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Bunuel
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Is x > 1? (1) |x − 4| < −2x + 3 (2) |x − 4| > |x + 4| 01 Dec 2021, 03:00
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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95% (hard)

Question Stats:

37% (02:26) correct 63% (02:36) wrong based on 100 sessions

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Is x > 1?

(1) |x − 4| < −2x + 3

(2) |x − 4| > |x + 4|
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D
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Is x > 1? (1) |x − 4| < −2x + 3 (2) |x − 4| > |x + 4| 01 Dec 2021, 20:14
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Bunuel
Is x > 1?

(1) |x − 4| < −2x + 3

(2) |x − 4| > |x + 4|
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(1) |x − 4| < −2x + 3
\(-2x+3>|x-4|\geq 0\)
So both sides are non negative, and we can square them without changing signs.
\((-2x+3)^2>(x-4)^2……..4x^2-12x+9>x^2-8x+16……..3x^2-4x-7>0……(3x-7)(x+1)>0\)
That is either x<-1 or x>7/3.
But we know that -2x+3>0 or x<3/2.
This, only range x<-1 is left and our answer is NO.
Sufficient

(2) |x − 4| > |x + 4|
Again square both sides.
\(x^2-8x+16>x^2+8x+16……..16x<0…….x<0\)
Sufficient


D
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Is x > 1? (1) |x − 4| < −2x + 3 (2) |x − 4| > |x + 4| 12 Jun 2023, 13:44
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chetan2u
Bunuel
Is x > 1?

(1) |x − 4| < −2x + 3

(2) |x − 4| > |x + 4|


(1) |x − 4| < −2x + 3
\(-2x+3>|x-4|\geq 0\)
So both sides are non negative, and we can square them without changing signs.
\((-2x+3)^2>(x-4)^2……..4x^2-12x+9>x^2-8x+16……..3x^2-4x-7>0……(3x-7)(x+1)>0\)
That is either x<-1 or x>7/3.
But we know that -2x+3>0 or x<3/2.
This, only range x<-1 is left and our answer is NO.
Sufficient

(2) |x − 4| > |x + 4|
Again square both sides.
\(x^2-8x+16>x^2+8x+16……..16x<0…….x<0\)
Sufficient


D
Show more


chetan2u x is not greater than 7/3, x is less than 7/3. Actually the range of solutions of the inequality in statement 1 is x<7/3 or x<-1

Bunuel in similar inequalities to |x − 4| <> −2x + 3 , we usually get as a solution either a conjonction (a situation where the eventual solutions will be inside a specific interval), or a disconjonction (solutions being in two opposite parts of the number line) but in this case we have both solutions looking in the same direction. Is that the reason why we have to check the correctness of each of the solutions?
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Is x > 1? (1) |x − 4| < −2x + 3 (2) |x − 4| > |x + 4| Updated on: 12 Jun 2023, 19:35
Originally posted by chetan2u on 12 Jun 2023, 19:24.
Last edited by chetan2u on 12 Jun 2023, 19:35, edited 1 time in total.
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Bunuel
Is x > 1?

(1) |x − 4| < −2x + 3

(2) |x − 4| > |x + 4|


(1) |x − 4| < −2x + 3
\(-2x+3>|x-4|\geq 0\)
So both sides are non negative, and we can square them without changing signs.
\((-2x+3)^2>(x-4)^2……..4x^2-12x+9>x^2-8x+16……..3x^2-4x-7>0……(3x-7)(x+1)>0\)
That is either x<-1 or x>7/3.
But we know that -2x+3>0 or x<3/2.
This, only range x<-1 is left and our answer is NO.
Sufficient

(2) |x − 4| > |x + 4|
Again square both sides.
\(x^2-8x+16>x^2+8x+16……..16x<0…….x<0\)
Sufficient


D


chetan2u x is not greater than 7/3, x is less than 7/3. Actually the range of solutions of the inequality in statement 1 is x<7/3 or x<-1

Bunuel in similar inequalities to |x − 4| <> −2x + 3 , we usually get as a solution either a conjonction (a situation where the eventual solutions will be inside a specific interval), or a disconjonction (solutions being in two opposite parts of the number line) but in this case we have both solutions looking in the same direction. Is that the reason why we have to check the correctness of each of the solutions?
Show more

Hi

The solution is correct as given. The solution mentions that x>7/3 is not a possible range.

Substitute x=1 in the equation and you will realise x<7/3 is incorrect

Actually whatever you get x<7/3 or x>7/3 is a flawed solution and therefore is discarded in the end to get range of x as only x<-1.
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Is x > 1? (1) |x − 4| < −2x + 3 (2) |x − 4| > |x + 4| 12 Jun 2023, 19:26
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(1) Only
Consider two scenarios.

If x-4>=0, or x >= 4, we have
x-4<-2x+3
=> 3x < 7, or x<7/3
Because x>=4 and x<7/3 cannot both be true, no solution when x-4>=0

If x-4<0, or x<4, we have
4-x<-2x+3
=> x<-1
Because x<4 and x<-1, its solution is x<-1

We can unequivocally answer "No" to the question.
Sufficient.

(2) Only
Consider three scenarios.

If x>=4
x-4>x+4 => -4 > 4, which has no solution.

If -4<=x<4
4-x>x+4 => x <0
Because -4<=x<4 and x<0, its solution is -4<=x<0

If x<-4
4-x>-x-4 => 4>-4, which is always true.
Its solution is thus x<-4

Combining -4<=x<0 and x<-4
We have x<0 as the final solution and can also answer the question with a resounding "No".

Sufficient.

The answer is thus (D).
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Is x > 1? (1) |x − 4| < −2x + 3 (2) |x − 4| > |x + 4| 13 Jun 2023, 21:00
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Chetan2u
Hi

The solution is correct as given. The solution mentions that x>7/3 is not a possible range.

Substitute x=1 in the equation and you will realise x<7/3 is incorrect

Actually whatever you get x<7/3 or x>7/3 is a flawed solution and therefore is discarded in the end to get range of x as only x<-1.
Show more

chetan2u I do not understand this statement. "Actually whatever you get x<7/3 or x>7/3 is a flawed solution and therefore is discarded in the end to get range of x as only x<-1"[/quote]

I too got x < 7/3. Will you please elaborate this statement above? I am not clear about the method you used. Appreciate an explanation. Thanks.
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Is x > 1? (1) |x − 4| < −2x + 3 (2) |x − 4| > |x + 4| 18 Jun 2023, 08:50
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Chetan2u
Hi

The solution is correct as given. The solution mentions that x>7/3 is not a possible range.

Substitute x=1 in the equation and you will realise x<7/3 is incorrect

Actually whatever you get x<7/3 or x>7/3 is a flawed solution and therefore is discarded in the end to get range of x as only x<-1.

chetan2u I do not understand this statement. "Actually whatever you get x<7/3 or x>7/3 is a flawed solution and therefore is discarded in the end to get range of x as only x<-1"
Show more


Hello Engineer1

I am not sure how chetan2u got x>7/3 but I guess he means that it doesn't matter if x>7/3 or x< 7/3 because in either case those ranges don't make the inequality true and only x<-1 holds true.

I have solved this question as following:

Statement 1 tells us that |x − 4| < −2x + 3

We get two cases: either x-4<-2x+3 or x-4>2x-3
After further simplification, we get either x<7/3 or x<-1

And at this point, what was alarming to me is to see that the two ranges point in the same direction, because usually when you have |something| <or> something, we either get a conjonction of values, which means a situation where the eventual values of x will be inside a specific interval or a disconjonction, with values of x being in two opposite parts of the number line, in a symmetrical manner and heading towards infinity.
Therefore, I leaned towards verifying the correctness of the obtained ranges.
When you pick for example x=1 which is within the range x<7/3, you find that it doesn't make the inequality true because 3 can never be less than 1. Therefore x<7/3 is not a valid range for this inequality.
On the other hand if you pick any value of x less than -1, the inequality will hold true.
Hence the range of solutions of the inequality in statement 1 is x<-1.
And since the question asks is x>1 you can answer with a definite No.
Statement 1 is sufficient.

Statement 2 tells us that |x − 4| > |x + 4|

Which translates to sqrt(x-4)^2 > sqrt(x+4)^2 (Always keep in mind the following rule: sqrt x^2 = |x|)
Squaring both sides we get, (x-4)^2 > (x+4)^2
Simplifying further, (x-4)^2 - (x+4)^2 > 0 (A quadratic in the form (a^2 - b^2)= (a-b) (a+b) )
(x-4+x+4) (x-4-x-4)>0
-16x>0
Therefore x<0
Again this tells us that the answer to our question is NO, no x is not greater than 1 because it is less than 0 as find out above.
Hence statement 2 is also sufficient.

Correct answer is D

Hope this helps!
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