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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

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 \(2A^2>1\).

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 \(A>\frac{1}{\sqrt{2}}\).

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

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:

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 .

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 .

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

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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

gmatclubot

Re: If AB/7 > 1/14 and A = B, which of the following must be gre
[#permalink]
23 Apr 2017, 23:46

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