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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

Answer is 8.

As already mentioned, if you just graph it out, this should be straight forward to see
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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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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.

https://gmatclub.com/forum/ps-triple-modul-1185.html#p807380
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shrouded1

|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,
R J
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|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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VeritasPrepKarishma
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
answer D
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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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@ 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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@ 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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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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Eshaninan
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

Yes, you can do it this way too. But you have to be careful of the values given e.g.
In instead of 2, we have 0, you will get 4 solutions since |B-5| = 0 will give you only 1 value of B. Each of the subsequent values will give you 2 values each. I am not a big fan of multiple variables so I don't use them. Graphical approach just seems more intuitive to me.
Anyway, as long as you know what you are doing, you can use whatever method you like.
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VeritasPrepKarishma
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.)


Hi Karishma,
Your sig. in this post contains a link that says "Download the Veritas Prep GMAT On Demand App Free https://itunes.apple.com/us/app/gmat-on- ... 60224?mt=8".

But I think veritas Prep on Demand course is NOT available FREE to download on apple product.

Your thoughts on this..!!
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debayan222

Hi Karishma,
Your sig. in this post contains a link that says "Download the Veritas Prep GMAT On Demand App Free https://itunes.apple.com/us/app/gmat-on- ... 60224?mt=8".

But I think veritas Prep on Demand course is NOT available FREE to download on apple product.

Your thoughts on this..!!

The app is free. You can download it. The first lesson which discusses Test structure, scoring, pacing, myths, Veritas methodologies, typical Veritas class etc is free. Thereafter, you can choose the modules you want to work on. You will obviously need to pay for them. You can buy the modules you want instead of buying the whole course.
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shrouded1
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

Answer is 8.

As already mentioned, if you just graph it out, this should be straight forward to see



Can you please draw a graph for this question. very weak in this topic :(
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shrouded1
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

Answer is 8.

As already mentioned, if you just graph it out, this should be straight forward to see



Can you please draw a graph for this question. very weak in this topic :(

Here is a graph of |||x - 5| -10| -5| = 2:
Attachment:
MSP55621c3692c156b65ec00004239b89491b925i9.gif
MSP55621c3692c156b65ec00004239b89491b925i9.gif [ 5.74 KiB | Viewed 5254 times ]

Theory on Abolute Values: math-absolute-value-modulus-86462.html
Absolute value tips: absolute-value-tips-and-hints-175002.html

DS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=58

Hard set on Abolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html

Hope it helps.
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Under 2min:
|||x - 5| -10| -5| = 2

||x - 5| -10| = 7 or -3 (two possible values)

|x - 5| = 17, -3, 7 or -13 (four possible values)

Now, the final answers, based on two possible outcomes:
|x - 5| > 0 and |x - 5| < 0

if |x - 5| > 0
x - 5 = (17 or - 3 or 7 or -13)
then x would be equal to = 22, 2, 12 and -8.

if |x - 5| < 0
- x + 5 = (17 or - 3 or 7 or -13)
then x would be equal to = -12, 8, -2 and 18.

So, we have 8 possible outcomes: -12, -8, -2, 2, 8, 12, 18 and 22.

Hope I could help.
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Squaring the terms to eliminate the modulus can yield the answer. I squared three times, so 2*2*2 = 8 will be the power of x. Thus, there are eight solutions.
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