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 Your Progress

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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

If 8x > 4 + 6x, what is the value of the integer x?

(1) 6 – 5x > -13

(2) 3 – 2x < -x + 4 < 7.2 – 2x

Could anybody tell me when it is appropriate to add up equations? That's what I did and it didn't work...

Given: \(x=integer\) and \(8x>4+6x\) --> \(2x>4\) --> \(x>2\). Question: \(x=?\)

(1) \(6-5x>-13\) --> \(19>5x\) --> \(\frac{19}{5}=3.8>x\) --> as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient.

(2) \(3-2x<-x+4<7.2-2x\) --> take only the following part: \(-x+4<7.2-2x\)--> \(x<3.2\) --> as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient.

Answer: D.

So no need to add inequality in this case. But if you are interested:

You can only add inequalities when their signs are in the same direction:

If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\). Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).

You can only apply subtraction when their signs are in the opposite directions:

If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

If 8x > 4 + 6x, what is the value of the integer x?

(1) 6 – 5x > -13

(2) 3 – 2x < -x + 4 < 7.2 – 2x

Could anybody tell me when it is appropriate to add up equations? That's what I did and it didn't work...

Given: \(x=integer\) and \(8x>4+6x\) --> \(2x>4\) --> \(x>2\). Question: \(x=?\)

(1) \(6-5x>-13\) --> \(19>5x\) --> \(\frac{19}{5}=3.8>x\) --> as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient.

(2) \(3-2x<-x+4<7.2-2x\) --> take only the following part: \(-x+4<7.2-2x\)--> \(x<3.2\) --> as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient.

Answer: D.

So no need to add inequality in this case. But if you are interested:

You can only add inequalities when their signs are in the same direction:

If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\). Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).

You can only apply subtraction when their signs are in the opposite directions:

If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

Hope it helps.

Hi,

can you explain stmt B please, why is that only the second part is considered?

Given: \(x=integer\) and \(8x>4+6x\) --> \(2x>4\) --> \(x>2\). Question: \(x=?\)

(1) \(6-5x>-13\) --> \(19>5x\) --> \(\frac{19}{5}=3.8>x\) --> as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient.

(2) \(3-2x<-x+4<7.2-2x\) --> take only the following part: \(-x+4<7.2-2x\)--> \(x<3.2\) --> as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient.

Answer: D.

Hi,

can you explain stmt B please, why is that only the second part is considered?

mission2009 wrote:

Hi Bunuel, why have to taken only the second equation in Statement B not the first one? Plz explain briefly.

Thanks..

Because to reach the answer we don't need the first part at all. The part which says \(-x+4<7.2-2x\) is enough to give necessary info: \(x<3.2\) --> as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient. _________________

If 8x > 4 + 6x, what is the value of the integer x?

(1) 6 – 5x > -13

(2) 3 – 2x < -x + 4 < 7.2 – 2x

Could anybody tell me when it is appropriate to add up equations? That's what I did and it didn't work...

Given: \(x=integer\) and \(8x>4+6x\) --> \(2x>4\) --> \(x>2\). Question: \(x=?\)

(1) \(6-5x>-13\) --> \(18>5x\) --> \(\frac{18}{5}=3.6>x\) --> as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient.

(2) \(3-2x<-x+4<7.2-2x\) --> take only the following part: \(-x+4<7.2-2x\)--> \(x<3.2\) --> as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient.

Answer: D.

So no need to add inequality in this case. But if you are interested:

You can only add inequalities when their signs are in the same direction:

If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\). Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).

You can only apply subtraction when their signs are in the opposite directions:

If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

Hope it helps.

Slightly off on statement 1, however the answer doesn't change.

(1) \(6-5x>-13\) --> \(19>5x\) --> \(\frac{19}{5}=3.8>x\) --> as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient.

If 8x > 4 + 6x, what is the value of the integer x?

(1) 6 – 5x > -13

(2) 3 – 2x < -x + 4 < 7.2 – 2x

Could anybody tell me when it is appropriate to add up equations? That's what I did and it didn't work...

Given: \(x=integer\) and \(8x>4+6x\) --> \(2x>4\) --> \(x>2\). Question: \(x=?\)

(1) \(6-5x>-13\) --> \(19>5x\) --> \(\frac{19}{5}=3.8>x\) --> as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient.

(2) \(3-2x<-x+4<7.2-2x\) --> take only the following part: \(-x+4<7.2-2x\)--> \(x<3.2\) --> as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient.

Answer: D.

So no need to add inequality in this case. But if you are interested:

You can only add inequalities when their signs are in the same direction:

If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\). Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).

You can only apply subtraction when their signs are in the opposite directions:

If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

Hope it helps.

One small doubt... As said X>2 and x>3.8 how can we say x=3 even it is integer.The 2nd one say X.3.8 so y can't we take x=4???? Thanks to help Thanks to help

Given: \(x=integer\) and \(8x>4+6x\) --> \(2x>4\) --> \(x>2\). Question: \(x=?\)

(1) \(6-5x>-13\) --> \(19>5x\) --> \(\frac{19}{5}=3.8>x\) --> as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient.

(2) \(3-2x<-x+4<7.2-2x\) --> take only the following part: \(-x+4<7.2-2x\)--> \(x<3.2\) --> as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient.

Answer: D.

Hi,

can you explain stmt B please, why is that only the second part is considered?

mission2009 wrote:

Hi Bunuel, why have to taken only the second equation in Statement B not the first one? Plz explain briefly.

Thanks..

Because to reach the answer we don't need the first part at all. The part which says \(-x+4<7.2-2x\) is enough to give necessary info: \(x<3.2\) --> as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient.

Hi Bunuel,

but if we consider the first part of the statement B we don't get the answer, right? if we have been given the equation with both the parts then we should be considering both the parts, right??? please explain...

Given: \(x=integer\) and \(8x>4+6x\) --> \(2x>4\) --> \(x>2\). Question: \(x=?\)

(1) \(6-5x>-13\) --> \(19>5x\) --> \(\frac{19}{5}=3.8>x\) --> as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient.

(2) \(3-2x<-x+4<7.2-2x\) --> take only the following part: \(-x+4<7.2-2x\)--> \(x<3.2\) --> as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient.

Answer: D.

Hi,

can you explain stmt B please, why is that only the second part is considered?

mission2009 wrote:

Hi Bunuel, why have to taken only the second equation in Statement B not the first one? Plz explain briefly.

Thanks..

Because to reach the answer we don't need the first part at all. The part which says \(-x+4<7.2-2x\) is enough to give necessary info: \(x<3.2\) --> as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient.

Hi Bunuel,

but if we consider the first part of the statement B we don't get the answer, right? if we have been given the equation with both the parts then we should be considering both the parts, right??? please explain...

Yes, if (2) were just \(3-2x<-x+4\) --> \(x>-1\), then it wouldn't be sufficient. As for your other question: can you please elaborate what you mean? Thank you. _________________

Given: \(x=integer\) and \(8x>4+6x\) --> \(2x>4\) --> \(x>2\). Question: \(x=?\)

(1) \(6-5x>-13\) --> \(19>5x\) --> \(\frac{19}{5}=3.8>x\) --> as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient.

(2) \(3-2x<-x+4<7.2-2x\) --> take only the following part: \(-x+4<7.2-2x\)--> \(x<3.2\) --> as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient.

Answer: D.

Hi,

can you explain stmt B please, why is that only the second part is considered?

mission2009 wrote:

Hi Bunuel, why have to taken only the second equation in Statement B not the first one? Plz explain briefly.

Thanks..

Because to reach the answer we don't need the first part at all. The part which says \(-x+4<7.2-2x\) is enough to give necessary info: \(x<3.2\) --> as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient.

So, Does it mean that in the questions where more than one IE are given such as in option B, we dont need to look at the first part? _________________

Given: \(x=integer\) and \(8x>4+6x\) --> \(2x>4\) --> \(x>2\). Question: \(x=?\)

(1) \(6-5x>-13\) --> \(19>5x\) --> \(\frac{19}{5}=3.8>x\) --> as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient.

(2) \(3-2x<-x+4<7.2-2x\) --> take only the following part: \(-x+4<7.2-2x\)--> \(x<3.2\) --> as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient.

Answer: D.

Hi,

can you explain stmt B please, why is that only the second part is considered?

mission2009 wrote:

Hi Bunuel, why have to taken only the second equation in Statement B not the first one? Plz explain briefly.

Thanks..

Because to reach the answer we don't need the first part at all. The part which says \(-x+4<7.2-2x\) is enough to give necessary info: \(x<3.2\) --> as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient.

So, Does it mean that in the questions where more than one IE are given such as in option B, we dont need to look at the first part?

Of course not! It was only for that particular question. _________________

but if we consider the first part of the statement B we don't get the answer, right? if we have been given the equation with both the parts then we should be considering both the parts, right??? please explain...[/quote]

Yes, if (2) were just \(3-2x<-x+4\) --> \(x>-1\), then it wouldn't be sufficient. As for your other question: can you please elaborate what you mean? Thank you.[/quote]

Hi Bunuel,

what I meant to say that in the statement B there are two equations and both gives us different answers for X, but you have taken the 2nd part only and discarded the first part..why is that..if both the parts are given then should not we take both the parts into consideration?? and if we take both the parts into consideration we do not get the desired result..hope I am able to explain myself....

what I meant to say that in the statement B there are two equations and both gives us different answers for X, but you have taken the 2nd part only and discarded the first part..why is that..if both the parts are given then should not we take both the parts into consideration?? and if we take both the parts into consideration we do not get the desired result..hope I am able to explain myself....

I did not discard anything. The point is that the first part does not give us any relevant info.

If we use both we'd have the same!

\(3-2x < -x + 4 < 7.2 - 2x\) --> \(-1<x<3.2\). We know that \(x>2\), so we have \(2<x<3.2\).

Re: If 8x > 4 + 6x, what is the value of the integer x? (1) 6 [#permalink]

Show Tags

25 Aug 2015, 04:42

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. _________________

So, my final tally is in. I applied to three b schools in total this season: INSEAD – admitted MIT Sloan – admitted Wharton – waitlisted and dinged No...

HBS alum talks about effective altruism and founding and ultimately closing MBAs Across America at TED: Casey Gerald speaks at TED2016 – Dream, February 15-19, 2016, Vancouver Convention Center...