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# Which of the following sets includes ALL of the solutions of

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Senior Manager
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Which of the following sets includes ALL of the solutions of [#permalink]

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17 Sep 2012, 07:29
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Which of the following sets includes ALL of the solutions of x that will satisfy the equation: ?

|x-2|-|x-3|=|x-5|

(A) {-6, -5, 0, 1, 7, 8}
(B) {-4, -2, 0, 10/3, 4, 5}
(C) {-4, 0, 1, 4, 5, 6}
(D) {-1, 10/3, 3, 5, 6, 8}
(E) {-2, -1, 1, 3, 4, 5}

Usually, I'm able to solve such questions, however I couldn't solve this one.
Can someone run pass me through using number line approach.

Thanks
H
[Reveal] Spoiler: OA

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17 Sep 2012, 07:41
I believe there is something wrong with question because -4,5 & 8 does not satisfy the inequality.

I believe testing values is the best solution for this question.

Let me know if i am wrong.
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Last edited by fameatop on 17 Sep 2012, 08:27, edited 1 time in total.
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17 Sep 2012, 08:20
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the equation has 3 intervals 2, 3 , and 5. i.e 4 segments x<2, 2<x<3, 3<x<5, 5<x.

1) When X>5 none of the brackets will change the sign and hence
(x-2)-(x-3) = (x-5)-------> x=6 ..We accept this solution as it satisfies the equation.

Now since we know x=6 is one of the solution we know the answer could be C or D

2) When 3<x<5 here (x-5) will change the sign
(x-2)-(x-3) = -(x-5) ------>x=4 ..We accept the solution as 3<4<5

We can stop solving here as the only set consisting both 6 and 4 is set C hence correct answer is C
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17 Sep 2012, 08:55
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imhimanshu wrote:
Which of the following sets includes ALL of the solutions of x that will satisfy the equation: ?

|x-2|-|x-3|=|x-5|

(A) {-6, -5, 0, 1, 7, 8}
(B) {-4, -2, 0, 10/3, 4, 5}
(C) {-4, 0, 1, 4, 5, 6}
(D) {-1, 10/3, 3, 5, 6, 8}
(E) {-2, -1, 1, 3, 4, 5}

Usually, I'm able to solve such questions, however I couldn't solve this one.
Can someone run pass me through using number line approach.

Thanks
H

The given equation can be rewritten as $$|x-2|=|x-3|+|x-5|,$$ which means that the distance between $$x$$ and $$2$$ is the sum of the distances between $$x$$ and $$3$$ plus the distance between $$x$$ and 5.
Using the number line and visualizing the points, we can easily see that $$x$$ cannot be smaller than $$3.$$

If $$x>5,$$ then $$x-2=x-3+x-5,$$ from which $$x=6.$$ We have to choose between answers C and D.
If $$3<x<5,$$ then $$x-2=x-3-x+5,$$ which yields $$x=4.$$

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17 Sep 2012, 09:06
EvaJager wrote:
imhimanshu wrote:
Which of the following sets includes ALL of the solutions of x that will satisfy the equation: ?

|x-2|-|x-3|=|x-5|

(A) {-6, -5, 0, 1, 7, 8}
(B) {-4, -2, 0, 10/3, 4, 5}
(C) {-4, 0, 1, 4, 5, 6}
(D) {-1, 10/3, 3, 5, 6, 8}
(E) {-2, -1, 1, 3, 4, 5}

Usually, I'm able to solve such questions, however I couldn't solve this one.
Can someone run pass me through using number line approach.

Thanks
H

The given equation can be rewritten as $$|x-2|=|x-3|+|x-5|,$$ which means that the distance between $$x$$ and $$2$$ is the sum of the distances between $$x$$ and $$3$$ plus the distance between $$x$$ and 5.
Using the number line and visualizing the points, we can easily see that $$x$$ cannot be smaller than $$3.$$

If $$x>5,$$ then $$x-2=x-3+x-5,$$ from which $$x=6.$$ We have to choose between answers C and D.
If $$3<x<5,$$ then $$x-2=x-3-x+5,$$ which yields $$x=4.$$

Hi Eva,

If i am not wrong then x will take any value which can satisfy this modulus equation |x-2|-|x-3|=|x-5| i.e. LHS = RHS
Now if take x=5 (as 5 is one of the value of x in option C), then
LHS--> |5-2| - |5-3| = |3|-|2| = 3-2 = 1
RHS--> |5-5| = |0| = 0

In this case LHS is not equal to RHS i.e. how can 1 equals 0

How can C be the correct option if it contains one value which does not satisfy the equation.

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17 Sep 2012, 09:15
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fameatop wrote:
EvaJager wrote:
imhimanshu wrote:
Which of the following sets includes ALL of the solutions of x that will satisfy the equation: ?

|x-2|-|x-3|=|x-5|

(A) {-6, -5, 0, 1, 7, 8}
(B) {-4, -2, 0, 10/3, 4, 5}
(C) {-4, 0, 1, 4, 5, 6}
(D) {-1, 10/3, 3, 5, 6, 8}
(E) {-2, -1, 1, 3, 4, 5}

Usually, I'm able to solve such questions, however I couldn't solve this one.
Can someone run pass me through using number line approach.

Thanks
H

The given equation can be rewritten as $$|x-2|=|x-3|+|x-5|,$$ which means that the distance between $$x$$ and $$2$$ is the sum of the distances between $$x$$ and $$3$$ plus the distance between $$x$$ and 5.
Using the number line and visualizing the points, we can easily see that $$x$$ cannot be smaller than $$3.$$

If $$x>5,$$ then $$x-2=x-3+x-5,$$ from which $$x=6.$$ We have to choose between answers C and D.
If $$3<x<5,$$ then $$x-2=x-3-x+5,$$ which yields $$x=4.$$

Hi Eva,

If i am not wrong then x will take any value which can satisfy this modulus equation |x-2|-|x-3|=|x-5| i.e. LHS = RHS
Now if take x=5 (as 5 is one of the value of x in option C), then
LHS--> |5-2| - |5-3| = |3|-|2| = 3-2 = 1
RHS--> |5-5| = |0| = 0

In this case LHS is not equal to RHS i.e. how can 1 equals 0

How can C be the correct option if it contains one value which does not satisfy the equation.

The question reads "Which of the following sets includes ALL of the solutions of x that will satisfy the equation" it doesn't say that all the elements of the set are solutions of the given equation. When solving, you can find that there are exactly two values only which are solutions: 4 and 6. And they are both included only in the set of answer C.
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17 Sep 2012, 09:17
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fameatop wrote:
EvaJager wrote:
imhimanshu wrote:
Which of the following sets includes ALL of the solutions of x that will satisfy the equation: ?

|x-2|-|x-3|=|x-5|

(A) {-6, -5, 0, 1, 7, 8}
(B) {-4, -2, 0, 10/3, 4, 5}
(C) {-4, 0, 1, 4, 5, 6}
(D) {-1, 10/3, 3, 5, 6, 8}
(E) {-2, -1, 1, 3, 4, 5}

Usually, I'm able to solve such questions, however I couldn't solve this one.
Can someone run pass me through using number line approach.

Thanks
H

The given equation can be rewritten as $$|x-2|=|x-3|+|x-5|,$$ which means that the distance between $$x$$ and $$2$$ is the sum of the distances between $$x$$ and $$3$$ plus the distance between $$x$$ and 5.
Using the number line and visualizing the points, we can easily see that $$x$$ cannot be smaller than $$3.$$

If $$x>5,$$ then $$x-2=x-3+x-5,$$ from which $$x=6.$$ We have to choose between answers C and D.
If $$3<x<5,$$ then $$x-2=x-3-x+5,$$ which yields $$x=4.$$

Hi Eva,

If i am not wrong then x will take any value which can satisfy this modulus equation |x-2|-|x-3|=|x-5| i.e. LHS = RHS
Now if take x=5 (as 5 is one of the value of x in option C), then
LHS--> |5-2| - |5-3| = |3|-|2| = 3-2 = 1
RHS--> |5-5| = |0| = 0

In this case LHS is not equal to RHS i.e. how can 1 equals 0

How can C be the correct option if it contains one value which does not satisfy the equation.

We need the set that INCLUDES all the solutions (which are 4 and 6), it's not necessary that all numbers of the set to satisfy the equation. Only option C contains all of the solutions, so C is the correct answer.

Hope it's clear.
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17 Sep 2012, 22:50
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imhimanshu wrote:
Which of the following sets includes ALL of the solutions of x that will satisfy the equation: ?

|x-2|-|x-3|=|x-5|

(A) {-6, -5, 0, 1, 7, 8}
(B) {-4, -2, 0, 10/3, 4, 5}
(C) {-4, 0, 1, 4, 5, 6}
(D) {-1, 10/3, 3, 5, 6, 8}
(E) {-2, -1, 1, 3, 4, 5}

Usually, I'm able to solve such questions, however I couldn't solve this one.
Can someone run pass me through using number line approach.

Thanks
H

Responding to a pm:

|x-2|-|x-3|=|x-5|

First and foremost, 'the difference between the distances' is a concept which is a little harder to work with as compared with 'the sum of distances'. So try to get rid of the negative sign, if possible.

|x-2| = |x-3|+|x-5|

This equation tells you the following: x is a point whose distance from 2 is equal to the sum of its distance from 3 and its distance from 5. Plot the 3 points on the number line and run through the 4 regions.

Attachment:

Ques3.jpg [ 3.83 KiB | Viewed 10372 times ]

Green: For every point to the left of 2, the distance of the point from 2 is less than the distance from 3. So there is no way any point here will satisfy the equation. Distance of x from 2 will always be less than the sum of distances from 3 and 5.

Blue: Between 2 and 3, 3 is farthest from 2 i.e. at a distance of 1 but 5 is at a distance of 2 from 3. So again, no point here can satisfy the equation. Distance of x from 2 will still be less than the sum of distances from 3 and 5.

Black: Now we are going a little farther from 2 and toward 5 so there might be a point where distance from 2 is equal to sum of distances from 3 and 5. Check what happens at 4. You see that 4 satisfies the equation. At 4, the distance of x from 2 is equal to the sum of distances from 3 and 5. After 4, distance of x from 2 will be more than the sum of distances from 3 and 5 till we reach 5.

Red: After 5, as x moves to the right, its distance from 3 as well as 5 increases. So it is possible that the distance of x from 2 is again equal to the sum of distances from 3 and 5. At 5, x is 3 units away from 2 and 2 units away from 3 and 5 together. Check what happens at 6. At 6, x is 4 away from 2 and 4 away from 3 and 5 together. Hence 6 satisfies too. As you move one step to the right of 6, distance from 2 increases by 1 unit and distance from 3 and 5 together increases by 2 units. So the distances will never be the same again.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 30 Sep 2009 Posts: 120 Re: |x-2|-|x-3|=|x-5| [#permalink] ### Show Tags 23 Sep 2012, 00:03 EvaJager wrote: imhimanshu wrote: Which of the following sets includes ALL of the solutions of x that will satisfy the equation: ? |x-2|-|x-3|=|x-5| (A) {-6, -5, 0, 1, 7, 8} (B) {-4, -2, 0, 10/3, 4, 5} (C) {-4, 0, 1, 4, 5, 6} (D) {-1, 10/3, 3, 5, 6, 8} (E) {-2, -1, 1, 3, 4, 5} Usually, I'm able to solve such questions, however I couldn't solve this one. Can someone run pass me through using number line approach. Thanks H The given equation can be rewritten as $$|x-2|=|x-3|+|x-5|,$$ which means that the distance between $$x$$ and $$2$$ is the sum of the distances between $$x$$ and $$3$$ plus the distance between $$x$$ and 5. Using the number line and visualizing the points, we can easily see that $$x$$ cannot be smaller than $$3.$$ If $$x>5,$$ then $$x-2=x-3+x-5,$$ from which $$x=6.$$ We have to choose between answers C and D. If $$3<x<5,$$ then $$x-2=x-3-x+5,$$ which yields $$x=4.$$ Answer C Hi, Please explain in detail why x cannot be smaller than 3... I am not able to understand this. thanks Director Joined: 22 Mar 2011 Posts: 612 WE: Science (Education) Re: |x-2|-|x-3|=|x-5| [#permalink] ### Show Tags 23 Sep 2012, 01:08 abhi398 wrote: EvaJager wrote: imhimanshu wrote: Which of the following sets includes ALL of the solutions of x that will satisfy the equation: ? |x-2|-|x-3|=|x-5| (A) {-6, -5, 0, 1, 7, 8} (B) {-4, -2, 0, 10/3, 4, 5} (C) {-4, 0, 1, 4, 5, 6} (D) {-1, 10/3, 3, 5, 6, 8} (E) {-2, -1, 1, 3, 4, 5} Usually, I'm able to solve such questions, however I couldn't solve this one. Can someone run pass me through using number line approach. Thanks H The given equation can be rewritten as $$|x-2|=|x-3|+|x-5|,$$ which means that the distance between $$x$$ and $$2$$ is the sum of the distances between $$x$$ and $$3$$ plus the distance between $$x$$ and 5. Using the number line and visualizing the points, we can easily see that $$x$$ cannot be smaller than $$3.$$ If $$x>5,$$ then $$x-2=x-3+x-5,$$ from which $$x=6.$$ We have to choose between answers C and D. If $$3<x<5,$$ then $$x-2=x-3-x+5,$$ which yields $$x=4.$$ Answer C Hi, Please explain in detail why x cannot be smaller than 3... I am not able to understand this. thanks The equation is $$|x-2|=|x-3|+|x-5|.$$ Draw the number line and place 2, 3 and 5 on it. If $$x$$ is between 2 and 3, the distance between $$x$$ and 2, which is $$|x-2|,$$ is less than 1, while the distance between $$x$$ and 5, which is $$|x-5|,$$ is greater than 2. So, the left hand side is for sure less than the right hand side of the equality. If $$x$$ is less than 2, the distance between $$x$$ and 2 is the smallest, is both less than distance between $$x$$ and 3 and less than the distance between $$x$$ and 5. Again, equality cannot hold. _________________ PhD in Applied Mathematics Love GMAT Quant questions and running. Senior Manager Joined: 13 Aug 2012 Posts: 464 Concentration: Marketing, Finance GPA: 3.23 Re: |x-2|-|x-3|=|x-5| [#permalink] ### Show Tags 05 Dec 2012, 02:49 9 This post received KUDOS 2 This post was BOOKMARKED 1) Get the checkpoints using the absolute value expressions. |x-2| tells us that x=2 as one checkpoint |x-3| tells us that x=3 as another checkpoint |x-5| tells us that x=5 as last checkpoint 2) Let us test for x where x < 2 |x-2| = |x-5| + |x-3| -(x-2)= -(x-5) - (x-3) -x +2 = -x +5 -x +3 x=6 Invalid since x=6 is not x<2 3) Let us test for x where 2 < x < 3 (x-2) = -(x-5) -(x-3) x-2 = -x+5 -x +3 3x = 10 x = 10/3 Invalid since x=3.3 is not within x = (2,3) 4) Let us test for x where 3 < x < 5 x-2 = x-3 - (x-5) x-2 = 2 x = 4 Valid 5) Let us test for x where x > 5 x-2 = x-3 + x - 5 -x = -6 x=6 Valid 6) Now let's look for 4 and 6 in the sets of answer choices. (A) {-6, -5, 0, 1, 7, 8} (B) {-4, -2, 0, 10/3, 4, 5} (C) {-4, 0, 1, 4, 5, 6} (D) {-1, 10/3, 3, 5, 6, 8} (E) {-2, -1, 1, 3, 4, 5} Only C has both x=4 and x=6. Answer: C _________________ Impossible is nothing to God. Intern Joined: 29 Aug 2012 Posts: 2 Re: |x-2|-|x-3|=|x-5| [#permalink] ### Show Tags 09 Feb 2013, 03:12 Guys.. help me out of this doubt.. |x-2| tells us that x=2 as one checkpoint Let us test for x where x < 2 |x-2| = |x-5| + |x-3| -(x-2)= -(x-5) - (x-3) -x +2 = -x +5 -x +3 x=6 Invalid since x=6 is not x<2 This shows obviously that value of x will not lie in the region x<2. But the Answer choice C contains (-1) as one of the solution in the set. (In fact all answer choices contains a negative value). Is my understanding wrong ?? Math Expert Joined: 02 Sep 2009 Posts: 39662 Re: |x-2|-|x-3|=|x-5| [#permalink] ### Show Tags 09 Feb 2013, 03:18 mani6389 wrote: Guys.. help me out of this doubt.. |x-2| tells us that x=2 as one checkpoint Let us test for x where x < 2 |x-2| = |x-5| + |x-3| -(x-2)= -(x-5) - (x-3) -x +2 = -x +5 -x +3 x=6 Invalid since x=6 is not x<2 This shows obviously that value of x will not lie in the region x<2. But the Answer choice C contains (-1) as one of the solution in the set. (In fact all answer choices contains a negative value). Is my understanding wrong ?? Your doubt is addressed here: x-2-x-3-x-139053.html#p1122254 Hope it helps. _________________ Current Student Joined: 24 Nov 2012 Posts: 176 Concentration: Sustainability, Entrepreneurship GMAT 1: 770 Q50 V44 WE: Business Development (Internet and New Media) Re: |x-2|-|x-3|=|x-5| [#permalink] ### Show Tags 21 Apr 2013, 01:33 fameatop wrote: I believe there is something wrong with question because -4,5 & 8 does not satisfy the inequality. I believe testing values is the best solution for this question. Let me know if i am wrong. Second that.. By test 8,6,5,4 you could arrive at the solution in less than 30 secs. _________________ You've been walking the ocean's edge, holding up your robes to keep them dry. You must dive naked under, and deeper under, a thousand times deeper! - Rumi http://www.manhattangmat.com/blog/index.php/author/cbermanmanhattanprep-com/ - This is worth its weight in gold Economist GMAT Test - 730, Q50, V41 Aug 9th, 2013 Manhattan GMAT Test - 670, Q45, V36 Aug 11th, 2013 Manhattan GMAT Test - 680, Q47, V36 Aug 17th, 2013 GmatPrep CAT 1 - 770, Q50, V44 Aug 24th, 2013 Manhattan GMAT Test - 690, Q45, V39 Aug 30th, 2013 Manhattan GMAT Test - 710, Q48, V39 Sep 13th, 2013 GmatPrep CAT 2 - 740, Q49, V41 Oct 6th, 2013 GMAT - 770, Q50, V44, Oct 7th, 2013 My Debrief - http://gmatclub.com/forum/from-the-ashes-thou-shall-rise-770-q-50-v-44-awa-5-ir-162299.html#p1284542 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7440 Location: Pune, India Re: |x-2|-|x-3|=|x-5| [#permalink] ### Show Tags 21 Apr 2013, 09:02 Transcendentalist wrote: fameatop wrote: I believe there is something wrong with question because -4,5 & 8 does not satisfy the inequality. I believe testing values is the best solution for this question. Let me know if i am wrong. Second that.. By test 8,6,5,4 you could arrive at the solution in less than 30 secs. Usually, plugging in numbers will be the best option but things are a little complicated here. Let me point out one thing: "Which of the following sets includes ALL of the solutions of x that will satisfy the equation: ?" means you need to find ALL the solutions and some members of the set may not be a solution. So say, you check for 8 and see that it doesn't satisfy the equation, it doesn't mean that options (A) and (D) are out of the running. It may be one of the numbers which do not satisfy the equation but the set (A) or (D) may still contain all the solutions. With a lot of different numbers, it could get confusing and complicated. Anyway, it all depends on your comfort with the various methods. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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22 Apr 2013, 02:41
VeritasPrepKarishma wrote:

This equation tells you the following: x is a point whose distance from 2 is equal to the sum of its distance from 3 and its distance from 5. Plot the 3 points on the number line and run through the 4 regions.

Attachment:
Ques3.jpg

Green: For every point to the left of 2, the distance of the point from 2 is less than the distance from 3. So there is no way any point here will satisfy the equation. Distance of x from 2 will always be less than the sum of distances from 3 and 5.

Blue: Between 2 and 3, 3 is farthest from 2 i.e. at a distance of 1 but 5 is at a distance of 2 from 3. So again, no point here can satisfy the equation. Distance of x from 2 will still be less than the sum of distances from 3 and 5.

Black: Now we are going a little farther from 2 and toward 5 so there might be a point where distance from 2 is equal to sum of distances from 3 and 5. Check what happens at 4. You see that 4 satisfies the equation. At 4, the distance of x from 2 is equal to the sum of distances from 3 and 5. After 4, distance of x from 2 will be more than the sum of distances from 3 and 5 till we reach 5.

Red: After 5, as x moves to the right, its distance from 3 as well as 5 increases. So it is possible that the distance of x from 2 is again equal to the sum of distances from 3 and 5. At 5, x is 3 units away from 2 and 2 units away from 3 and 5 together. Check what happens at 6. At 6, x is 4 away from 2 and 4 away from 3 and 5 together. Hence 6 satisfies too. As you move one step to the right of 6, distance from 2 increases by 1 unit and distance from 3 and 5 together increases by 2 units. So the distances will never be the same again.

I didnt get this part at all
I am sorry, could you please elaborate a bit more, or give a link regarding this method ?
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22 Apr 2013, 22:39
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LalaB wrote:
VeritasPrepKarishma wrote:

This equation tells you the following: x is a point whose distance from 2 is equal to the sum of its distance from 3 and its distance from 5. Plot the 3 points on the number line and run through the 4 regions.

Attachment:
Ques3.jpg

Green: For every point to the left of 2, the distance of the point from 2 is less than the distance from 3. So there is no way any point here will satisfy the equation. Distance of x from 2 will always be less than the sum of distances from 3 and 5.

Blue: Between 2 and 3, 3 is farthest from 2 i.e. at a distance of 1 but 5 is at a distance of 2 from 3. So again, no point here can satisfy the equation. Distance of x from 2 will still be less than the sum of distances from 3 and 5.

Black: Now we are going a little farther from 2 and toward 5 so there might be a point where distance from 2 is equal to sum of distances from 3 and 5. Check what happens at 4. You see that 4 satisfies the equation. At 4, the distance of x from 2 is equal to the sum of distances from 3 and 5. After 4, distance of x from 2 will be more than the sum of distances from 3 and 5 till we reach 5.

Red: After 5, as x moves to the right, its distance from 3 as well as 5 increases. So it is possible that the distance of x from 2 is again equal to the sum of distances from 3 and 5. At 5, x is 3 units away from 2 and 2 units away from 3 and 5 together. Check what happens at 6. At 6, x is 4 away from 2 and 4 away from 3 and 5 together. Hence 6 satisfies too. As you move one step to the right of 6, distance from 2 increases by 1 unit and distance from 3 and 5 together increases by 2 units. So the distances will never be the same again.

I didnt get this part at all
I am sorry, could you please elaborate a bit more, or give a link regarding this method ?

Go through this link: http://www.veritasprep.com/blog/2011/01 ... s-part-ii/
It might help.
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06 Jun 2013, 04:59
Manhnip wrote:
Which of the following sets includes ALL of the solutions of x that will satisfy the equation:

$$|x - 2| - |x - 3| = |x - 5|?$$

$$A) {-6, -5, 0, 1, 7, 8}$$
$$B) {-4, -2, 0, 10/3, 4, 5}$$
$$C) {-4, 0, 1, 4, 5, 6}$$
$$D) {-1, 10/3, 3, 5, 6, 8}$$
$$E) {-2, -1, 1, 3, 4, 5}$$

How to approach this kind of problem ???

Merging similar topics.
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13 Jun 2013, 09:52
Why can we not plug in the answer choices into the question stem to solve that way?
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13 Jun 2013, 09:56
WholeLottaLove wrote:
Why can we not plug in the answer choices into the question stem to solve that way?

You can, but you have to be extremely cautious. (and the process would be quite time counsiming)

Which of the following sets includes ALL of the solutions of x that will satisfy the equation: ?

The question asks for ALL the solutions, even if a statement gives you valid options, you cannot be 100% sure that they are ALL the possible options.
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Re: |x-2|-|x-3|=|x-5|   [#permalink] 13 Jun 2013, 09:56

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