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Re: Math: Absolute value (Modulus) [#permalink]
15 Jan 2011, 02:36

I've understood everything about Absolute Values in these posts except for Example #1 in walker's first post. I was hoping someone could help me better understand the logic behind the solution.

First, I think what i'm struggling with most is the fact that we have absolute values on both the left and right hand side of the equation. How do you deal with those?

Second, I noticed that the three key points mentioned here are -8, -3, 4. I get that these three values for x equate the value inside each modulus to 0. But why do we need the value of each modulus to be 0?

Finally, I don't understand how we derived the 4 conditions. With the first one, for example, if x<-8, let's say x=-10, then |8+(-10)|=2 not 0. I'm completely lost here.

Re: Math: Absolute value (Modulus) [#permalink]
15 Jan 2011, 22:52

1

This post received KUDOS

Expert's post

rahul321 wrote:

I've understood everything about Absolute Values in these posts except for Example #1 in walker's first post. I was hoping someone could help me better understand the logic behind the solution.

First, I think what i'm struggling with most is the fact that we have absolute values on both the left and right hand side of the equation. How do you deal with those?

Second, I noticed that the three key points mentioned here are -8, -3, 4. I get that these three values for x equate the value inside each modulus to 0. But why do we need the value of each modulus to be 0?

Finally, I don't understand how we derived the 4 conditions. With the first one, for example, if x<-8, let's say x=-10, then |8+(-10)|=2 not 0. I'm completely lost here.

I am assuming that you have understood that mod is nothing but distance on the number line. If so, then you can easily solve the question without equations. I re arrange the eg to get |x + 3| = |x + 8| + |x - 4| (Inside the mod, 4-x is same as x-4 since mod will always be non-negative) What I get from this question is: "I want the point on the number line whose distance from -3 is equal to the sum of its distances from -8 and from 4." i.e. the green distance is equal to red distance + blue distance.

Attachment:

Ques5.jpg [ 3.7 KiB | Viewed 2171 times ]

Is there any such point possible? If x is between -8 and 4, you see above that it is not possible. If x < -8 or x > 4, see the diagrams below and it will be obvious that there is no such point. In the first case, the blue distance itself is greater than the green distance and in the second case, the red distance itself is greater than the green distance. Since there is no such point on the number line, no such value of x exists.

Re: Math: Absolute value (Modulus) [#permalink]
15 Jan 2011, 23:08

3

This post received KUDOS

Expert's post

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

Now, x can be either greater than 2, between -3 and 2 or less than -3. So you solve for these 3 cases: Case 1: x >= 2 (x - 2) = (x + 3) -2 = 3 No solution

Case 2: -3 <= x < 2 -(x - 2) = (x + 3) x = -1/2 which lies between -3 and 2 So this is a solution to the equation

Case 3: x <= -3 -(x - 2) = -(x + 3) 2 = -3 No solution

Similarly you can solve for as many terms as you want. _________________

Re: Math: Absolute value (Modulus) [#permalink]
17 Jan 2011, 00:25

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

Now, x can be either greater than 2, between -3 and 2 or less than -3. So you solve for these 3 cases: Case 1: x >= 2 (x - 2) = (x + 3) -2 = 3 No solution

Case 2: -3 <= x < 2 -(x - 2) = (x + 3) x = -1/2 which lies between -3 and 2 So this is a solution to the equation

Case 3: x <= -3 -(x - 2) = -(x + 3) 2 = -3 No solution

Similarly you can solve for as many terms as you want.

Re: Math: Absolute value (Modulus) [#permalink]
09 Feb 2011, 03:41

i have a question related to Example 1:

|x+3| - |4-x| = |8+x|

-3 \leq x < 4 x \geq 4. if we take /4-x/ than we have 4-x >=0 so x >=4 and 4-x<0 4<=0 i see in the solution that you switched x-4 and wrote x-4>= 0 and x-4 <o i understand that it is perfectly ok to do so. my question: when do we have to know when to switch ? because if not switched than we shall have x >=4 if switched x<=4 i hope i made myself clear thanks

Re: Math: Absolute value (Modulus) [#permalink]
09 Feb 2011, 04:47

1

This post received KUDOS

Expert's post

tinki wrote:

i have a question related to Example 1:

|x+3| - |4-x| = |8+x|

-3 \leq x < 4 x \geq 4. if we take /4-x/ than we have 4-x >=0 so x >=4 and 4-x<0 4<=0 i see in the solution that you switched x-4 and wrote x-4>= 0 and x-4 <o i understand that it is perfectly ok to do so. my question: when do we have to know when to switch ? because if not switched than we shall have x >=4 if switched x<=4 i hope i made myself clear thanks

You can switch inside the mod when you like! When you feel switching makes it easier for you to handle. We are used to getting terms in the form |x-4|, not |4-x|... so you switch.. it doesn't matter at all... let me show with a simpler example:

Question: |4-x| > 5

Case 1: When 4 - x >= 0 i.e. when x <= 4 4 - x > 5 or x < -1 So solution is x < -1

Case 2: When 4 - x <= 0 i.e. when x >= 4 -(4-x) > 5 x > 9 So solution is x > 9

Answer: x is either less than -1 or greater than 9.

Now switch: Question: |x-4| > 5

Case 1: When x - 4 >= 0 i.e. when x >= 4 x - 4 > 5 or x > 9 Solution is x > 9

Case 2: When x - 4 <= 0 i.e. when x <= 4 -(x - 4) > 5 x < -1 So solution is x < -1

Answer: x is either less than -1 or greater than 9.

The same two cases in both the questions.. same answer in both... _________________

Re: Math: Absolute value (Modulus) [#permalink]
09 Feb 2011, 10:54

VeritasPrepKarishma wrote:

tinki wrote:

i have a question related to Example 1:

|x+3| - |4-x| = |8+x|

-3 \leq x < 4 x \geq 4. if we take /4-x/ than we have 4-x >=0 so x >=4 and 4-x<0 4<=0 i see in the solution that you switched x-4 and wrote x-4>= 0 and x-4 <o i understand that it is perfectly ok to do so. my question: when do we have to know when to switch ? because if not switched than we shall have x >=4 if switched x<=4 i hope i made myself clear thanks

You can switch inside the mod when you like! When you feel switching makes it easier for you to handle. We are used to getting terms in the form |x-4|, not |4-x|... so you switch.. it doesn't matter at all... let me show with a simpler example:

Question: |4-x| > 5

Case 1: When 4 - x >= 0 i.e. when x <= 4 4 - x > 5 or x < -1 So solution is x < -1

Case 2: When 4 - x <= 0 i.e. when x >= 4 -(4-x) > 5 x > 9 So solution is x > 9

Answer: x is either less than -1 or greater than 9.

Now switch: Question: |x-4| > 5

Case 1: When x - 4 >= 0 i.e. when x >= 4 x - 4 > 5 or x > 9 Solution is x > 9

Case 2: When x - 4 <= 0 i.e. when x <= 4 -(x - 4) > 5 x < -1 So solution is x < -1

Answer: x is either less than -1 or greater than 9.

The same two cases in both the questions.. same answer in both...

Re: Math: Absolute value (Modulus) [#permalink]
14 Feb 2011, 12:27

Very Helpful notes.Thanks

But can you pls explain the following third Trick

III. Thinking about absolute values as distance between points at the number line.

For example, Problem: A<X<Y<B. Is |A-X| <|X-B|? 1) |Y-A|<|B-Y| Solution: Image We can think about absolute values here as distance between points. Statement 1 means than distance between Y and A is less than Y and B. Because X is between A and Y, distance between |X-A| < |Y-A| and at the same time distance between X and B will be larger than that between Y and B (|B-Y|<|B-X|). Therefore, statement 1 is sufficient. _________________

The proof of understanding is the ability to explain it.

Re: Math: Absolute value (Modulus) [#permalink]
14 Feb 2011, 18:03

Expert's post

GMATD11 wrote:

Hi

one small question can we write |x|>=0

when 0 is neither positive nor negative but neutral.

I am not sure I got your question.. |x|>=0 implies |x| is positive(when x is positive or negative) or 0 (when x = 0). |x| is of course never negative. _________________

Re: Math: Absolute value (Modulus) [#permalink]
19 Feb 2011, 22:40

Hi, In the original post, in the section GMAC books, there are some prob numbers stated. Can you please explain how they relate to the post? How one will use those numbers is what I am asking.

gmatclubot

Re: Math: Absolute value (Modulus)
[#permalink]
19 Feb 2011, 22:40