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Is x > 0? (1) |2x - 12| < 10 (2) x^2 - 10x >= -21

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Bunuel
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Is x > 0? (1) |2x - 12| < 10 (2) x^2 - 10x >= -21 [#permalink] New post   06 Nov 2019, 01:19
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Is \(x > 0\)?

(1) \(|2x - 12| < 10\)
(2) \(x^2 - 10x \geq {-21}\)


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Re: Is x > 0? (1) |2x - 12| < 10 (2) x^2 - 10x >= -21 [#permalink] New post   06 Nov 2019, 02:54
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(1) |2x - 12| < 10
—> -10 < 2x - 12 < 10
—> - 10 + 12 < 2x - 12 + 12 < 10 + 12
—> 2 < 2x < 22
—> 1 < x < 11
—> x > 0 always —> Sufficient

(2) x^2 - 10x ≥ -21
—> x^2 - 10x + 21 ≥ 0
—> (x - 7)(x - 3) ≥ 0
—> x ≤ 3 or x ≥ 7

Since x ≤ 3, it can take values less than zero also —> Insufficient

IMO Option A

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Re: Is x > 0? (1) |2x - 12| < 10 (2) x^2 - 10x >= -21 [#permalink] New post   07 Nov 2019, 02:10
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Raxit85 wrote:
Can anyone please explain regarding x <=3 rather than x >=3??

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I also got this wrong for the same reason. Looking back makes sense:

(x—3)(x—7) >=0

LHS will be positive only when (x-3) and (x-7) have same signs. So when we take x>=3 but less than 7 i.e let’s say x = 4 then LHS is (4-3)*(4-7)= -3 which isn’t correct. Hence the only set of nos which satisfy the above equation is either x>=7 or x<=3.
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Is x > 0? (1) |2x - 12| < 10 (2) x^2 - 10x >= -21 [#permalink] New post   Updated on: 07 Nov 2019, 01:55
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Is x>0?

(1) |2x−12|<10
(2) x2−10x≥−21

#1
2x-12<10
2x<22
x<11
and
-2x+12<10
-2x<-2
x>1
sufficient

1<x<11
yes insufficient
#2
x2−10x≥−21
(x-7)(x-3)>=0
x>=7 and x<=3
insufficient
IMO A

Originally posted by Archit3110 on 06 Nov 2019, 02:37.
Last edited by Archit3110 on 07 Nov 2019, 01:55, edited 1 time in total.
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Re: Is x > 0? (1) |2x - 12| < 10 (2) x^2 - 10x >= -21 [#permalink] New post   06 Nov 2019, 03:36
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(1) |2x-12|<10

From this we can get 11>x>1

Sufficient


(2) x^2-10x >= -21

We get (x-7)(x-3)>=0

x>=7 or x<=3
Not sufficient

OA:A

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Re: Is x > 0? (1) |2x - 12| < 10 (2) x^2 - 10x >= -21 [#permalink] New post   06 Nov 2019, 04:38
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Quote:
Is \(x>0\)?

(1) \(|2x−12|<10\)
(2) \(x^2−10x≥−21\)


(1) \(|2x−12|<10\) sufic.

\(|2x−12|<10…(2x-12)^2<10^2…4x^2+44-48x<0…x^2-12x+11<0…(x-11)(x-1)<0\)
\((x-11)(x-1)<0…[less.than.sign=inside.range]…1<x<11\)

(2) \(x^2−10x≥−21\) insufic.

\(x^2−10x≥−21…x^2-10x+21≥0…(x-7)(x-3)≥0\)
\((x-7)(x-3)≥0…[greater.than.sign=outside.range]…x≥3…or…x≤7\)
\(x=[-9,-1,0,1,2,3,7,8,100…]\)

Answer (A)
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Re: Is x > 0? (1) |2x - 12| < 10 (2) x^2 - 10x >= -21 [#permalink] New post   06 Nov 2019, 09:41
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We are to determine if x>0?

1. |2x-12|<10
when x>0, 2x-12<10
hence x<11

when x<0
-(2x-12)<10
2x-12>-10
x>1
Hence x>1
This results in a range of 1<x<11
Statement 1 alone is therefore sufficient since x>1 in the range above.

2. x^2 -10x ≥-21
x^2 -10x + 21≥0
(x-7)(x-3)≥0
This results in the range x≥7 and x≤3
This is insufficient because when x≥7 then x is always greater than 0. But when x≤3, then x is not always greater than 0, because x can be -1, which is less than 0.

The answer is therefore A.
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Re: Is x > 0? (1) |2x - 12| < 10 (2) x^2 - 10x >= -21 [#permalink] New post   06 Nov 2019, 11:11
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Is x >0?

(Statement1): -10 < 2x—12< 10
—> 2< 2x< 22
1 < x < 11
Sufficient

(Statement2): \(x^{2} —10x+ 21>=0\)
(x—3)(x—7) >=0

x <=3 and x>=7
—> If x=2, then YES
—> If x= —1, then NO
Insufficient

The answer is A

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Re: Is x > 0? (1) |2x - 12| < 10 (2) x^2 - 10x >= -21 [#permalink] New post   06 Nov 2019, 15:52
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Question : is X>0

(1) Given : |2x−12|<10
i.e |x-6|<5
This implies 1<x<11

Statement 1 is sufficient

(2) x^2−10x≥−21
i.e x^2−10x+21≥0
This imples x<3 or x>7
According to this X could take positive and negative values.

Therefore answer is A.
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Re: Is x > 0? (1) |2x - 12| < 10 (2) x^2 - 10x >= -21 [#permalink] New post   07 Nov 2019, 01:48
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Can anyone please explain regarding x <=3 rather than x >=3??

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Re: Is x > 0? (1) |2x - 12| < 10 (2) x^2 - 10x >= -21 [#permalink] New post   25 Jan 2023, 22:55
Bunuel please post the official solution
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Is x > 0? (1) |2x - 12| < 10 (2) x^2 - 10x >= -21 [#permalink] New post   26 Jan 2023, 03:13
Expert Reply
Is \(x > 0\)?

(1) \(|2x - 12| < 10\).

Get rid of the modulus sign:
    \(-10<2x - 12 < 10\);

Add 12 to all three parts:
    \(2<2x < 22\);

Divide by 2:
    \(1<x < 11\).

Sufficient.


(2) \(x^2 - 10x \geq {-21}\)

Here I'd use number plugging:

    If x is large enough, say if x = 1,000, the inequality holds. So, x can be more than 0.
    If x = 0, the inequality holds too. So, in this case x > 0 is not true.

Two different answers. Not sufficient.

Another approach:

Transitions points are x = 3 and x = 7. "\(\geq\)" sign indicates that the solution is to the left of the smaller root and to the right of the larger root. Thus \(x \leq 3\) and \(x \geq 7\). Check Solving Quadratic Inequalities - Graphic Approach: https://gmatclub.com/forum/solving-quad ... 70528.html

So, x may or may not be greater than 0. Not sufficient.

Answer: A.

Hope it's clear.
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Re: Is x > 0? (1) |2x - 12| < 10 (2) x^2 - 10x >= -21 [#permalink] New post   16 Apr 2024, 20:22
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Re: Is x > 0? (1) |2x - 12| < 10 (2) x^2 - 10x >= -21 [#permalink]
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