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# For how many values of x, is |||x - 5| -10| -5| = 2?

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Veritas Prep GMAT Instructor
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For how many values of x, is |||x - 5| -10| -5| = 2?  [#permalink]

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Updated on: 02 Feb 2014, 23:42
11
52
00:00

Difficulty:

95% (hard)

Question Stats:

49% (02:27) correct 51% (02:14) wrong based on 689 sessions

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Question of the Day:

For how many values of x, is |||x - 5| -10| -5| = 2?

(A) 0
(B) 2
(C) 4
(D) 8
(E) More than 8

(Those ls are mods)
(Earn kudos if you solve it in under 2 mins.)
Note: This question is beyond GMAT level. Nevertheless, it helps you understand mods and graphs and hence, is great for practice.)

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Karishma
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Originally posted by VeritasKarishma on 27 Oct 2010, 08:45.
Last edited by Bunuel on 02 Feb 2014, 23:42, edited 2 times in total.
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Re: Question of the Day  [#permalink]

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28 Oct 2010, 05:48
21
7
Here goes the solution.

Once you are comfortable with graphs, it should not take more than 2-3 mins to figure out the answer.
Attachment:
Mod Ques.pdf [206.83 KiB]

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Karishma
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Re: Question of the Day  [#permalink]

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30 Oct 2010, 01:45
10
8
|x-5|=0 .. Exactly 1 solution at x=5

This is strt fwd, |Expression|=0 means Expression=0

||x-5|-10|=0 .. Exactly 2 solutions at x=5-10=-5 & x=5+10=15

Now ||Expression|-10|=0, implies Expression must be either +10 or -10
Hence x-5=+10,-10
Hence x=15,-5

|||x-5|-10|-5|=0 .. Exactly 4 solutions at x=-5-5,-5+5,15-5,15+5 OR x=-10,0,10,20

Same concept as above
x=15+5,15-5 & x=-5-5,-5+5
Hence x=-10,0,10,20

Finally to say |Expression|=2, means either expression is +2, or -2
Hence x=-10+2,-10-2,0+2,0-2,10+2,10-2,20+2,20-2

Thats how you get 8 solutions algebraicly w/o graphs
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Re: Question of the Day  [#permalink]

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27 Oct 2010, 14:32
9
1
Easiest way is to graph it, which will immediately show exactly 8 solutions. Alternatively :
|x-5|=0 .. Exactly 1 solution at x=5
||x-5|-10|=0 .. Exactly 2 solutions at x=5-10=-5 & x=5+10=15
|||x-5|-10|-5|=0 .. Exactly 4 solutions at x=-5-5,-5+5,15-5,15+5 OR x=-10,0,10,20

At this point it is very easy to imagine what the graph looks like, it'll be 3 triangular humps touching the x-axis at the above points and going to +inf as x goes below -10 or above 20.
All the lines are at 45 degrees, so the peaks will be at y=5

Now |||x-5|-10|-5|=2 is just equivalent to shifting the x-axis up by 2 units, or counting the number of intersections with the y=2 line. Given the peaks are at y=5, there will be the maximum possible number of intersections, which is 8 in the following regions :

Below -10
Between -10 & -5
Between -5 and 0
Between 0 & 5
Between 5 & 10
Betweem 10 & 15
Between 15 & 20
Above 20

As already mentioned, if you just graph it out, this should be straight forward to see
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Re: Question of the Day  [#permalink]

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27 Oct 2010, 15:24
3
2
graphs are the best approach.

$$|||x-5|-10|-5|= y$$

for y = 0 we have x = -10,0,10,20

also y>0 => the graph lies above x axis and the graph touches x axis 4 times.

=> y=2 will touch the graph 2*4 times = 8 times. ( we just have to check whether the peaks of y values are below y=2 or not. Eg if the question is $$|||x-5|-10|-5|= 6$$ then the answer will be 2)

Hence 8
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Re: Question of the Day  [#permalink]

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27 Oct 2010, 18:52
1
2
shrouded1 and gurpreet: Excellent work! There will be 8 values of x.
shrouded1: I like your alternate solution as well.
gurpreet: You are right! If it were 6 instead of 2, there would be only 2 solutions.

I will provide the graphical solution in the morning. Meanwhile, if someone else wants to try too, but is unsure of the graphical approach, check out my yesterday's link where I have solved a similar question using graphs. See if you get the method on your own.

http://gmatclub.com/forum/ps-triple-modul-1185.html#p807380
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Karishma
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Re: Question of the Day  [#permalink]

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29 Oct 2010, 11:37
shrouded1 wrote:
|x-5|=0 .. Exactly 1 solution at x=5
||x-5|-10|=0 .. Exactly 2 solutions at x=5-10=-5 & x=5+10=15
|||x-5|-10|-5|=0 .. Exactly 4 solutions at x=-5-5,-5+5,15-5,15+5 OR x=-10,0,10,20

Will really appreciate if u can tell us how you got the number of solutions in each step mathematically (and not graphically)??

Cheers,
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Re: Question of the Day  [#permalink]

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14 Jun 2011, 02:21
10
1
|x-5| = a
|y-10|=b
|b-5|= 2
thus b = 7 or 3.

y = 17,3,13,7.

x = 2 values for each of the values of y. Hence 8.
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Re: Question of the Day  [#permalink]

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14 Jun 2011, 03:02
1
+1 Thanks for a good question
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Re: Question of the Day  [#permalink]

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14 Jun 2011, 07:17
Hey, I got this one in 1:08
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Re: Question of the Day  [#permalink]

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26 May 2012, 12:33
3
1
VeritasPrepKarishma wrote:
Question of the Day:

Q. For how many values of x, is lllx - 5l -10l -5l = 2?
(A) 0
(B) 2
(C) 4
(D) 8
(E) More than 8

(Those ls are mods)
(Earn kudos if you solve it in under 2 mins.)
Note: This question is beyond GMAT level. Nevertheless, it helps you understand mods and graphs and hence, is great for practice.)

I think its 8

|||x-5|-10|-5| =2
let |x-5| = a which makes above
||a-10|-5| =2

let |a-10| = b which makes
|b-5| = 2

now for the above b can take 3, 7
for every b =3 a can have 13, 7
and for b = 7 a can have 17 and 3

so 'a' has four solutions 13, 7, 17 and 3

for a = 13; x has 18 or -8 thus has 2 for every combination hence 4x2 = 8
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Re: Question of the Day  [#permalink]

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26 May 2012, 12:55
1
For equation mentioned by Gurpreet i.e. |||x-5|-10|-5| = 6 can you please explain, without graph, how you were able to get answer as 2.

Following one of the post with mathematical values

|||x-5|-10|-5| =6
let |x-5| = a which makes above
||a-10|-5| =6

let |a-10| = b which makes
|b-5| = 6

now for the above b can take 1, 11
for every b =1 a can have 9, 11
and for b = 11 a can have 1 and 21

so 'a' has four solutions 1, 9, 11 and 21

Thus for a as 1 x can take 4,6
Thus for a as 9 x can take 4 , 14
Thus for a as 11 x can take 6 , 14
Thus for a as 21 x can take 16 , 24

Total - 5 different values.
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Re: Question of the Day  [#permalink]

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27 May 2012, 10:56
hey really super question! especially like the substitution approaches discussed in this thread. Also this question exposed me to a totally different question type.
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Re: Question of the Day  [#permalink]

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27 May 2012, 11:00
just studied the graphical method, its a perfect compliment to the numberline methods to solve absolute value equations i think!
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Re: Question of the Day  [#permalink]

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27 May 2012, 11:09
@ Karishma in problems such as |x-2| >4 we plot the equation on the number line (at x= +2, mark units to the left and to the right) and we get the ans. When do you recommend we opt for this graphical method that you suggest? Could you give us a thumb rule so to say?
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Re: Question of the Day  [#permalink]

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31 May 2012, 07:08
I got this in less than 2 min, without graphs.

nice question.
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Re: Question of the Day  [#permalink]

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05 Jun 2012, 05:35
vibhav wrote:
@ Karishma in problems such as |x-2| >4 we plot the equation on the number line (at x= +2, mark units to the left and to the right) and we get the ans. When do you recommend we opt for this graphical method that you suggest? Could you give us a thumb rule so to say?

Number line approach, which you can use in some situations, is just a short cut to the graphical approach. Think of the number line as the x axis and that there is the y axis at x = 0. You can solve |x-2| >4 using the graphical approach too but to solve |||x-5|-10|-5| =2, you need to use the complete graphical approach.
You can use the number line only in cases where simple mod terms are added/subtracted (though subtraction is more complicated).
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Re: Question of the Day  [#permalink]

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06 Jun 2012, 19:04
Hi Karishma,
I can solve such questions using Algebraic approach, however, I am keen to understand the graphical approach as well. I have also studied your response in the attached pdf, but I am not too sure whether I have understood the whole process of graphs or not. Could you please point me to your forum posts, where you have detailed the graphical reasoning, or any good resource of understanding graphs.
The expected graphical solution is too complex for me to draw.

Thanks
H

VeritasPrepKarishma wrote:
vibhav wrote:
@ Karishma in problems such as |x-2| >4 we plot the equation on the number line (at x= +2, mark units to the left and to the right) and we get the ans. When do you recommend we opt for this graphical method that you suggest? Could you give us a thumb rule so to say?

Number line approach, which you can use in some situations, is just a short cut to the graphical approach. Think of the number line as the x axis and that there is the y axis at x = 0. You can solve |x-2| >4 using the graphical approach too but to solve |||x-5|-10|-5| =2, you need to use the complete graphical approach.
You can use the number line only in cases where simple mod terms are added/subtracted (though subtraction is more complicated).
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Re: Question of the Day  [#permalink]

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02 Jul 2012, 10:36
1
1
Not using graphs:
Take |x-5| as "A" , |A-10| as "B"
then |B-5| = 2 would have 2 answers 7 and 3 for B
Now, take |A-10|=7, would give two solutions for A and |A-10|=3 would give 2 solutions for A
For each of these 4 solutions equate for |x-5| which would give 2 solutions for each equation and hence a total of 4*2 = 8 solution for x
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Re: Question of the Day  [#permalink]

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04 Jul 2012, 00:10
imhimanshu wrote:
Could you please point me to your forum posts, where you have detailed the graphical reasoning, or any good resource of understanding graphs.

Check this out: http://www.veritasprep.com/blog/2011/01 ... h-to-mods/
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Re: Question of the Day   [#permalink] 04 Jul 2012, 00:10

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