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Re: Inequalities - Challenging and tricky One [#permalink]
16 Jan 2010, 11:54

6

This post received KUDOS

Expert's post

Hussain15 wrote:

Thanks for the reply Bunue!

But there is one more issue.

In range A, you have changed the signs of both modules, in range B you have done it only for 2x+5 and in third case you haven't changed the signs. What's the logic behind this one??

I got your point. I'm not "changing" the signs, I'm expanding the absolute values in each range.

In range A, when \(x<\frac{2}{3}\): \(|3x-2|=-3x+2\) and \(|2x-5|=-2x+5\), so we get \(-3x+2\leq-2x+5\).

In range B, when \(\frac{2}{3}\leq{x}\leq\frac{5}{2}\): \(|3x-2|=3x-2\) and \(|2x-5|=-2x+5\), so we get \(3x-2\leq-2x+5\).

In range C, when \(x>\frac{5}{2}\): \(|3x-2|=3x-2\) and \(|2x-5|=2x-5\), so we get \(3x-2\leq2x-5\). _________________

Re: Inequalities - Challenging and tricky One [#permalink]
16 Jan 2010, 06:34

5

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

GMATMadeeasy wrote:

What is the solution set for \(|3x-2|\leq|2x-5|\)

One way to solve is to square both the terms of course , but what is other way of solving it.

First you should determine the check points (key points): \(\frac{2}{3}\) and \(\frac{5}{2}\). Hence we'll have three ranges to check:

A. \(x<\frac{2}{3}\) --> \(-3x+2\leq-2x+5\) --> \(-3\leq{x}\), as \(x<\frac{2}{3}\), then \(-3\leq{x}<\frac{2}{3}\);

B. \(\frac{2}{3}\leq{x}\leq\frac{5}{2}\) --> \(3x-2\leq-2x+5\) --> -\(x\leq\frac{7}{5}\), as \(\frac{2}{3}\leq{x}\leq\frac{5}{2}\) , then \(\frac{2}{3}\leq{x}\leq\frac{7}{5}\);

C. \(x>\frac{5}{2}\) --> \(3x-2\leq2x-5\) --> \(x\leq{-3}\), as \(x>\frac{5}{2}\), then in this range we have no solution;

Ranges from A and B give us the solution as: \(-3\leq{x}\leq\frac{7}{5}\). _________________

Re: Inequalities - Challenging and tricky One [#permalink]
11 Mar 2011, 12:51

5

This post received KUDOS

Expert's post

fluke wrote:

Bunuel wrote:

Hussain15 wrote:

Thanks for the reply Bunue!

But there is one more issue.

In range A, you have changed the signs of both modules, in range B you have done it only for 2x+5 and in third case you haven't changed the signs. What's the logic behind this one??

I got your point. I'm not "changing" the signs, I'm expanding the absolute values in each range.

In range A, when \(x<\frac{2}{3}\): \(|3x-2|=-3x+2\) and \(|2x-5|=-2x+5\), so we get \(-3x+2\leq-2x+5\).

In range B, when \(\frac{2}{3}\leq{x}\leq\frac{5}{2}\): \(|3x-2|=3x-2\) and \(|2x-5|=-2x+5\), so we get \(3x-2\leq-2x+5\).

In range C, when \(x>\frac{5}{2}\): \(|3x-2|=3x-2\) and \(|2x-5|=2x-5\), so we get \(3x-2\leq2x-5\).

My slow brain is just not getting it!!! How both LHS and RHS get negated for the range x<2/3 and only the RHS gets negated for the range B and nothing for range C. What if there were 4 or 5 ranges; what is the deciding factor as to what gets negated.

Could you please help me understand?

Absolute value properties: When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|\leq{-(some \ expression)}\). For example: \(|-5|=5=-(-5)\);

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|\leq{some \ expression}\). For example: \(|5|=5\);

In range A, when \(x<\frac{2}{3}\): \(3x-2<0\) so \(|3x-2|=-(3x-2)\) and \(2x-5<0\) so \(|2x-5|=-(2x-5)\), and we get \(-3x+2\leq-2x+5\).

In range B, when \(\frac{2}{3}\leq{x}\leq\frac{5}{2}\): \(3x-2>0\) so \(|3x-2|=3x-2\) and \(2x-5<0\) so \(|2x-5|=-(2x-5)\), so we get \(3x-2\leq-2x+5\).

Re: Inequalities - Challenging and tricky One [#permalink]
16 Jan 2010, 11:04

1

This post received KUDOS

Expert's post

Hussain15 wrote:

@ Bunuel:

I didn't pick your strategy.

My way of doing is simple but it is giving different answer.

|3x-2| <= |2x-5| We will consider two scenarios 3x-2 <= 2x-5 & -(3x-2) <= 2x-5 x<= -3 & x >= 7/5

So the range will be x<= -3 & x >= 7/5

Kindly let me know where did I go wrong???

If you plug the numbers from the ranges you got, you'll see that the inequality doesn't hold true.

As for the solution: we have two absolute values \(|3x-2|\) and \(|2x-5|\). \(|3x-2|\) changes sign at \(\frac{2}{3}\) and \(|2x-5|\) changes sign at \(\frac{5}{2}\).

Re: Inequalities - Challenging and tricky One [#permalink]
03 Sep 2012, 04:04

1

This post received KUDOS

Expert's post

Amogh wrote:

Bunuel, when you say as -3<= X and as x< \(\frac{2}{3}\) then -3<=x<=\(\frac{2}{3}\) does this mean that the range of x that satisfies the condition x<\(\frac{2}{3}\) is all those points on the number line which satisfy both inequalities.

Also you say ranges from A and B give us solution as -3<= x<= \(\frac{7}{5}\) . Can you please explain this to me. Does this mean that the final solution to this problem would be all those ranges that have a solution to their respective conditions i.e the conditions x<\(\frac{2}{3}\) , \(\frac{2}{3}\)<=x<\(\frac{5}{2}\) and x>=\(\frac{5}{2}\).

Also if for example ranges from A and B were -5<=x<2 and 4<=x<10, then would the final solution be both these ranges or do they have to have an overlap?

I know this is a supremely dumb and elementary question, but unfortunately inequalities and modulus happen to be my weakest areas. I cannot express in words my gratitude to you for having put up the GMAT Math Book. That thing is my Math Bible. You've made math so easy to understand.

Best wishes and many many thanks !!! Amogh.

We consider three ranges: A. \(x<\frac{2}{3}\); B. \(\frac{2}{3}\leq{x}\leq\frac{5}{2}\); C. \(x>\frac{5}{2}\).

In range A we get: \(-3\leq{x}<\frac{2}{3}\); In range B we get: \(\frac{2}{3}\leq{x}\leq\frac{7}{5}\); In range C there is no solution.

So, the given inequality holds true for \(-3\leq{x}<\frac{2}{3}\) and \(\frac{2}{3}\leq{x}\leq\frac{7}{5}\). Now, we can combine these two ranges and write: \(-3\leq{x}\leq\frac{7}{5}\).

Next, the solution set is \(-3\leq{x}\leq\frac{7}{5}\), means that any x from \(-3\leq{x}\leq\frac{7}{5}\) will satisfy \(|3x-2|\leq|2x-5|\).

If for some question we had -5<=x<2 from one range and 4<=x<10 from another, then the solution will be both ranges (no need for overlap).

Re: Inequalities - Challenging and tricky One [#permalink]
02 Nov 2012, 04:58

1

This post received KUDOS

Bunuel wrote:

GMATMadeeasy wrote:

What is the solution set for \(|3x-2|\leq|2x-5|\)

One way to solve is to square both the terms of course , but what is other way of solving it.

First you should determine the check points (key points): \(\frac{2}{3}\) and \(\frac{5}{2}\). Hence we'll have three ranges to check:

A. \(x<\frac{2}{3}\) --> \(-3x+2\leq-2x+5\) --> \(-3\leq{x}\), as \(x<\frac{2}{3}\), then \(-3\leq{x}<\frac{2}{3}\);

B. \(\frac{2}{3}\leq{x}\leq\frac{5}{2}\) --> \(3x-2\leq-2x+5\) --> -\(x\leq\frac{7}{5}\), as \(\frac{2}{3}\leq{x}\leq\frac{5}{2}\) , then \(\frac{2}{3}\leq{x}\leq\frac{7}{5}\);

C. \(x>\frac{5}{2}\) --> \(3x-2\leq2x-5\) --> \(x\leq{-3}\), as \(x>\frac{5}{7}\), then in this range we have no solution;

Ranges from A and B give us the solution as: \(-3\leq{x}\leq\frac{7}{5}\).

Hi Bunnuel excellent approach My doubt in range c is whether the fraction is 5/2 or 5/7 if it is the latter can you pls explain how.

Re: Inequalities - Challenging and tricky One [#permalink]
16 Jan 2010, 11:39

Thanks for the reply Bunue!

But there is one more issue.

In range A, you have changed the signs of both modules, in range B you have done it only for 2x+5 and in third case you haven't changed the signs. What's the logic behind this one?? _________________

Re: Inequalities - Challenging and tricky One [#permalink]
11 Mar 2011, 10:22

Bunuel wrote:

GMATMadeeasy wrote:

What is the solution set for \(|3x-2|\leq|2x-5|\)

One way to solve is to square both the terms of course , but what is other way of solving it.

First you should determine the check points (key points): \(\frac{2}{3}\) and \(\frac{5}{2}\). Hence we'll have three ranges to check:

A. \(x<\frac{2}{3}\) --> \(-3x+2\leq-2x+5\) --> \(-3\leq{x}\), as \(x<\frac{2}{3}\), then \(-3\leq{x}<\frac{2}{3}\);

B. \(\frac{2}{3}\leq{x}\leq\frac{5}{2}\) --> \(3x-2\leq-2x+5\) --> -\(x\leq\frac{7}{5}\), as \(\frac{2}{3}\leq{x}\leq\frac{5}{2}\) , then \(\frac{2}{3}\leq{x}\leq\frac{7}{5}\);

C. \(x>\frac{5}{2}\) --> \(3x-2\leq2x-5\) --> \(x\leq{-3}\), as \(x>\frac{5}{7}\), then in this range we have no solution;

Ranges from A and B give us the solution as: \(-3\leq{x}\leq\frac{7}{5}\).

Re: Inequalities - Challenging and tricky One [#permalink]
11 Mar 2011, 12:42

Bunuel wrote:

Hussain15 wrote:

Thanks for the reply Bunue!

But there is one more issue.

In range A, you have changed the signs of both modules, in range B you have done it only for 2x+5 and in third case you haven't changed the signs. What's the logic behind this one??

I got your point. I'm not "changing" the signs, I'm expanding the absolute values in each range.

In range A, when \(x<\frac{2}{3}\): \(|3x-2|=-3x+2\) and \(|2x-5|=-2x+5\), so we get \(-3x+2\leq-2x+5\).

In range B, when \(\frac{2}{3}\leq{x}\leq\frac{5}{2}\): \(|3x-2|=3x-2\) and \(|2x-5|=-2x+5\), so we get \(3x-2\leq-2x+5\).

In range C, when \(x>\frac{5}{2}\): \(|3x-2|=3x-2\) and \(|2x-5|=2x-5\), so we get \(3x-2\leq2x-5\).

My slow brain is just not getting it!!! How both LHS and RHS get negated for the range x<2/3 and only the RHS gets negated for the range B and nothing for range C. What if there were 4 or 5 ranges; what is the deciding factor as to what gets negated.

Could you please help me understand? _________________

Re: Inequalities - Challenging and tricky One [#permalink]
02 Sep 2012, 07:33

Bunuel, when you say as -3<= X and as x< \(\frac{2}{3}\) then -3<=x<=\(\frac{2}{3}\) does this mean that the range of x that satisfies the condition x<\(\frac{2}{3}\) is all those points on the number line which satisfy both inequalities.

Also you say ranges from A and B give us solution as -3<= x<= \(\frac{7}{5}\) . Can you please explain this to me. Does this mean that the final solution to this problem would be all those ranges that have a solution to their respective conditions i.e the conditions x<\(\frac{2}{3}\) , \(\frac{2}{3}\)<=x<\(\frac{5}{2}\) and x>=\(\frac{5}{2}\).

Also if for example ranges from A and B were -5<=x<2 and 4<=x<10, then would the final solution be both these ranges or do they have to have an overlap?

I know this is a supremely dumb and elementary question, but unfortunately inequalities and modulus happen to be my weakest areas. I cannot express in words my gratitude to you for having put up the GMAT Math Book. That thing is my Math Bible. You've made math so easy to understand.

Re: What is the solution set for |3x-2|<=|2x-5| [#permalink]
01 Nov 2012, 12:32

Expert's post

Ousmane wrote:

Hello Bunuel. Does this approach always work?: |3x-2| <= |2x-5| ===> (3x-2)^2<= (2x-5)^2 ===> 5x^2+ 8x -21 <= 0 solution of the ineq (-3, 7/5) Brother Karamazov

Do you mean squaring? If both parts of an inequality are non-negative (as in our case), then you can safely raise both parts to an even power (for example square).

A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality). For example: \(2<4\) --> we can square both sides and write: \(2^2<4^2\); \(0\leq{x}<{y}\) --> we can square both sides and write: \(x^2<y^2\);

But if either of side is negative then raising to even power doesn't always work. For example: \(1>-2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we can not square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are non-negative.

B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality). For example: \(-2<-1\) --> we can raise both sides to third power and write: \(-2^3=-8<-1=-1^3\) or \(-5<1\) --> \(-5^2=-125<1=1^3\); \(x<y\) --> we can raise both sides to third power and write: \(x^3<y^3\).

Re: Inequalities - Challenging and tricky One [#permalink]
02 Nov 2012, 05:22

Bunuel wrote:

Hussain15 wrote:

Thanks for the reply Bunue!

But there is one more issue.

In range A, you have changed the signs of both modules, in range B you have done it only for 2x+5 and in third case you haven't changed the signs. What's the logic behind this one??

I got your point. I'm not "changing" the signs, I'm expanding the absolute values in each range.

In range A, when \(x<\frac{2}{3}\): \(|3x-2|=-3x+2\) and \(|2x-5|=-2x+5\), so we get \(-3x+2\leq-2x+5\).

In range B, when \(\frac{2}{3}\leq{x}\leq\frac{5}{2}\): \(|3x-2|=3x-2\) and \(|2x-5|=-2x+5\), so we get \(3x-2\leq-2x+5\).

In range C, when \(x>\frac{5}{2}\): \(|3x-2|=3x-2\) and \(|2x-5|=2x-5\), so we get \(3x-2\leq2x-5\).

While solving for A i understand that since (3x-2)<0 so you have multiplied the (3x-5) with negetive. But (2x-5) is >0 so i didnt understand why did you multiply it with negetive. Is it that you plugged a value 1/2 which is less than 2/3 and than analysed that we have to multiply negetive value.

Re: Inequalities - Challenging and tricky One [#permalink]
02 Nov 2012, 07:47

Expert's post

Archit143 wrote:

Bunuel wrote:

GMATMadeeasy wrote:

What is the solution set for \(|3x-2|\leq|2x-5|\)

One way to solve is to square both the terms of course , but what is other way of solving it.

First you should determine the check points (key points): \(\frac{2}{3}\) and \(\frac{5}{2}\). Hence we'll have three ranges to check:

A. \(x<\frac{2}{3}\) --> \(-3x+2\leq-2x+5\) --> \(-3\leq{x}\), as \(x<\frac{2}{3}\), then \(-3\leq{x}<\frac{2}{3}\);

B. \(\frac{2}{3}\leq{x}\leq\frac{5}{2}\) --> \(3x-2\leq-2x+5\) --> -\(x\leq\frac{7}{5}\), as \(\frac{2}{3}\leq{x}\leq\frac{5}{2}\) , then \(\frac{2}{3}\leq{x}\leq\frac{7}{5}\);

C. \(x>\frac{5}{2}\) --> \(3x-2\leq2x-5\) --> \(x\leq{-3}\), as \(x>\frac{5}{7}\), then in this range we have no solution;

Ranges from A and B give us the solution as: \(-3\leq{x}\leq\frac{7}{5}\).

Hi Bunnuel excellent approach My doubt in range c is whether the fraction is 5/2 or 5/7 if it is the latter can you pls explain how.

Regards Archit

Yes, it should be 5/2. Typo edited. Thank you. +1. _________________

Re: Inequalities - Challenging and tricky One [#permalink]
02 Nov 2012, 07:50

Expert's post

Archit143 wrote:

Bunuel wrote:

Hussain15 wrote:

Thanks for the reply Bunue!

But there is one more issue.

In range A, you have changed the signs of both modules, in range B you have done it only for 2x+5 and in third case you haven't changed the signs. What's the logic behind this one??

I got your point. I'm not "changing" the signs, I'm expanding the absolute values in each range.

In range A, when \(x<\frac{2}{3}\): \(|3x-2|=-3x+2\) and \(|2x-5|=-2x+5\), so we get \(-3x+2\leq-2x+5\).

In range B, when \(\frac{2}{3}\leq{x}\leq\frac{5}{2}\): \(|3x-2|=3x-2\) and \(|2x-5|=-2x+5\), so we get \(3x-2\leq-2x+5\).

In range C, when \(x>\frac{5}{2}\): \(|3x-2|=3x-2\) and \(|2x-5|=2x-5\), so we get \(3x-2\leq2x-5\).

While solving for A i understand that since (3x-2)<0 so you have multiplied the (3x-5) with negetive. But (2x-5) is >0 so i didnt understand why did you multiply it with negetive. Is it that you plugged a value 1/2 which is less than 2/3 and than analysed that we have to multiply negetive value.

Pls explain

For A: if \(x<\frac{2}{3}\) then both 3x-2 and 2x-5 are negative. Just substitute x=0<2/3 to check.

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