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Nikhil30
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If |x + 5| <= |2x - 5|, which of the following specifies all the value Updated on: 04 Apr 2021, 00:02
Originally posted by Nikhil30 on 03 Apr 2021, 23:20.
Last edited by Bunuel on 04 Apr 2021, 00:02, edited 1 time in total.
Renamed the topic and edited the question.
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If |x + 5| <= |2x - 5|, which of the following specifies all the value of x ?

A. x >= 10
B. -5 <= x<= 0 or x >= 10
C. x <= -5 or x >= 10
D. x <= 0 or x >= 10
E. none.
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D
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If |x + 5| <= |2x - 5|, which of the following specifies all the value 04 Apr 2021, 02:46
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If |x + 5| <= |2x - 5|, which of the following specifies all the value of x ?

A. x >= 10
B. -5 <= x<= 0 or x >= 10
C. x <= -5 or x >= 10
D. x <= 0 or x >= 10
E. none.
Show more

APPROACH #1

\(|x + 5| \leq |2x - 5|\);

Square both sides: \(x^2 + 10x + 25 \leq 4x^2 - 20x + 25\)

Re-arrange and simplify: \( x^2 - 10x \geq 0\)

Factor x: \( x(x - 10) \geq 0\);

\(x \leq 0\) or \(x \geq 10\)

Answer: D.

APPROACH #2

Transition points of \(|x + 5| \leq |2x - 5|\) are -5 and 5/2. Consider three ranges:

When \(x \leq -5\), then \(x + 5 \leq 0\) and \(2x - 5 \leq 0\), so \(|x + 5| \leq |2x - 5|\) becomes \(-(x + 5) \leq -(2x -5)\). This gives \(x \leq 10\). Since we consider the case when \(x \leq -5\) then the actual solution for this range is \(x \leq -5\) (the overlap of \(x \leq -5\) and \(x \leq 10\) is \(x \leq -5\));

When \( -5 < x < \frac{5}{2}\), then \(x + 5 > 0\) and \(2x - 5 < 0\), so \(|x + 5| \leq |2x - 5|\) becomes \(x + 5 \leq -(2x -5)\). This gives \(x \leq 0\). Since we consider the case when \( -5 < x < \frac{5}{2}\) then the actual solution for this range is \( -5 < x \leq 0\) (the overlap of \(x \leq 0\) and \( -5 < x < \frac{5}{2}\) is \( -5 < x \leq 0\));


When \(x \geq \frac{5}{2}\), then \(x + 5 > 0\) and \(2x - 5 \geq 0\), so \(|x + 5| \leq |2x - 5|\) becomes \(x + 5 \leq 2x -5\). This gives \(x \leq 10\). Since we consider the case when \(x \leq -5\) then the actual solution for this range is \(x \geq 10\) (the overlap of \(x \geq 10\) and \(x \geq \frac{5}{2}\) is \(x \geq 10\)).

The three ranges \(x \leq -5\), \( -5 < x \leq 0\) and \(x \geq 10\) give the final range of x as \(x \leq 0\) or \(x \geq 10\).

Answer: D.

Hope it's clear.
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DreamerHBS
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If |x + 5| <= |2x - 5|, which of the following specifies all the value 04 Apr 2021, 00:31
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Use options. It would hardly take a minute.
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If |x + 5| <= |2x - 5|, which of the following specifies all the value 04 Apr 2021, 00:38
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Use options. It would hardly take a minute.
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hey , i did solved the question , i am getting only x>=10 , not getting x<=0 , thats why i have posted.

please come forward with your answer , rather just asking me to solve with the help pf option , if i could have solved the question at first , i wouldn't have posted
here . :)
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If |x + 5| <= |2x - 5|, which of the following specifies all the value 04 Apr 2021, 00:47
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Nikhil30
DreamerHBS
Use options. It would hardly take a minute.


hey , i did solved the question , i am getting only x>=10 , not getting x<=0 , thats why i have posted.

please come forward with your answer , rather just asking me to solve with the help pf option , if i could have solved the question at first , i wouldn't have posted
here . :)
Show more

|x+5|<=|2x-5|
case 1) both sides +
x+5 <=2x-5
x-2x+5+5<=0
-x+10<=0---> x>=10

case 2) both sides negative
-x-5<= -2x+5
-x+2x-5-5<=0
x-10<=0
x<=10

case 3) LHS positive and RHS negative
x+5<= -2x+5
x+2x+5-5<=0
3x<=0---> x<=0
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If |x + 5| <= |2x - 5|, which of the following specifies all the value 04 Apr 2021, 01:53
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Nikhil30
DreamerHBS
Use options. It would hardly take a minute.


hey , i did solved the question , i am getting only x>=10 , not getting x<=0 , thats why i have posted.

please come forward with your answer , rather just asking me to solve with the help pf option , if i could have solved the question at first , i wouldn't have posted
here . :)

|x+5|<=|2x-5|
case 1) both sides +
x+5 <=2x-5
x-2x+5+5<=0
-x+10<=0---> x>=10

case 2) both sides negative
-x-5<= -2x+5
-x+2x-5-5<=0
x-10<=0
x<=10

case 3) LHS positive and RHS negative
x+5<= -2x+5
x+2x+5-5<=0
3x<=0---> x<=0
Show more


hey , can we solve this question by squaring both side ,
as |a| <= |B| , can also be write as a^2<=b^2

and than proceed , by that i am getting only x>=10. can you please tell me whats wrong with that method.
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If |x + 5| <= |2x - 5|, which of the following specifies all the value 04 Apr 2021, 03:26
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Asked: If |x + 5| <= |2x - 5|, which of the following specifies all the value of x ?

|x + 5| <= |2x - 5|

Case 1: x<=-5
|x + 5| <= |2x - 5|
-x-5 < = 5-2x
x <= 10
x<=-5<=10
The inequality is valid for all x such that x<=-5

Case 2: -5<x<=2.5
|x + 5| <= |2x - 5|
x+5 <= 5-2x
3x <=0
The inequality is valid for all x such that -5<x<=0

Case 3: x>2.5
|x + 5| <= |2x - 5|
x+5 <= 2x-5
x >=10
The inequality is valid for all x such that x >=10

Combing the 3 results
x<=0 or x>=10

IMO D
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If |x + 5| <= |2x - 5|, which of the following specifies all the value 04 Apr 2021, 03:34
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Nikhil30
If |x + 5| <= |2x - 5|, which of the following specifies all the value of x ?

A. x >= 10
B. -5 <= x<= 0 or x >= 10
C. x <= -5 or x >= 10
D. x <= 0 or x >= 10
E. none.

APPROACH #1

\(|x + 5| \leq |2x - 5|\);

Square both sides: \(x^2 + 10x + 25 \leq 4x^2 - 20x + 25\)

Re-arrange and simplify: \( x^2 - 10x \geq 0\)

Factor x: \( x(x - 10) \geq 0\);

\(x \leq 0\) or \(x \geq 10\)

Answer: D.

APPROACH #2

Transition points of \(|x + 5| \leq |2x - 5|\) are -5 and 5/2. Consider three ranges:

When \(x \leq -5\), then \(x + 5 \leq 0\) and \(2x - 5 \leq 0\), so \(|x + 5| \leq |2x - 5|\) becomes \(-(x + 5) \leq -(2x -5)\). This gives \(x \leq 10\). Since we consider the case when \(x \leq -5\) then the actual solution for this range is \(x \leq -5\) (the overlap of \(x \leq -5\) and \(x \leq 10\) is \(x \leq -5\));

When \( -5 < x < \frac{5}{2}\), then \(x + 5 > 0\) and \(2x - 5 < 0\), so \(|x + 5| \leq |2x - 5|\) becomes \(x + 5 \leq -(2x -5)\). This gives \(x \leq 0\). Since we consider the case when \( -5 < x < \frac{5}{2}\) then the actual solution for this range is \( -5 < x \leq 0\) (the overlap of \(x \leq 0\) and \( -5 < x < \frac{5}{2}\) is \( -5 < x \leq 0\));


When \(x \geq \frac{5}{2}\), then \(x + 5 > 0\) and \(2x - 5 \geq 0\), so \(|x + 5| \leq |2x - 5|\) becomes \(x + 5 \leq 2x -5\). This gives \(x \leq 10\). Since we consider the case when \(x \leq -5\) then the actual solution for this range is \(x \geq 10\) (the overlap of \(x \geq 10\) and \(x \geq \frac{5}{2}\) is \(x \geq 10\)).

The three ranges \(x \leq -5\), \( -5 < x \leq 0\) and \(x \geq 10\) give the final range of x as \(x \leq 0\) or \(x \geq 10\).

Answer: D.

Hope it's clear.
Show more

hey i got your point thanks for your reply,
i am again getting confused here ,
x(x-10) >=0,
it means , x>=0 and x>=10,
would you please explain how you have written x<=0. (in approach 1)
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If |x + 5| <= |2x - 5|, which of the following specifies all the value 04 Apr 2021, 05:13
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Nikhil30
If |x + 5| <= |2x - 5|, which of the following specifies all the value of x ?

A. x >= 10
B. -5 <= x<= 0 or x >= 10
C. x <= -5 or x >= 10
D. x <= 0 or x >= 10
E. none.

APPROACH #1

\(|x + 5| \leq |2x - 5|\);

Square both sides: \(x^2 + 10x + 25 \leq 4x^2 - 20x + 25\)

Re-arrange and simplify: \( x^2 - 10x \geq 0\)

Factor x: \( x(x - 10) \geq 0\);

\(x \leq 0\) or \(x \geq 10\)

Answer: D.

APPROACH #2

Transition points of \(|x + 5| \leq |2x - 5|\) are -5 and 5/2. Consider three ranges:

When \(x \leq -5\), then \(x + 5 \leq 0\) and \(2x - 5 \leq 0\), so \(|x + 5| \leq |2x - 5|\) becomes \(-(x + 5) \leq -(2x -5)\). This gives \(x \leq 10\). Since we consider the case when \(x \leq -5\) then the actual solution for this range is \(x \leq -5\) (the overlap of \(x \leq -5\) and \(x \leq 10\) is \(x \leq -5\));

When \( -5 < x < \frac{5}{2}\), then \(x + 5 > 0\) and \(2x - 5 < 0\), so \(|x + 5| \leq |2x - 5|\) becomes \(x + 5 \leq -(2x -5)\). This gives \(x \leq 0\). Since we consider the case when \( -5 < x < \frac{5}{2}\) then the actual solution for this range is \( -5 < x \leq 0\) (the overlap of \(x \leq 0\) and \( -5 < x < \frac{5}{2}\) is \( -5 < x \leq 0\));


When \(x \geq \frac{5}{2}\), then \(x + 5 > 0\) and \(2x - 5 \geq 0\), so \(|x + 5| \leq |2x - 5|\) becomes \(x + 5 \leq 2x -5\). This gives \(x \leq 10\). Since we consider the case when \(x \leq -5\) then the actual solution for this range is \(x \geq 10\) (the overlap of \(x \geq 10\) and \(x \geq \frac{5}{2}\) is \(x \geq 10\)).

The three ranges \(x \leq -5\), \( -5 < x \leq 0\) and \(x \geq 10\) give the final range of x as \(x \leq 0\) or \(x \geq 10\).

Answer: D.

Hope it's clear.

hey i got your point thanks for your reply,
i am again getting confused here ,
x(x-10) >=0,
it means , x>=0 and x>=10,
would you please explain how you have written x<=0. (in approach 1)
Show more

The product x(x - 10) will be non-negative (\(\geq 0\)) in the following two cases:

  • both x and x - 10 are non-negative (\(\geq 0\)). Here notice that x - 10 is always less than x, so if x - 10 is positive, x will also be positive. Thus, the product will be non-negative if x -10 >= 0, or if x >= 10.
  • both x and x - 10 are non-positive (\(\leq 0\)). Here notice that x is always more than x - 10, so if x is negative, x - 10 will also be negative. Thus, the product will be non-positive if x <= 0.

Else you can use the method explained here: https://gmatclub.com/forum/solving-quad ... 70528.html

Also, you might find the following links useful:

9. Inequalities



10. Absolute Value



For more check Ultimate GMAT Quantitative Megathread



Hope it helps.
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If |x + 5| <= |2x - 5|, which of the following specifies all the value 05 Apr 2021, 12:05
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APPROACH #1

\(|x + 5| \leq |2x - 5|\);

Square both sides: \(x^2 + 10x + 25 \leq 4x^2 - 20x + 25\)

Re-arrange and simplify: \( x^2 - 10x \geq 0\)

Factor x: \( x(x - 10) \geq 0\);

\(x \leq 0\) or \(x \geq 10\)

Answer: D.

Hey Bunuel, in approach 1 when you rearrange wouldn't it be \( 3x^2 - 10x \geq 0\) and wouldn't answer choices change a bit because of it ?
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If |x + 5| <= |2x - 5|, which of the following specifies all the value 05 Apr 2021, 13:37
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APPROACH #1

\(|x + 5| \leq |2x - 5|\);

Square both sides: \(x^2 + 10x + 25 \leq 4x^2 - 20x + 25\)

Re-arrange and simplify: \( x^2 - 10x \geq 0\)

Factor x: \( x(x - 10) \geq 0\);

\(x \leq 0\) or \(x \geq 10\)

Answer: D.

Hey Bunuel, in approach 1 when you rearrange wouldn't it be \( 3x^2 - 10x \geq 0\) and wouldn't answer choices change a bit because of it ?
Show more

Here is how it goes:

\(x^2 + 10x + 25 \leq 4x^2 - 20x + 25\);

\(0 \leq 3x^2 - 30x\)

\(0 \leq x^2 - 10x\)

So, you should get \(0 \leq x^2 - 10x\) (\( x^2 - 10x \geq 0\)) not \( 3x^2 - 10x \geq 0\)
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If |x + 5| <= |2x - 5|, which of the following specifies all the value 17 Jun 2026, 06:02
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Can anyone explain why in approach 1, after factorizing x(x-10) =>0, we took x <=0?
Bunuel


APPROACH #1

\(|x + 5| \leq |2x - 5|\);

Square both sides: \(x^2 + 10x + 25 \leq 4x^2 - 20x + 25\)

Re-arrange and simplify: \( x^2 - 10x \geq 0\)

Factor x: \( x(x - 10) \geq 0\);

\(x \leq 0\) or \(x \geq 10\)

Answer: D.

APPROACH #2

Transition points of \(|x + 5| \leq |2x - 5|\) are -5 and 5/2. Consider three ranges:

When \(x \leq -5\), then \(x + 5 \leq 0\) and \(2x - 5 \leq 0\), so \(|x + 5| \leq |2x - 5|\) becomes \(-(x + 5) \leq -(2x -5)\). This gives \(x \leq 10\). Since we consider the case when \(x \leq -5\) then the actual solution for this range is \(x \leq -5\) (the overlap of \(x \leq -5\) and \(x \leq 10\) is \(x \leq -5\));

When \( -5 < x < \frac{5}{2}\), then \(x + 5 > 0\) and \(2x - 5 < 0\), so \(|x + 5| \leq |2x - 5|\) becomes \(x + 5 \leq -(2x -5)\). This gives \(x \leq 0\). Since we consider the case when \( -5 < x < \frac{5}{2}\) then the actual solution for this range is \( -5 < x \leq 0\) (the overlap of \(x \leq 0\) and \( -5 < x < \frac{5}{2}\) is \( -5 < x \leq 0\));


When \(x \geq \frac{5}{2}\), then \(x + 5 > 0\) and \(2x - 5 \geq 0\), so \(|x + 5| \leq |2x - 5|\) becomes \(x + 5 \leq 2x -5\). This gives \(x \leq 10\). Since we consider the case when \(x \leq -5\) then the actual solution for this range is \(x \geq 10\) (the overlap of \(x \geq 10\) and \(x \geq \frac{5}{2}\) is \(x \geq 10\)).

The three ranges \(x \leq -5\), \( -5 < x \leq 0\) and \(x \geq 10\) give the final range of x as \(x \leq 0\) or \(x \geq 10\).

Answer: D.

Hope it's clear.
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If |x + 5| <= |2x - 5|, which of the following specifies all the value 17 Jun 2026, 06:41
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Can anyone explain why in approach 1, after factorizing x(x-10) =>0, we took x <=0?

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This is explained HERE.

Another approach. We have:

\(x(x-10) \geq 0\)

The transition points, in ascending order, are \(x = 0\) and \(x = 10\). This gives us three ranges:

\(x < 0\)

\(0 < x < 10\)

\(x > 10\)

Next, test an extreme value for x: if x is a large enough number, say 100, then both factors will be positive, resulting in a positive value for the whole expression. Therefore, when \(x > 10\), the expression is positive.

Now, here’s the trick: since the expression is positive in the 3rd range, it will be negative in the 2nd range and positive again in the 1st range, following the pattern: +, -, +.

Thus, the expression is positive for \(x < 0\) and \(x > 10\). Since the inequality is \(\geq 0\), we also include \(x = 0\) and \(x = 10\).

Therefore, \(x \leq 0\) or \(x \geq 10\).
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