If AB/7 > 1/14 and A = B, which of the following must be gre : GMAT Problem Solving (PS)
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# If AB/7 > 1/14 and A = B, which of the following must be gre

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

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19 Feb 2014, 01:10
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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, 02:52, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Inequalities MGMAT [#permalink]

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19 Feb 2014, 02:51
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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.

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.
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Re: Inequalities MGMAT [#permalink]

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19 Feb 2014, 04:01
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 $$A<-\frac{1}{\sqrt{2}}$$ ?
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Re: Inequalities MGMAT [#permalink]

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19 Feb 2014, 04:06
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.
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Re: Inequalities MGMAT [#permalink]

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19 Feb 2014, 06:56
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 .
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Re: Inequalities MGMAT [#permalink]

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19 Feb 2014, 07:03
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 .

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

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15 Nov 2014, 09:06
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*($$\frac{1}{4}$$) = $$\frac{1}{2}$$

What have I done wrong? thank you in advance for your answer
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If AB/7 > 1/14 and A = B, which of the following must be gre [#permalink]

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15 Nov 2014, 09:49
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*($$\frac{1}{4}$$) = $$\frac{1}{2}$$

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

You picked A = -1/2. Is -1/2 less than -sqrt(1/2)? In addition, does A = -1/2 satisfy the original condition?
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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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20 Nov 2014, 01:08
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*($$\frac{1}{4}$$) = $$\frac{1}{2}$$

What have I done wrong? thank you in advance for your 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
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Re: If AB/7 > 1/14 and A = B, which of the following must be gre   [#permalink] 20 Nov 2014, 01:08
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