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If 8x > 4 + 6x, what is the value of the integer x? (1) 6 [#permalink]
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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...
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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... Given: \(x=integer\) and \(8x>4+6x\) > \(2x>4\) > \(x>2\). Question: \(x=?\) (1) \(65x>13\) > \(19>5x\) > \(\frac{19}{5}=3.8>x\) > as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient. (2) \(32x<x+4<7.22x\) > take only the following part: \(x+4<7.22x\)> \(x<3.2\) > 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 \(<\)) > \(ac>bd\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) > \(35<41\). Hope it helps.
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Re: Inequality [#permalink]
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19 Aug 2010, 00:55
Bunuel wrote: 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... Given: \(x=integer\) and \(8x>4+6x\) > \(2x>4\) > \(x>2\). Question: \(x=?\) (1) \(65x>13\) > \(19>5x\) > \(\frac{19}{5}=3.8>x\) > as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient. (2) \(32x<x+4<7.22x\) > take only the following part: \(x+4<7.22x\)> \(x<3.2\) > 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 \(<\)) > \(ac>bd\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) > \(35<41\). Hope it helps. Hi, can you explain stmt B please, why is that only the second part is considered?



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Re: Inequality [#permalink]
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19 Aug 2010, 03:15
Hi Bunuel, why have to taken only the second equation in Statement B not the first one? Plz explain briefly.
Thanks..



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Re: Inequality [#permalink]
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19 Aug 2010, 07:24
ramana wrote: Bunuel wrote: Given: \(x=integer\) and \(8x>4+6x\) > \(2x>4\) > \(x>2\). Question: \(x=?\)
(1) \(65x>13\) > \(19>5x\) > \(\frac{19}{5}=3.8>x\) > as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient.
(2) \(32x<x+4<7.22x\) > take only the following part: \(x+4<7.22x\)> \(x<3.2\) > 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 \(x+4<7.22x\) is enough to give necessary info: \(x<3.2\) > as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient.
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Re: Inequality [#permalink]
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19 Aug 2010, 23:44
Bunuel wrote: 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... Given: \(x=integer\) and \(8x>4+6x\) > \(2x>4\) > \(x>2\). Question: \(x=?\) (1) \(65x>13\) > \(18>5x\) > \(\frac{18}{5}=3.6>x\) > as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient. (2) \(32x<x+4<7.22x\) > take only the following part: \(x+4<7.22x\)> \(x<3.2\) > 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 \(<\)) > \(ac>bd\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) > \(35<41\). Hope it helps. Slightly off on statement 1, however the answer doesn't change. (1) \(65x>13\) > \( 19>5x\) > \(\frac{19}{5}=3.8>x\) > as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient.



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Re: Inequality [#permalink]
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27 Sep 2011, 12:24
Bunuel wrote: 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... Given: \(x=integer\) and \(8x>4+6x\) > \(2x>4\) > \(x>2\). Question: \(x=?\) (1) \(65x>13\) > \(19>5x\) > \(\frac{19}{5}=3.8>x\) > as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient. (2) \(32x<x+4<7.22x\) > take only the following part: \(x+4<7.22x\)> \(x<3.2\) > 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 \(<\)) > \(ac>bd\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) > \(35<41\). 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



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Re: Inequality [#permalink]
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27 Sep 2011, 20:37
8x>4+6x =>2x>4 => x>2 x=?
1. Sufficient
65x>13 5x>19 => x<19/5 => x<3.2 => x ={..........1,0,1,2,3} as x is an integer as x>2 and x<3.2=> x can only be 3
2. Sufficient
32x <  x+4 <7.22x
32x<x+4 => 1 < x x+4 < 7.22x => x<3.2
=> 1<x<3.2 , but we know x>2 = 2<x<3.2 then x can only be 3 as it has to be an integer.
Answer is D.



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Re: Inequality [#permalink]
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28 Sep 2011, 10:33
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... Given equation can be simplified to 8x > 4 + 6x => x < 2 Given x is an integer 1) 6 – 5x > 13 => x < 3.2 and since x > 2 It means x = 3 Sufficient 2) x + 4 < 7.2 – 2x => x < 3.2 3 – 2x < x + 4 => x > 1 And we know x > 2 and x is an integer So x can be only 3 Sufficient. So D is the answer.
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Re: Inequality [#permalink]
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29 Aug 2013, 08:21
Bunuel wrote: ramana wrote: Bunuel wrote: Given: \(x=integer\) and \(8x>4+6x\) > \(2x>4\) > \(x>2\). Question: \(x=?\)
(1) \(65x>13\) > \(19>5x\) > \(\frac{19}{5}=3.8>x\) > as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient.
(2) \(32x<x+4<7.22x\) > take only the following part: \(x+4<7.22x\)> \(x<3.2\) > 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 \(x+4<7.22x\) 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...



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Re: Inequality [#permalink]
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29 Aug 2013, 09:47
Reetabrata Ghosh wrote: Bunuel wrote: ramana wrote: Given: \(x=integer\) and \(8x>4+6x\) > \(2x>4\) > \(x>2\). Question: \(x=?\)
(1) \(65x>13\) > \(19>5x\) > \(\frac{19}{5}=3.8>x\) > as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient.
(2) \(32x<x+4<7.22x\) > take only the following part: \(x+4<7.22x\)> \(x<3.2\) > 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 \(x+4<7.22x\) 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 \(32x<x+4\) > \(x>1\), then it wouldn't be sufficient. As for your other question: can you please elaborate what you mean? Thank you.
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Re: Inequality [#permalink]
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29 Aug 2013, 13:40
Bunuel wrote: ramana wrote: Bunuel wrote: Given: \(x=integer\) and \(8x>4+6x\) > \(2x>4\) > \(x>2\). Question: \(x=?\)
(1) \(65x>13\) > \(19>5x\) > \(\frac{19}{5}=3.8>x\) > as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient.
(2) \(32x<x+4<7.22x\) > take only the following part: \(x+4<7.22x\)> \(x<3.2\) > 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 \(x+4<7.22x\) 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?
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Re: Inequality [#permalink]
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29 Aug 2013, 13:43
swati007 wrote: Bunuel wrote: ramana wrote: Given: \(x=integer\) and \(8x>4+6x\) > \(2x>4\) > \(x>2\). Question: \(x=?\)
(1) \(65x>13\) > \(19>5x\) > \(\frac{19}{5}=3.8>x\) > as \(x=integer\) and \(x>2\), then \(x=3\). Sufficient.
(2) \(32x<x+4<7.22x\) > take only the following part: \(x+4<7.22x\)> \(x<3.2\) > 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 \(x+4<7.22x\) 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.
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Re: Inequality [#permalink]
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29 Aug 2013, 21:15
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...[/quote]
Yes, if (2) were just \(32x<x+4\) > \(x>1\), 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....



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Re: Inequality [#permalink]
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30 Aug 2013, 04:34
Reetabrata Ghosh wrote: 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! \(32x < x + 4 < 7.2  2x\) > \(1<x<3.2\). We know that \(x>2\), so we have \(2<x<3.2\). Hope it's clear now.
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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 8x6x > 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. 32x<x+4<7.22x 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. answer is D.




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