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

Question

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

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

Bunuel · Accepted Answer

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 . Answer: C. 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. P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to rules 3, 5, and 7. Also, please hide the OA under the spoiler. Thank you.

Conquistador22 · Answer

Thanks Bunuel, Your explanation is awesome as usual.

But I am a bit confused about below part. 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}} ?

Bunuel · Answer

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: Theory on Inequalities: x2-4x-94661.html#p731476 inequalities-trick-91482.html data-suff-inequalities-109078.html range-for-variable-x-in-a-given-inequality-109468.html everything-is-less-than-zero-108884.html graphic-approach-to-problems-with-inequalities-68037.html All DS Inequalities Problems to practice: search.php?search_id=tag&tag_id=184 All PS Inequalities Problems to practice: search.php?search_id=tag&tag_id=189 700+ Inequalities problems: inequality-and-absolute-value-questions-from-my-collection-86939.html Hope this helps.

Conquistador22 · Answer

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

.

Bunuel · Answer

Guess you did not follow the links I proposed...

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.

pietropietro · Answer

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}

What have I done wrong? thank you in advance for your answer

jwang27 · Answer

You picked A = -1/2. Is -1/2 less than -sqrt(1/2)? In addition, does A = -1/2 satisfy the original condition?

PareshGmat · Answer

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

Nunuboy1994 · Answer

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

Basshead · Answer

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

Since AB must be positive we can cross multiply:

14AB > 7

2AB > 1

2A^2 > 1

Answer is C.

pierjoejoe · Answer

we know that 
A*B/7 > 1/14
and A=B
meaning that 
A^2 > 1/2
and thus A<-1/radq(2) and A>1/radq(2)

A and B can be either both positive or both negative because A = B 
we can exclude A for this reason.
B can be excluded too because A can be positive
D can be excluded because A^2 is 1/2 if A is positive or negative, thus the value would be less than 1
E cannot be the answer, because A can be positive or negative

C is the only left. infact A^2 is positive, no matter what is the value of A. moreover A is only GREATER than 1/radq(2) , meaning that multiplying by 2 will yield to a value greater than 1 for sure.

bumpbot · Answer

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

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