrt{A^2} > \sqrt{\frac{1}{2}}$$

$$|A| > \sqrt{\frac{1}{2}}$$

so it's either

$$A > \sqrt{\frac{1}{2}}$$

or

$$A < -(\sqrt{\frac{1}{2}})$$

following this logic i find that answer C does not necessarily produce a number greater than 1, for example:

A=-1/2

then:

2*($$\frac{-1}{2}^2$$) = 2*($$\frac{1}{4}$$) = $$\frac{1}{2}$$

There is no need to go along in simplification of the equation. Upto the step highlighted should be fine.

Just place values in OA to get the answer
_________________

Kindly press "+1 Kudos" to appreciate

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15413
Followers: 649

Kudos [?]: 206 [0], given: 0

Re: If AB/7 > 1/14 and A = B, which of the following must be gre [#permalink]

### Show Tags

03 Apr 2017, 00:13
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Manager
Joined: 12 Nov 2016
Posts: 191
Followers: 0

Kudos [?]: 7 [0], given: 135

Re: If AB/7 > 1/14 and A = B, which of the following must be gre [#permalink]

### Show Tags

23 Apr 2017, 23:46
If AB/7 > 1/14 and A = B, which of the following must be greater than 1?

A. A+B
B. 1-A
C. 2A^2
D. A^2 - 1/2
E. A

[Reveal] Spoiler:
OA is C but It seems A also satisfies this condition...

I calculated it as below;
AB/7 > 1/14
multiplying both sides by 14
2AB > 1
as A=B
2A^2 >1 and 2 B^2 > 1
Option (C) is true as per this logic
While analyzing other options this is what I got...
2A^2>1
A^2>1/2
A>1/sqrt(2)
as A=B, B >1/sqrt(2)

so A+B > 1/sqrt(2) + 1/sqrt(2)
A+B > 2/sqrt(2)
A+B > sqrt(2)
This is also going to be greater than 1

The key to solving this problem is remembering to consider that although the product of A and B must be positive their individuals values can be negative- a GMAT trap
Re: If AB/7 > 1/14 and A = B, which of the following must be gre   [#permalink] 23 Apr 2017, 23:46
Similar topics Replies Last post
Similar
Topics:
2 If a > b > 1, which of the following must be less than 1 but more than 1 04 May 2017, 22:53
4 Which of the following must be true => 3 11 Aug 2016, 05:01
4 If a > 0, b > 0, and (a^2)b = a, then which of the following MUST equa 4 24 Feb 2016, 03:14
8 If a > b and if c > d , then which of the following must be 7 03 Oct 2016, 02:34
14 If |a| > |b|, which of the following must be true? 6 06 Mar 2017, 18:43
Display posts from previous: Sort by