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Rakesh1987
Current Student
Joined: 17 Jul 2018
Last visit: 31 Dec 2024
Posts: 66
Given Kudos: 100
Location: India
Concentration: Finance, Leadership
Schools: Stern '23 (M$)
GMAT 1: 760 Q50 V44
GPA: 4
Updated on: 12 Oct 2023, 00:42
Originally posted by Rakesh1987 on 03 May 2019, 02:43.
Last edited by Bunuel on 12 Oct 2023, 00:42, edited 2 times in total.
Last edited by Bunuel on 12 Oct 2023, 00:42, edited 2 times in total.
Renamed the topic and edited the question.
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
37% (02:01) correct
63%
(01:52)
wrong
based on 99
sessions
History
Date
Time
Result
Not Attempted Yet
06 May 2019, 23:26
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If \(x\neq{0}\), is \(|x|\)>1?
(1) \(\frac{x}{|x|}<x\)
CASE 1: if x > 0, then the above will give \(\frac{x}{x}<x\), which leads to \(1 < x\). This gives an YES answer to the question.
CASE 2: if x < 0, then the above will give \(\frac{x}{-x}<x\), which leads to \(-1 < x\). As we considering the range when x < 0, then \(-1<x<0\). This gives a NO answer to the question.
(2) \(\sqrt{{x^2}}=x\).
Since, \(\sqrt{{x^2}}=|x|\), then the above gives \(|x|=x\). This, on the other hand, means that x > 0. Not sufficient.
(1)+(2) From (2) we got that x > 0, thus we'd have CASE 1 from (1): \(1 < x\). This gives an YES answer to the question. Sufficient.
Answer: C.
Hope it helps.
(1) \(\frac{x}{|x|}<x\)
CASE 1: if x > 0, then the above will give \(\frac{x}{x}<x\), which leads to \(1 < x\). This gives an YES answer to the question.
CASE 2: if x < 0, then the above will give \(\frac{x}{-x}<x\), which leads to \(-1 < x\). As we considering the range when x < 0, then \(-1<x<0\). This gives a NO answer to the question.
(2) \(\sqrt{{x^2}}=x\).
Since, \(\sqrt{{x^2}}=|x|\), then the above gives \(|x|=x\). This, on the other hand, means that x > 0. Not sufficient.
(1)+(2) From (2) we got that x > 0, thus we'd have CASE 1 from (1): \(1 < x\). This gives an YES answer to the question. Sufficient.
Answer: C.
Hope it helps.
Rakesh1987
Current Student
Joined: 17 Jul 2018
Last visit: 31 Dec 2024
Posts: 66
Given Kudos: 100
Location: India
Concentration: Finance, Leadership
Schools: Stern '23 (M$)
GMAT 1: 760 Q50 V44
GPA: 4
06 May 2019, 23:38
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Rakesh1987
\(|x|\)>1 when x>1 or x<-1. For all values -1<x<1, \(|x|\) will not be greater than 1. As it is a DS question we need to find whether it is |x|>1 or not, we dont need the exact value.
1) \(\frac{x}{|x|}<x\)
As |x| is always positive we can multiply both sides by |x| without changing sign of inequality.
x<x*|x|
=> x-x*|x|<0
=> x(1-|x|)<0.
Therefore one of the term is positive and another negative.
Scenario 1.1) If x is negative, then 1-|x| must be positive which will only happen if |x|<1. |x|<1 can be written as -1<x<1. As in this scenario x is negative we can trim it down to -1<x<0.
Scenario 1.2) If x is positive, then 1-|x| must be negative which will only happen if |x|>1. As x is positive and |x|>1, x>1.
In Scenario 1.1, |x|<1 and in Scenario 1.2, |x|>1. Therefore Insufficient.
2) \(\sqrt{{x^2}}=x\) gives us that the number is positive. As square is always positive, \(x^2\) is positive and its square root, which is x is also positive. Therefore, we can only write x>0.
Now if x falls in this range 0<x<1, then |x|<1 and if it falls in this range x>1, then |x|>1, therefore Insufficient.
1) and 2) combined. We know x>0 (or x is positive), therefore from 1) above, we are left with only scenario 1.2 "x>1". Therefore we can say |x|>1. Sufficient. Hence, the answer is C.
Please Hit +1 Kudos, if you liked the explanation.
