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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]

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01 Nov 2016, 06:43

heyholetsgo wrote:

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...

from the given information 8x-6x > 4 2x > 4 x>2.

1. 19>5x -> 3.8 > x -> since x must be an integer, and since x>2, we know for sure that X is 3. Suficient.

2. 3-2x<-x+4<7.2-2x add 2x to each side: 3<x+4<7.2 subtract 4 -1<x<3.2 since x>2, must be and integer, and since x<3.2, x must be equal to 3 sufficient.

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