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Updated on: 09 Jul 2024, 00:18
Originally posted by QuantMadeEasy on 15 Feb 2020, 00:15.
Last edited by QuantMadeEasy on 09 Jul 2024, 00:18, edited 2 times in total.
Last edited by QuantMadeEasy on 09 Jul 2024, 00:18, edited 2 times in total.
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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:
75%
(hard)
Question Stats:
43% (01:52) correct
57%
(02:04)
wrong
based on 47
sessions
History
Date
Time
Result
Not Attempted Yet
Is –1 < a < 0 ?
(1) a^3 > a
(2) a^2 > a^3
For complete pocket friendly GMAT Quant video course:
https://www.udemy.com/course/best-gmat- ... 37602183BB
(1) a^3 > a
(2) a^2 > a^3
For complete pocket friendly GMAT Quant video course:
https://www.udemy.com/course/best-gmat- ... 37602183BB
16 Feb 2020, 17:55
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1)+2):
a^2 > a^3 > a
For any positive number .. this can’t be true ...a^2 has to be the middle number (except when a=1,all equal)
So a must be negative
If a<-1 ..then a2>a>a^3 ...wrong
When -1<a<0... a^2 > a^3 > a ....... perfect
Hence C
Posted from my mobile device
a^2 > a^3 > a
For any positive number .. this can’t be true ...a^2 has to be the middle number (except when a=1,all equal)
So a must be negative
If a<-1 ..then a2>a>a^3 ...wrong
When -1<a<0... a^2 > a^3 > a ....... perfect
Hence C
Posted from my mobile device
19 Feb 2020, 00:39
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1) \(a^3\) > a
Step 1:
Finding critical points \(a^3\) – a >0 -> a (\(a^2\)-1) >0 -> a (a+1)(a-1)>0
Critical points are a = -1, 0 , 1
Step 2: Validating the regions on the number line
i) -∞ to –1: when a=-2, inequlaity is invalid
ii) -1 to 0; when a=-1/2, inequlaity is valid
iii) 0 to –1; when a=½, it is invalid
iv) 1 to ∞, when a =2, it is valid
Step 3: Solution set is –1<a<0 and a>1
No concrete answer to the question
2) \(a^2\) > \(a^3\)
Step 1:
Finding critical points
\(a^3 – a^2\) < 0
\(a^2\) (a-1) <0
Critical points are a = 0, 1
Step 2:
Validating the regions
-∞ to 0, when a=-1, valid
i) 0 to 1, when a = ½ , valid
ii)1 to ∞, when a =2, invalid
Step 3: Solution set -∞ <a<1
No concrete answer to the question
Combining statements (1) and (2)
Common solution set is –1<a<0
C is correct
Step 1:
Finding critical points \(a^3\) – a >0 -> a (\(a^2\)-1) >0 -> a (a+1)(a-1)>0
Critical points are a = -1, 0 , 1
Step 2: Validating the regions on the number line
i) -∞ to –1: when a=-2, inequlaity is invalid
ii) -1 to 0; when a=-1/2, inequlaity is valid
iii) 0 to –1; when a=½, it is invalid
iv) 1 to ∞, when a =2, it is valid
Step 3: Solution set is –1<a<0 and a>1
No concrete answer to the question
2) \(a^2\) > \(a^3\)
Step 1:
Finding critical points
\(a^3 – a^2\) < 0
\(a^2\) (a-1) <0
Critical points are a = 0, 1
Step 2:
Validating the regions
-∞ to 0, when a=-1, valid
i) 0 to 1, when a = ½ , valid
ii)1 to ∞, when a =2, invalid
Step 3: Solution set -∞ <a<1
No concrete answer to the question
Combining statements (1) and (2)
Common solution set is –1<a<0
C is correct
