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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

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 _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

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 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.

Waiting for reply _________________

If you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS. Kudos always maximizes GMATCLUB worth-Game Theory

If you have any question regarding my post, kindly pm me or else I won't be able to reply

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 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.

Waiting for reply

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. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

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 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.

Waiting for reply

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.

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 5702 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. _________________

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.

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.

Re: |x-2|-|x-3|=|x-5| [#permalink]
05 Dec 2012, 01:49

6

This post received KUDOS

1

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.

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).

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).

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

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. _________________

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 ? _________________

Happy are those who dream dreams and are ready to pay the price to make them come true

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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