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Is x > 0? (1) 2x  12 < 10 (2) x^2  10x >= 21
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06 Nov 2019, 00:19
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Competition Mode Question Is \(x > 0\)? (1) \(2x  12 < 10\) (2) \(x^2  10x \geq {21}\) Are You Up For the Challenge: 700 Level Questions
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Re: Is x > 0? (1) 2x  12 < 10 (2) x^2  10x >= 21
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06 Nov 2019, 00:54
Is x>0x>0?
(1) 2x−12<102x−12<10 a. Solving for x: 2x12<10 2x<22 x<11
Or 2x12>10 2x>2 x>1
From above we get 1<x<11. Ans Yes. Sufficient (2) x2−10x≥−21 (x3)(x7)≥0 x≥3 or x≥7. Yes. Sufficient.
Ans D



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Re: Is x > 0? (1) 2x  12 < 10 (2) x^2  10x >= 21
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06 Nov 2019, 01:08
Is x>0? (1) 2x−12<10 Ix6I<5 1<x<11 x=0.5 No X=6 yes (2) x^2−10x≥−21 x^210x+21≥0 x≤3; x≥7 X=0 no X=7 yes combining both the options: 1<x≤3; 7≤x<11 E is the correct option!
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Is x > 0? (1) 2x  12 < 10 (2) x^2  10x >= 21
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Updated on: 07 Nov 2019, 00:55
Is x>0?
(1) 2x−12<10 (2) x2−10x≥−21
#1 2x12<10 2x<22 x<11 and 2x+12<10 2x<2 x>1 sufficient
1<x<11 yes insufficient #2 x2−10x≥−21 (x7)(x3)>=0 x>=7 and x<=3 insufficient IMO A
Originally posted by Archit3110 on 06 Nov 2019, 01:37.
Last edited by Archit3110 on 07 Nov 2019, 00:55, edited 1 time in total.



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Re: Is x > 0? (1) 2x  12 < 10 (2) x^2  10x >= 21
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06 Nov 2019, 01:54
(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
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06 Nov 2019, 02:36
(1) 2x12<10
From this we can get 11>x>1
Sufficient
(2) x^210x >= 21
We get (x7)(x3)>=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
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06 Nov 2019, 03:38
Quote: Is \(x>0\)?
(1) \(2x−12<10\) (2) \(x^2−10x≥−21\)
(1) \(2x−12<10\) sufic.\(2x−12<10…(2x12)^2<10^2…4x^2+4448x<0…x^212x+11<0…(x11)(x1)<0\) \((x11)(x1)<0…[less.than.sign=inside.range]…1<x<11\) (2) \(x^2−10x≥−21\) insufic.\(x^2−10x≥−21…x^210x+21≥0…(x7)(x3)≥0\) \((x7)(x3)≥0…[greater.than.sign=outside.range]…x≥3…or…x≤7\) \(x=[9,1,0,1,2,3,7,8,100…]\) Answer (A)



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Re: Is x > 0? (1) 2x  12 < 10 (2) x^2  10x >= 21
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06 Nov 2019, 08:41
We are to determine if x>0?
1. 2x12<10 when x>0, 2x12<10 hence x<11
when x<0 (2x12)<10 2x12>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 (x7)(x3)≥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
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06 Nov 2019, 09:16
Is x>0? Is x +ve ?
(1) 2x−12<10, 2x12 <10 or 2x+12<10, 2x<22 or 2x>2, x<11 or x>1, x = +ve. Sufficient. 2) x^2−10x≥−21, x^210x+21≥0, x7≥0 or x3≥0, x≥7 or x≥3, x = +ve, sufficient.
Imo. D



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Re: Is x > 0? (1) 2x  12 < 10 (2) x^2  10x >= 21
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06 Nov 2019, 10:04
Is x > 0? (1) 2x − 12 < 10 2x − 6 < 10 x − 6 < 5 x  6 < 5 or  x + 6 < 5 1 < x < 11 SUFFICIENT. (2) \(x^2−10x ≥ − 21\) \(x^2−10x + 21 ≥ − 21 + 21\) (x  3)(x  7) ≥ 0 x ≥ 3 or x ≥ 7 x ≥ 7 SUFFICIENT. IMO Answer D.
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Re: Is x > 0? (1) 2x  12 < 10 (2) x^2  10x >= 21
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06 Nov 2019, 10:11
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
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06 Nov 2019, 14:52
Question : is X>0
(1) Given : 2x−12<10 i.e x6<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
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07 Nov 2019, 00:48
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
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07 Nov 2019, 01:10
Raxit85 wrote: Can anyone please explain regarding x <=3 rather than x >=3??
Posted from my mobile device I also got this wrong for the same reason. Looking back makes sense: (x—3)(x—7) >=0 LHS will be positive only when (x3) and (x7) have same signs. So when we take x>=3 but less than 7 i.e let’s say x = 4 then LHS is (43)*(47)= 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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Re: Is x > 0? (1) 2x  12 < 10 (2) x^2  10x >= 21
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07 Nov 2019, 01:19
Thank you vg18!
I missed the concept while solving the problem that product of two numbers will be >= to 0 only when two numbers have the same sign.
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Re: Is x > 0? (1) 2x  12 < 10 (2) x^2  10x >= 21
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07 Nov 2019, 01:29
vg18 wrote: Raxit85 wrote: Can anyone please explain regarding x <=3 rather than x >=3??
Posted from my mobile device I also got this wrong for the same reason. Looking back makes sense: (x—3)(x—7) >=0 LHS will be positive only when (x3) and (x7) have same signs. So when we take x>=3 but less than 7 i.e let’s say x = 4 then LHS is (43)*(47)= 3 which isn’t correct. Hence the only set of nos which satisfy the above equation is either x>=7 or x<=3. I should have rechecked it..!! Fell for the 'Timer' which always takes the worst out of me.
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Re: Is x > 0? (1) 2x  12 < 10 (2) x^2  10x >= 21
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14 Nov 2019, 07:28
Bunuel wrote: BunuelPlease categorize the question in DS section.



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Re: Is x > 0? (1) 2x  12 < 10 (2) x^2  10x >= 21
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14 Nov 2019, 07:42
Rohit2015 wrote: Bunuel wrote: BunuelPlease categorize the question in DS section. _________________________ Moved to DS forum. Thank you.
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Re: Is x > 0? (1) 2x  12 < 10 (2) x^2  10x >= 21
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03 Jan 2020, 10:09
vg18 wrote: Raxit85 wrote: Can anyone please explain regarding x <=3 rather than x >=3??
Posted from my mobile device I also got this wrong for the same reason. Looking back makes sense: (x—3)(x—7) >=0 LHS will be positive only when (x3) and (x7) have same signs. So when we take x>=3 but less than 7 i.e let’s say x = 4 then LHS is (43)*(47)= 3 which isn’t correct. Hence the only set of nos which satisfy the above equation is either x>=7 or x<=3. Thanks this once slipped out of my mind...clock gets the worst out of me




Re: Is x > 0? (1) 2x  12 < 10 (2) x^2  10x >= 21
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03 Jan 2020, 10:09




