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# [#21] 2 Min. Challenge : Solutions

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CEO
Joined: 15 Aug 2003
Posts: 3377
[#21] 2 Min. Challenge : Solutions  [#permalink]

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13 Mar 2004, 15:55
2
Rules
1. Time yourself
2. Solve the problem seperately

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

|x тАУ 2| - |x тАУ 3| = |x тАУ 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}

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Director
Joined: 13 Nov 2003
Posts: 766
Location: BULGARIA

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14 Mar 2004, 04:56
I am getting D) as an answer.
Intern
Joined: 25 Jan 2004
Posts: 28
Location: Redwood City, USA

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14 Mar 2004, 06:41
I may be but I am not finding a single set where ALL values satisfy the equation.

D is certainly not right. take 3 for example
| 3- 2| - |3 - 3| =/ |3-5|
1 - 0 =/ 2

The best way then is to see which other sets contain 3 and cross them off. Then try another number such as 0 (shows up in three answer choices) same thing -1 =/ 5.
So this invalidates the other 3 sets because they all hold 0 as a value of X.

Therefore no one set contains values that satisfy the equation.

Time less than 2.. but again, if I did not get an answer for this question then it does not matter -- I still got it wrong:) - unless the question is wrong.

Though I do have a question - should this be solved in another way other than picking numbers?
SVP
Joined: 30 Oct 2003
Posts: 1739
Location: NewJersey USA

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14 Mar 2004, 07:10
Sorry I had to edit the answer

I get D as an aswer.
I did it using process of elimination.

4 6 and 10/3 satisfy the equation. I think the question is worded tricky.
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4163

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14 Mar 2004, 10:46
Does 10/3 really satisfy the equation? Only 4 and 6 do as a matter of fact and only C comprises both of them. Question is sure worded trickily. Why would the solution include a bunch of invalid answers anyways? I don't get it...
_________________

Best Regards,

Paul

Manager
Joined: 10 Mar 2004
Posts: 63
Location: Dallas,TX

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14 Mar 2004, 10:57
|x тАУ 2| - |x тАУ 3| = |x тАУ 5|?

I too think that the answer is 'd'.

Ques. says that all the possible answers should be there in one of the sets.
it does not say that all the elements of one of the sets should satisfy the equation.

I did it like this:

case 1: x<2 . eq. becomes

2-x + x-3 = 5-x => x=6

case 2: 2<=x<3. eq. becomes

x-2 + x-3 = 5-x => x=10/3

case 3: 3<=x<5. eq. becomes

x-2 - x+3 = 5-x => x=4

case 4: x>=5. eq. becomes

x-2-x+3 = x-5 => x=6

I think it covers all the cases.

Time > 2 minutes
cheers
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4163

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14 Mar 2004, 11:02
Plug in 10/3 in the formula

|10/3 - 2| - |10/3 - 3| = |10/3 - 5|
4/3 - 1/3 = 5/3
3/3 = 5/3
Wrong.
10/3 cannot be answer.
Only 4 and 6 work and those are only included in C
Besides, why was 4 not included in set D?
_________________

Best Regards,

Paul

Manager
Joined: 10 Mar 2004
Posts: 63
Location: Dallas,TX

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14 Mar 2004, 11:23
Sorry paul - my bad here.

Though my first 2 options gave an answer but they do not fall in the range I had selected. So they are not the valid answer.

I hope it is fine now.

4 and 6 from case 3 and 4 of my solution.
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4163

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14 Mar 2004, 11:26
Just to add that it took me more than 5 min. to figure out what this question was asking. Found out that possible roots were 4, 6 and 10/3 but when I plugged them in the formula, only 4 and 6 worked and only those were in C.
_________________

Best Regards,

Paul

CEO
Joined: 15 Aug 2003
Posts: 3377
Re: [#21] 2 Min. Challenge : Solutions  [#permalink]

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14 Mar 2004, 16:53
1
1
praetorian123 wrote:
Rules
1. Time yourself
2. Solve the problem seperately

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

|x тАУ 2| - |x тАУ 3| = |x тАУ 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}

Just a little twist All the solutions to the equation are present in ONLY ONE of the answer choices.

Paul wins this one

Well, the reason other solutions were included was to force you guys to consider if there are any other solutions. If you understand absolute values perfectly, you should be able to solve ANY Gmat problem based on those concepts.

We may remove the absolute value if the signs of the expressions are the same ok..lets see..for that, we need to look at intervals.

1. x<2
for all values x<2 , all the three "expressions" are negative.
so, we can remove the absolute value sign
|x тАУ 2| - |x тАУ 3| = |x тАУ 5|
x-2 - x+3 = x-5
x = 1+5 = 6 ;
Now this solution is not possible because we are looking solutions for x<2. so we need to discard this.

2. 2<x<3
For all values 2<x<3 , |x тАУ 2| is positive, but the other two are negative..
So, x- 2 + x-3 = -x+5 [ hope you see what i did here]
3x = 10
x=10/3
Again, 10/3 is not within range of our interval. so discard.

3. 3<x<5
For all values, 3<x<5, |x тАУ 2| and |x тАУ 3| are positive , but |x тАУ 5| is negative.
x-2 - x+3 = -x+5
x = 4 , ok 4 is within the interval ...so its one of our solutions.

4. x>5
Here all the expressions are positive. so, we can directly write
|x тАУ 2| - |x тАУ 3| = |x тАУ 5|
x-2 - x+3 = x-5
x = 1+5 = 6 ; Again, 6 is within the interval, so its our second and final solution.

Finally, our solutions are 4 and 6, which are both present ONLY in Choice C. Thus, C is our answer.
Director
Joined: 03 Jul 2003
Posts: 642
Re: [#21] 2 Min. Challenge : Solutions  [#permalink]

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14 Mar 2004, 18:26
praetorian123 wrote:
...
....
....
Finally, our solutions are 4 and 6, which are both present ONLY in Choice C. Thus, C is our answer.

Nice explanation. Thanks Preat!
Non-Human User
Joined: 09 Sep 2013
Posts: 9429
Re: [#21] 2 Min. Challenge : Solutions  [#permalink]

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31 Dec 2017, 00:49
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Re: [#21] 2 Min. Challenge : Solutions &nbs [#permalink] 31 Dec 2017, 00:49
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# [#21] 2 Min. Challenge : Solutions

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