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Can someone please elaborate about the part wherein we ascertain x>20..?

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|x - 10| > |x - 30| ???

essentially question means distance b/w a point x and 10 is greater than distance b/w x and 30. => x is closer to 30 than it is to 10. For that to be true, x needs to be greater than the midpoint of 10 and 30, so x needs to be greater than 20.
So question can be inferred as is x > 20?
statement 1 - x can be less that 20 & greater than 20. Not sufficient (cancel DA)
Statement 2 states x>25. in all cases sufficient

Answer B

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Is |x-10| > |x-30|

Here's mathematical try
Square both side (since both sides are positive)
(x-10)^2 > (x-30)^2
(x^2+100-20x)>(x^2+900-60x)
(60x-20x)>(900-100)
40x>800
x>20

Hence, only St. 2 satisfy i.e x>25
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aj3001
Can someone please elaborate about the part wherein we ascertain x>20..?

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  • Draw a number line and mark the points 0, 10 and 30 on the number line.
  • Pick any arbitrary point on the number line and mark it as x.
  • The distance of this point x from point 10 would be |x - 10| and distance of this point x from point 30 would be |x - 30|.
Check Statement 1:
\(x>10\)
  • The arbitrary point x can be anywhere on right side of the 10
  • If point x lies between 10 and the midpoint of 10 and 30 (i.e., 20), we can confidently say |x-10| < |10-30|
  • If point x lies anywhere on the right side of this midpoint (i.e, 20), we can confidently say |x-10| > |10-30|
Therefore, insufficient.

Check Statement 2:
\(x>25\)
  • The arbitrary point x lies anywhere on right side of the 25
  • Earlier, we have deduced that for all points on right side of 20, |x-10| > |x-30|
Therefore, sufficient.  ­
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a191291r

aj3001
Can someone please elaborate about the part wherein we ascertain x>20..?

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  • Draw a number line and mark the points 0, 10 and 30 on the number line.
  • Pick any arbitrary point on the number line and mark it as x.
  • The distance of this point x from point 10 would be |x - 10| and distance of this point x from point 30 would be |x - 30|.
Check Statement 1:
\(x>10\)

  • The arbitrary point x can be anywhere on right side of the 10
  • If point x lies between 10 and the midpoint of 10 and 30 (i.e., 20), we can confidently say |x-10| < |10-30|
  • If point x lies anywhere on the right side of this midpoint (i.e, 20), we can confidently say |x-10| > |10-30|
Therefore, insufficient.

Check Statement 2:
\(x>25\)

  • The arbitrary point x lies anywhere on right side of the 25
  • Earlier, we have deduced that for all points on right side of 20, |x-10| > |x-30|
Therefore, sufficient.  ­
­Great explaination, I tend to struggle whenever ineuqalities and modulus are together in any question. this is neat approach
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AnuK2222

a191291r

aj3001
Can someone please elaborate about the part wherein we ascertain x>20..?

Posted from my mobile device
 



  • Draw a number line and mark the points 0, 10 and 30 on the number line.
  • Pick any arbitrary point on the number line and mark it as x.
  • The distance of this point x from point 10 would be |x - 10| and distance of this point x from point 30 would be |x - 30|.
Check Statement 1:
\(x>10\)



  • The arbitrary point x can be anywhere on right side of the 10
  • If point x lies between 10 and the midpoint of 10 and 30 (i.e., 20), we can confidently say |x-10| < |10-30|
  • If point x lies anywhere on the right side of this midpoint (i.e, 20), we can confidently say |x-10| > |10-30|
Therefore, insufficient.

Check Statement 2:
\(x>25\)



  • The arbitrary point x lies anywhere on right side of the 25
  • Earlier, we have deduced that for all points on right side of 20, |x-10| > |x-30|
Therefore, sufficient.  ­
­Great explaination, I tend to struggle whenever ineuqalities and modulus are together in any question. this is neat approach
­Same! I struggle a lot with moduli and inequalities as well. Youtube playlists of Khan Academy and PatrickJMT on Linear inequalities helped me a lot. Just type it in the search box on Youtube. I could share the links, but I am new to this forum so I am not allowed to post urls.
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Easiest approach:-

Whenever you see |a|=|b| ,
either a=b or a=-b ;
Apply this here ,
Either x -10 > x-30 ( ignore as -10 is always greater than-30)
or x -10 > - (x-30)
x -10 > -x + 30
x> 20
Hence the question asks
is x >20 ?

Option A does not give a definite answer as x can be 15 or 25 .
Option B does give a definite answer as x is greater than 25 means it will obviously be greater than 20.
B is the answer.

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