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If x is not equal to 0, is x less than 1? [#permalink]
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Hi I got these questions whilst doing the MGMAT. I got it wrong and do not understand the answer;
a. If x is not equal to 0, is x less than 1?
(1) x/x < x
(2) x > x
b. Is x > 0?
(1) x + 3 = 4x – 3
(2) x + 1 = 2x – 1
Hope to hear from u guys soon!



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Re: Some inequalities questions [#permalink]
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06 May 2010, 04:53
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xianster wrote: Hi I got these questions whilst doing the MGMAT. I got it wrong and do not understand the answer;
a. If x is not equal to 0, is x less than 1?
(1) x/x < x
(2) x > x
b. Is x > 0?
(1) x + 3 = 4x – 3
(2) x + 1 = 2x – 1
Hope to hear from u guys soon! Welcome to the Gmat Club xianster. Below are the solutions to you questions. Given: \(x\neq{0}\), is \(x<1\)? Which means is \(1<x<1\)? (\(x\neq{0}\)) (1) \(\frac{x}{x}< x\) Two cases: A. \(x<0\) > \(\frac{x}{x}<x\) > \(1<x\). But remember that \(x<0\), so \(1<x<0\) B. \(x>0\) > \(\frac{x}{x}<x\) > \(1<x\). Two ranges \(1<x<0\) or \(x>1\). Which says that x either in the first range or in the second. Not sufficient to answer whether \(1<x<1\). (For instance \(x\) can be \(0.5\) or \(3\)) (2) \(x > x\) Well this basically tells that \(x\) is negative. But still if we want to see how it works: Two cases again: \(x<0\)> \(x>x\)> \(x<0\). \(x>0\) > \(x>x\): never correct. Only one range: \(x<0\), but still insufficient to say whether \(1<x<1\). (For instance \(x\) can be \(0.5\) or \(10\)) (1)+(2) \(x<0\) (from 2) and \(1<x<0\) or \(x>1\) (from 1), hence \(1<x<0\). Every \(x\) from this range is definitely in the range \(1<x<1\). Sufficient. Answer: C.
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Re: Some inequalities questions [#permalink]
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06 May 2010, 05:03
Is x > 0? (1) \(x + 3 = 4x3\) As absolute value is NEVER NEGATIVE (in our case x + 3), thus RHS (right hand side) in our case 4x3 must also be \(\geq{0}\). \(4x3\geq{0}\) > \(x\geq{\frac{3}{4}}\). Sufficient. We should check if the equation \(x + 3 = 4x3\), with the condition that \(x\geq{\frac{3}{4}}\) has real roots (not an issue on GMAT, so for GMAT we could already stop on the previous step and say that statement is sufficient). As \(x\geq{\frac{3}{4}}\), then \(x + 3=x+3\) > hence \(x + 3 = 4x3\) becomes \(x+3=4x3\) > \(x=2\). Only ONE solution: \(x=2>0\). Sufficient. (2) \(x + 1 = 2x1\) The same here: \(2x1\geq{0}\) > \(x\geq{\frac{1}{2}}\). Sufficient. Just to check: as \(x\geq{\frac{1}{2}}\) then \(x + 1=x+1\) > hence \(x + 1 = 2x1\) becomes \(x+1=2x1\), \(x=2>0\). Answer: D.
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Re: Some inequalities questions [#permalink]
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06 May 2010, 05:29
Thanks Bunuel! I had a bit of a problem trying to figure out wat MGMAT was saying. In any case here is their ans for the 2nd question;
x + 3 = +(4x – 3) x + 3 = 4x – 3 6 = 3x 2 = x
and
x + 3 = –(4x – 3) x + 3 = –4x + 3 5x = 0 x = 0
They went on to sub x=2 and x=0 back into the original equation. And subsequently found out that x will always be > 0 The problem that I am having is why is there a need to substitute it back? I figured if there are 2 values, it will not satisfy x>0? Maybe you can help to explain?



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Re: Some inequalities questions [#permalink]
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06 May 2010, 05:46
xianster wrote: Thanks Bunuel! I had a bit of a problem trying to figure out wat MGMAT was saying. In any case here is their ans for the 2nd question;
x + 3 = +(4x – 3) x + 3 = 4x – 3 6 = 3x 2 = x
and
x + 3 = –(4x – 3) x + 3 = –4x + 3 5x = 0 x = 0
They went on to sub x=2 and x=0 back into the original equation. And subsequently found out that x will always be > 0 The problem that I am having is why is there a need to substitute it back? I figured if there are 2 values, it will not satisfy x>0? Maybe you can help to explain? They solved this question with different approach but what they did is right. x=0 can not be the solution as when x=0 RHS becomes 4x3=3 but as LHS is absolute value, it can not equal to negative number. That is why x=0 is not valid solution Another way: when \(x=0\) > \(x + 3=3=3\neq{4x3=3}\) The above part can be solved also as follows: We have \(x + 3=4x3\). Check point is \(x=3\) (check point for absolute value is the value of the variable when absolute value equals to zero, so in our case x+3=0 gives the check point x=3) We should consider the following two cases: A. \(x\leq{3}\) > \(x + 3=x3\) > \(x3=4x3\) > \(x=0\), but this solution is not valid as we are checking the range \(x\leq{3}\). B. \(x\>{3}\) > \(x + 3=x+3\) > \(x+3=4x3\) > \(x=2\), this solution is valid as \(x=2\) is in the range \(x>3\). Hence only one solution \(x=2\). Hope it's clear.
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Re: Some inequalities questions [#permalink]
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07 May 2010, 02:51
Hi Bunuel, I was just looking thru some of the qn and this came up; from ur compilation; 5. What is the value of y? (1) 3x^2 4 = y  2 (2) 3  y = 11 (1) As we are asked to find the value of y, from this statement we can conclude only that y>=2, as LHS is absolute value which is never negative, hence RHS als can not be negative. Not sufficient. (2) 3  y = 11:
y<3 > 3y=11 > y=8 y>=3 > 3+y=11 > y=14
Two values for y. Not sufficient.(1)+(2) y>=2, hence y=14. Sufficient. Answer: C. In this case, there were 2 values of y as well and it was not subbed back into the original eqn. That's why Im confused



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Re: Some inequalities questions [#permalink]
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07 May 2010, 03:16
xianster wrote: Hi Bunuel, I was just looking thru some of the qn and this came up; from ur compilation; 5. What is the value of y? (1) 3x^2 4 = y  2 (2) 3  y = 11 (1) As we are asked to find the value of y, from this statement we can conclude only that y>=2, as LHS is absolute value which is never negative, hence RHS als can not be negative. Not sufficient. (2) 3  y = 11:
y<3 > 3y=11 > y=8 y>=3 > 3+y=11 > y=14
Two values for y. Not sufficient.(1)+(2) y>=2, hence y=14. Sufficient. Answer: C. In this case, there were 2 values of y as well and it was not subbed back into the original eqn. That's why Im confused When you solve these kind of questions with check point method, you test validity of solution on the stage of obtaining the values and if it's OK at this stage you don't need to substitute it afterwards. For example: \(3  y = 11\): Check point is y=3. So we should test two ranges: A. \(y<3\) > \(3y=11\) > \(y=8\). Now you should check whether the obtained solution is in the range you are considering. Is \(y=8\) in the range \(y<3\)? YES. So this value is OK. No need to substitute it in the equation \(3  y = 11\). B. \(y\geq{3}\) > \(3+y=11\) > \(y=14\). Is \(y=14\) in the range \(y\geq{3}\)? YES. So this value is OK. No need to substitute it in the equation \(3  y = 11\). So two solutions for this statement \(y=8\) and \(y=14\). Another example is in previous post: \(x + 3=4x3\). Check point is \(x=3\) We should consider the following two cases: A. \(x\leq{3}\) > \(x + 3=x3\) > \(x3=4x3\) > \(x=0\). Is \(x=0\) in the range \(x\leq{3}\)? NO. So this value is not OK. \(x=0\) is not the solution of equation \(x + 3=4x3\). B. \(x\>{3}\) > \(x + 3=x+3\) > \(x+3=4x3\) > \(x=2\), this solution is valid as \(x=2\) is in the range \(x>3\). Hence only one solution \(x=2\). Check Math Book chapter of Absolute value (link in my signature) for more. Hope it helps.
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Re: Some inequalities questions [#permalink]
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09 May 2010, 07:56
Thanks! I tink I sort of understand what you said. Can I also make this assumption that there is always a need to check back (substitute or use the check pt method). If I use the substitution method for this question,
3  y = 11:
y<3 > 3y=11 > y=8. When we put 8 back into the eqn 11 =11 hence ok. y>=3 > 3+y=11 > y=14 When we put 14 back into the eqn 11 =11 hence ok.
Same for this question,
x + 3 = 4x – 3
When we sub x=0, 3 is not = 3 hence not ok When we sub x=2, 5 = 5 hence ok
I guess whatever it is, we will need to do a validity check correct? Does this apply to inequalities like < or > as well? From the trend I dont see a need to do so. Am I right?



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Re: Some inequalities questions [#permalink]
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09 May 2010, 08:13
xianster wrote: Thanks! I tink I sort of understand what you said. Can I also make this assumption that there is always a need to check back (substitute or use the check pt method). If I use the substitution method for this question,
3  y = 11:
y<3 > 3y=11 > y=8. When we put 8 back into the eqn 11 =11 hence ok. y>=3 > 3+y=11 > y=14 When we put 14 back into the eqn 11 =11 hence ok.
Same for this question,
x + 3 = 4x – 3
When we sub x=0, 3 is not = 3 hence not ok When we sub x=2, 5 = 5 hence ok
I guess whatever it is, we will need to do a validity check correct? Does this apply to inequalities like < or > as well? From the trend I dont see a need to do so. Am I right? Again: When you solve these kind of questions with check point method (used in my previous post), you test validity of solution on the stage of obtaining the values (by checking whether the value is in the range you are testing at the moment) and if the value obtained IS in the range you are testing at the moment, you don't need to substitute it afterwards to check, you've already done the checking and if the value obtained IS NOT in the range you are testing at the moment you also don't need to substitute it afterwards to check, you've already done the checking. As for inequalities: they are whole different story. Usually you don't do substitutions while solving them.
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Re: Some inequalities questions [#permalink]
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09 May 2010, 08:32
Ok thanks for the clarification! Appreciate it!



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Re: Some inequalities questions [#permalink]
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16 May 2010, 12:23
I appreciate all yours efforts, thanks



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Re: Some inequalities questions [#permalink]
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19 Feb 2013, 09:03
Hi Bunuel, I have been going through all the materials you have given for the check point/key point approach on absolute expressions. While I get how the check points are selected and how conditions are laid, I am not getting where the equal sign would come. For instance: For 3y why is it y<3 and y>=3 and why not y<=3 and y>3 ?



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Re: Some inequalities questions [#permalink]
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19 Feb 2013, 09:10



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Re: If x is not equal to 0, is x less than 1? [#permalink]
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22 Apr 2014, 06:32
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Bunuel, I think I have been persistent on asking about this but I find it to be essential to my GMAT prep In which cases do we need to test for the two cases when expression inside an absolute value is >=0 and <0 and in whcih cases can we just use RHS >=0 solve and check whether the solution for 'x' is in the given range? Could you please elaborate on this shortcut that you use oftenly for these type of questions for the benefit of others? Thanks!! Cheers J



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Re: If x is not equal to 0, is x less than 1? [#permalink]
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10 May 2014, 11:02
Question rephrased: Is x between 1 and 1?
Statement 1: x/x < x x < xx
0 < xx  x
0 < x (x  1)
The CRITICAL POINTS are 1, 0 and 1. These are the only values where x(x1) = 0. To determine the ranges where x(x1) > 0, test one value to the left and right of each critical point.
Case 1: x<1 Plug x = 2 into x/x < x: 2/ 2 < 2 1 < 2. Doesn't work. Thus, x < 1 is not a valid range.
Case 2: 1<x<0 Plug x = 1/2 into x/x < x: 1/2/ 1/2 < 1/2 1 < 1/2. This works. Thus, 1<x<0 is a valid range.
Case 3: 0<x<1 Plug x = 1/2 into x/x < x: (1/2)/ 1/2 < 1/2 1 < 1/2 Doesn't work. Thus, 0<x<1 is not a valid range.
Case 4: x>1 Plug x = 2 into x/x < x: 2/ 2 < 2 1 < 2. This works. Thus, x > 1 is a valid range.
Thus, 1<x<0 or x>1. INSUFFICIENT.
Statement 2: x > x Any negative value will satisfy this inequality. If x=1/2, then x is between 1 and 1. If x=2, then x is NOT between 1 and 1. INSUFFICIENT.
Statements combined: The only range that satisfies both statements is 1<x<0. Thus, x is between 1 and 1. SUFFICIENT.
The correct answer is C.




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