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17 Sep 2012, 07:29
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
47% (02:38) correct
53%
(02:18)
wrong
based on 1137
sessions
History
Date
Time
Result
Not Attempted Yet
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}
|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}
05 Dec 2012, 02:49
Kudos
Bookmarks
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
|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
17 Sep 2012, 22:50
Kudos
Bookmarks
imhimanshu
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 37021 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.
