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.

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:

Re: Math: Absolute value (Modulus) [#permalink]
14 May 2013, 20:15

2

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

nikhil007 wrote:

VeritasPrepKarishma wrote:

Of course it can be done using algebra as well. It doesn't matter how many mods there are. you always deal with them in the same way. |x|= x when x is >= 0, |x|= -x when x < 0

|x - 2|= (x - 2) when x - 2 >= 0 (or x >= 2), |x - 2|= -(x - 2) when (x-2) < 0 (or x < 2)

Then you solve the equations using both conditions given above. That is the importance of the points. So if you have: |x - 2|= |x + 3|

You say, |x - 2|= (x - 2) when x >= 2. |x - 2|= -(x - 2) when x < 2 |x + 3| = (x + 3) when x >= -3 |x + 3| = -(x + 3) when x < -3

Ok, Now after literally banging my head for 3 hrs and reading you blog articles back and forth, I get it that to make an EQ in (X-K) format we manipulate it by taking -tive sign out but I guess in this example its this concept that we need to understand

|x|= x when x is >= 0, |x|= -x when x < 0

ok, so based on this understanding I will take a fresh shot, please let me know what's wrong

Quote:

a) x < -8. -(x+3) - (4-x) = -(8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not less than -8)

In this test case, since we will always have |x+3| negative we put a -tive sign outside because modulus will turn it into non negative, so to do that we take multiply it by (-1), is this understanding correct? and since we are ok with -(4-x), because we will again get |4-x| positive with a negative x, the -tive sign outside the bracket will make sure its always -tive when out of the Modulus. However to be frank, a little confusion here is, as you mentioned in the blog, why don't we try to convert it into (x-k) format? in RHS we have -(8-x) because again we want |8-x| to turn out a negative number so we put -(8-x) to make it always negative, let me know if I got it correctly.

Quote:

b) -8 \leq x < -3. -(x+3) - (4-x) = (8+x) --> x = -15. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)

again, I get it why LHS is that way, however I still don't get it why we don't have -(8-X) as we need to make sure that the result of this bracket is -tive so |8-x| = -(8-x)

Quote:

c) -3 \leq x < 4. (x+3) - (4-x) = (8+x) --> x = 9. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)

I still dont get it, if we test x against both -tive and positive scenario, why is that we just have 1 equation? in my view we should split it in 2 eq. to test against both negative and positive value.

Quote:

d) x \geq 4. (x+3) + (4-x) = (8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not more than 4)

(x+3) since we don't have a negative value of X this bracket will always be positive, we don't need a -tive sign outside, is this the reason? (4-x) again since a >4 will always make it positive we don't need a -tive sign outside the bracket, is this the reason? (8+x) again same reason as above for this?

x(-2x^2 + 17x - 30) > 0 (just took out x common) ok x(2x – 5)(6 – x) > 0(factoring the quadratic) ok 2x(x – 5/2)(-1)(x – 6) > 0 (take 2 common) ---------> I think in this you took out -1 common to make the second bracket = (x-k) format?

First of all, if you do get a question with multiple mods and if you want to be prepared for it, using algebra will be far more time consuming than the approaches discussed in my blog. But nevertheless, you should understand it properly.

When you have an equation with x in it, you solve by taking x to one side and everything else to the other. What happens when you have mods in it? Say |x| = 4, you still haven't got the value of x. You have the value of |x| only. So you need to remove the mod. Now there are rules to remove the mod.

|x|= x (mod removed) when x is >= 0, |x|= -x (mod removed) when x < 0

So |x| = 4 to remove the mod, I need to know whether x is positive or negative. If x >= 0, |x| = x so |x| = 4 = x We get that x is 4

If x < 0, |x| = -x so |x| = 4 = -x hence x = -4

So if we are looking for a positive value, then it is 4 and if we are looking for a negative value, it is -4.

Similarly, when you have |x+4| + |x - 3| = 10 (just an example), you need to remove the mods to solve for x. But to remove mods (which are around the entire factors x-4 and x-3 and not just around x), you need to know whether (x+ 4) and (x - 3) (the thing inside the mod) are positive/negative.

So you split it into ranges:

x > 3 Put any value greater than 3 in (x+4), (x+4) will remain positive. Put any value greater than 3 in (x - 3), (x - 3) will remain positive.

So when x > 3, we can remove the mods without any modification: (x + 4) + (x-3) = 10 x = 9/2 Since 9/2 is greater than 3, this value of x is acceptable.

-4 < x< 3 For these values of x, (x+4) will always be positive but (x-3) will be negative. So |x - 3| = -(x-3) (x + 4) - (x-3) = 10 You don't have any such value for x

x < -4 For these values of x, (x+4) and (x-3) will be negative. So |x - 3| = -(x-3) and |x+4| = -(x+4) -(x + 4) - (x-3) = 10 x = -11/2 Since -11/2 is less than -4, this value of x is also acceptable.

As for question with factors that are multiplied (discussed in the 3 links given above), We know how to deal with (x-a)(x-b)(x-c) > 0 type of questions so we try to bring it that form.

2(x–0)(x–5/2)(x–6) < 0 (multiply both sides by -1)-------> how did you arrive at 2(x-0)? i think it should be just 2x(x-\frac{5}{2})(x-6) <0

(x-0) is nothing but x. I put as (x-0) to make it consistent to the (x-a)(x-b).... form to help you remember that you have to take 0 as a transition point too. _________________

Re: Math: Absolute value (Modulus) [#permalink]
15 May 2013, 05:30

Karishma,

Kudos given for the post, thanks for explaining in detail, I agree that we would be better off plugging number on such a ques, but things get tricky when it comes to DS

I basically covered this from MGmat guides and I can handle a simple Mod like |x-2|>5 what I learnt is simply take 2 conditions, x-2>5 and 2-x>5 and solve for 2 set of x, however the book never taught me this 3 step method. so I have to dig it in here. _________________

Life is very similar to a boxing ring. Defeat is not final when you fall down… It is final when you refuse to get up and fight back!

Re: |x^2-4| = 1. What is x? [#permalink]
04 Jun 2013, 11:46

1

This post received KUDOS

The first method is the correct one and will always give you the correct results.

Consider however the following case

|x+5|=-4, at glance this equation has no solution because |x+5| cannot be less than 0. But I wanna take it as example:

With the first method you'll find if x>-5 x+5=-4, x=-9, out of the interval => it's not a solution

if x<-5 -x-5=-4, x=-1 out of the interval => it's not a solution

With the second method x+5=-4, x=-9 -(x+5)=-4, or x=-1 those seem valid... but the equation we know that has no solution.

Main point: the first method works always, do not rely on the other one. The second one does not take into consideration the intervals, so it might not work

Hope it's clear _________________

It is beyond a doubt that all our knowledge that begins with experience.

So, I believe that is the right answer but I'm still not sure about how the greater than/less than signs come into play. Sorry for the thick head!

Zarrolou wrote:

The first method is the correct one and will always give you the correct results.

Consider however the following case

|x+5|=-4, at glance this equation has no solution because |x+5| cannot be less than 0. But I wanna take it as example:

With the first method you'll find if x>-5 x+5=-4, x=-9, out of the interval => it's not a solution

if x<-5 -x-5=-4, x=-1 out of the interval => it's not a solution

With the second method x+5=-4, x=-9 -(x+5)=-4, or x=-1 those seem valid... but the equation we know that has no solution.

Main point: the first method works always, do not rely on the other one. The second one does not take into consideration the intervals, so it might not work

So, I believe that is the right answer but I'm still not sure about how the greater than/less than signs come into play. Sorry for the thick head!

The original solution used \geq{} and \leq{} to define the intervals. It solved the cases: x^2-4=1 so x=+-\sqrt{5}, and then it check weather those numbers are in the interval -2<x<2. Both are inside so both are valid solutions Then the other case -x^2+4=1 so x=+-\sqrt{3}, and then check the interval it is considering in this scenario x<-2 and x>2, both are inside the intervals so both are valid solutions

As I said before the correct method to solve abs values always checks if the result obtained is inside the interval is considering at that moment. If you do not double check weather the solution you find is inside the interval, you are likely to commit errors in the problem. What I am trying to say is that your method is incomplete: it lacks the last passage - you find the solutions, but you do not check if they are possible or not.

In this example all 4 solutions are possible, so the last step does nothing; but if one of the solution were not valid, you "incomplete" method would not be albe to detect it. ( as I showed you in the |x+5|=-4 example) Hope that what I mean is clear _________________

It is beyond a doubt that all our knowledge that begins with experience.

So, I believe that is the right answer but I'm still not sure about how the greater than/less than signs come into play. Sorry for the thick head!

The original solution used \geq{} and \leq{} to define the intervals. It solved the cases: x^2-4=1 so x=+-\sqrt{5}, and then it check weather those numbers are in the interval -2<x<2. Both are inside so both are valid solutions Then the other case -x^2+4=1 so x=+-\sqrt{3}, and then check the interval it is considering in this scenario x<-2 and x>2, both are inside the intervals so both are valid solutions

As I said before the correct method to solve abs values always checks if the result obtained is inside the interval is considering at that moment. If you do not double check weather the solution you find is inside the interval, you are likely to commit errors in the problem. What I am trying to say is that your method is incomplete: it lacks the last passage - you find the solutions, but you do not check if they are possible or not.

In this example all 4 solutions are possible, so the last step does nothing; but if one of the solution were not valid, you "incomplete" method would not be albe to detect it. ( as I showed you in the |x+5|=-4 example) Hope that what I mean is clear

Re: |x^2-4| = 1. What is x? [#permalink]
04 Jun 2013, 16:16

1

This post received KUDOS

WholeLottaLove wrote:

So, in other words, if I were to plug in -/+ √5 into x^2 = 5 it will yield me a result between -2<x<2? And where does -2<x<2 come from???

-2<x<2 is the interval in which the function is negative, bare with me:

Take the function |x^2-4|=1 1) Define where it is positive and where is negative => x^2-4>0 if x<-2 and x>2 So if x<-2 or x>2 is positive, if -2<x<2 is negative

2)Study each case on its own: x^2-4=1x=+-\sqrt{5}, are those results valid? Are they in the interval we are considering? Are they in the x<-2 or x>2 interval? -\sqrt{5} is less than -2, and +\sqrt{5} is more than 2. So they are valid solutions because they are in the intervals we are considering -x^2+4=1x=+-\sqrt{3}, are those results valid? same as above Yes they are valid because they are numbers between -2 and 2(the interval we are considering now, in which |abs| is negative => -x^2+4) _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: |x^2-4| = 1. What is x? [#permalink]
04 Jun 2013, 16:37

So, I need to find the positive and negative values of |x^2-4|=1 (i.e. x^2-4=1 and -x^2+4=1) and which x values make x^2-4 positive and -x^2+4 negative?

Thanks for putting up with my slowness in picking up these concepts!

Zarrolou wrote:

WholeLottaLove wrote:

So, in other words, if I were to plug in -/+ √5 into x^2 = 5 it will yield me a result between -2<x<2? And where does -2<x<2 come from???

-2<x<2 is the interval in which the function is negative, bare with me:

Take the function |x^2-4|=1 1) Define where it is positive and where is negative => x^2-4>0 if x<-2 and x>2 So if x<-2 or x>2 is positive, if -2<x<2 is negative

2)Study each case on its own: x^2-4=1x=+-\sqrt{5}, are those results valid? Are they in the interval we are considering? Are they in the x<-2 or x>2 interval? -\sqrt{5} is less than -2, and +\sqrt{5} is more than 2. So they are valid solutions because they are in the intervals we are considering -x^2+4=1x=+-\sqrt{3}, are those results valid? same as above Yes they are valid because they are numbers between -2 and 2(the interval we are considering now, in which |abs| is negative => -x^2+4)

Re: |x^2-4| = 1. What is x? [#permalink]
04 Jun 2013, 23:27

1

This post received KUDOS

WholeLottaLove wrote:

So, I need to find the positive and negative values of |x^2-4|=1 (i.e. x^2-4=1 and -x^2+4=1) and which x values make x^2-4 positive and -x^2+4 negative?

Thanks for putting up with my slowness in picking up these concepts!

Yes the first thing to do is finding the values of x that make the finction positive/negative.

So first you need to split the function into two cases: where is positive and where is negative (as you say), then you solve each case and find the result(s), then you check if the result is INSIDE the interval you are considering. So x^2-4 is positive for x<-2 and for x>2, so you solve x^2-4=1=>x=+-\sqrt{5}, are those value inside the interval x<-2 and x>2? YES, they are valid results. Then you repeat the same operation for the negative part -x^2+4 _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: Math: Absolute value (Modulus) [#permalink]
11 Aug 2013, 20:30

Expert's post

rrsnathan wrote:

hi,

i have a doubt in the example

Q.: |x^2-4| = 1. What is x? Solution: There are 2 conditions:

a) (x^2-4)\geq0

how the range for this example is (X^2-4)>=0 and not (x^2-4)>0?

regards, RRSNATHAN

You can ignore "=0" part. (x^2 - 4) cannot be 0 since its mod is 1. So you can certainly write the range as (x^2 - 4) > 0. In the overall scheme, it doesn't change anything. People usually write the ranges are ">=0" and "<0" to cover everything. Here "=0" is not needed. _________________

Re: Math: Absolute value (Modulus) [#permalink]
13 Apr 2014, 23:42

I have a doubt.

In this problem Problem: 1<x<9. What inequality represents this condition? A. |x|<3 B. |x+5|<4 C. |x-1|<9 D. |-5+x|<4 E. |3+x|<5

can anyone please explain how can I determine in 10-20 sec that B and D are contenders among the options. please put some light on this - D had left site 0 at x=5.

Re: Math: Absolute value (Modulus) [#permalink]
14 Apr 2014, 19:54

Expert's post

honey86 wrote:

I have a doubt.

In this problem Problem: 1<x<9. What inequality represents this condition? A. |x|<3 B. |x+5|<4 C. |x-1|<9 D. |-5+x|<4 E. |3+x|<5

can anyone please explain how can I determine in 10-20 sec that B and D are contenders among the options. please put some light on this - D had left site 0 at x=5.

Thanks

|x - a| = b implies that x is at a distance b from a.

|x - a| < b implies x is at a distance less than b from a.

|x - 5| < 4 implies x is at a distance less than 4 from 5. A distance of 4 from 5 gives two points: 1 and 9. Since we need 1 < x < 9, this is the answer.

|x + 5| < 4 implies x is at a distance less than 4 from -5. A distance of 4 from -5 gives points -9 and -1.

Re: Math: Absolute value (Modulus) [#permalink]
20 May 2014, 13:04

Dear Honey

Here's a quick shortcut to solving this problem:

You know that 1<x<9. So, pick x=6 and substitute in each of the options. Only options C and D remain.

But, if you analyze option C, you'll find that it'll also hold true for x=0 or x=-1, which are outside the given domain of x. So, the given domain cannot be for option C.

So, the solution must be the only option remaining- D. _________________

Re: Math: Absolute value (Modulus) [#permalink]
14 Jun 2014, 18:05

Can somebody help me solve the inequality:|x+5| > 3|x - 5|

This is what I did..... \frac{|x+5|}{|x - 5|}>3

==>\frac{x+5}{x - 5}>3 & \frac{x+5}{x - 5}<-3 ==>x+5>3x-15 & x+5<-3x+15 ==>-2x>-20 & 4x<10 ==>x<10 & x<5/2 BUT answer is 5/2<x<10. What am I doing wrong?? In the red highlighted portion I changed the inequality sign because I multiplied the equation with minus. Is it correct??

Re: Math: Absolute value (Modulus) [#permalink]
15 Jun 2014, 21:24

1

This post received KUDOS

Expert's post

madn800 wrote:

Can somebody help me solve the inequality:|x+5| > 3|x - 5|

This is what I did..... \frac{|x+5|}{|x - 5|}>3

==>\frac{x+5}{x - 5}>3 & \frac{x+5}{x - 5}<-3 ==>x+5>3x-15 & x+5<-3x+15 ==>-2x>-20 & 4x<10 ==>x<10 & x<5/2 BUT answer is 5/2<x<10. What am I doing wrong?? In the red highlighted portion I changed the inequality sign because I multiplied the equation with minus. Is it correct??

In an inequality, you cannot multiply/divide by a variable until and unless you know the sign of the variable. Also, you cannot divide by |x - 5| since x could be 5. Even if we assume that x is not 5, still you cannot multiply the inequality by (x - 5) as you have done here: ==>x+5>3x-15 & x+5<-3x+15

If x < 5 i.e. x - 5 is negative, when you multiply the inequality by x - 5, the sign of the inequality with flip.

If you would want to use algebra, you need to check in the regions between the transition points:

Case 1: x < -5 In this case x+5 is negative and x-5 is negative, so -(x+5) > -3(x - 5) 2x > 20 x > 10 Since x should be less than -5, x > 10 doesn't give us any value for x

Case 2: -5 <= x < 5 In this case x+5 is positive (or 0) and x-5 is negative, so (x+5) > -3(x - 5) 4x > 10 x > 2.5 Since x should be less than 5, the range we get is 2.5 < x< 5

Case 3: x >= 5 In this case x+5 is positive and x-5 is positive, so (x+5) > 3(x - 5) 2x < 20 x < 10 Since x should be greater than or equal to 5, the range we get is 5 <= x < 10

Combining case 2 and case 3, we get 2.5 < x < 10 _________________

Re: Math: Absolute value (Modulus) [#permalink]
15 Jun 2014, 21:36

1

This post received KUDOS

Expert's post

madn800 wrote:

Can somebody help me solve the inequality:|x+5| > 3|x - 5|

This is what I did..... \frac{|x+5|}{|x - 5|}>3

==>\frac{x+5}{x - 5}>3 & \frac{x+5}{x - 5}<-3 ==>x+5>3x-15 & x+5<-3x+15 ==>-2x>-20 & 4x<10 ==>x<10 & x<5/2 BUT answer is 5/2<x<10. What am I doing wrong?? In the red highlighted portion I changed the inequality sign because I multiplied the equation with minus. Is it correct??

You can use the number line approach too.

|x+5| > 3|x - 5| The distance of x from -5 is greater than 3 times the distance of x from 5.

At 0, the distance of x from -5 is same as distance of x from 5. So you must move to the right to make distance from -5 greater. As you keep moving to the right, distance of x from -5 will keep increasing but distance from 5 will keep decreasing. Mid way between 0 and 5, distance of x from -5 is 7.5 and distance of x from 5 is 2.5. So x = 2.5 is a point where distance of x from -5 is 3 times distance of x from 5.

As you keep moving further to the right and reach x = 5, distance from -5 will be 10. Moving further to the right 5 steps will give you a distance of 15 from -5 and 5 from 5. So x = 10 is a point where distance of x from -5 is 3 times distance of x from 5.

In between these two points, distance of x from -5 is more than 3 times distance of x from 5.