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12 Apr 2026, 19:00
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
38% (01:43) correct
62%
(02:03)
wrong
based on 101
sessions
History
Date
Time
Result
Not Attempted Yet
If √a > b^2 > c^3, which of the following may be true?
I. a > b > c
II. c > b > a
III. a > c > b
A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III
I. a > b > c
II. c > b > a
III. a > c > b
A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III
12 Apr 2026, 22:40
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I am going with E since there is no constraint btw a,b,c and no constraint like positive integers etc.
Also this is a may be true question. Definitely values would exist that would satisfy all of the three.
Also this is a may be true question. Definitely values would exist that would satisfy all of the three.
The question is a "could be/may be" true, so the goal is to find at least one case for each option in which both the inequality considered in the option and the inequality in the question stem are verified.
Lets start with the first option: a>b>c
In my opinion the fastest way to find a case for this option is to start from c: since it's the smallest value once you find c then you just have to find bigger enough values for b and a.
If we take c = 2 , b= 3 and a= 25 we have that \( \sqrt{a} \)= 5, \(b^2\) = 9 and \(c^3\)= 8 so we have both a>b>c and \(\sqrt{a}\)>\(b^2\)>\(c^3\)
Second option: c > b > a
For this case the inequality is "opposite" compared to the one in the question stem so the best way to think of this is to find a value for a that when rooted gets bigger and a value for b and c that when squared and cubed get smaller. This values are all fractions so it's like saying that \(c^{-1}\)>\(b^{-1}\)>\(a^{-1}\), we can easily see that this inequality turns into the first one were a>b>c and we already found a case for this so we know it's true for at least one case.
Third option: a>c>b
In this case we can just think about b and c (since for a we can take whatever bigger value after we choose b and c and both inequalities will be verified). We already seen that there is a case in which \(b^2\)>\(c^3\) while c>b so we already know that even this option is verified for at least one case.
Answer: E
Lets start with the first option: a>b>c
In my opinion the fastest way to find a case for this option is to start from c: since it's the smallest value once you find c then you just have to find bigger enough values for b and a.
If we take c = 2 , b= 3 and a= 25 we have that \( \sqrt{a} \)= 5, \(b^2\) = 9 and \(c^3\)= 8 so we have both a>b>c and \(\sqrt{a}\)>\(b^2\)>\(c^3\)
Second option: c > b > a
For this case the inequality is "opposite" compared to the one in the question stem so the best way to think of this is to find a value for a that when rooted gets bigger and a value for b and c that when squared and cubed get smaller. This values are all fractions so it's like saying that \(c^{-1}\)>\(b^{-1}\)>\(a^{-1}\), we can easily see that this inequality turns into the first one were a>b>c and we already found a case for this so we know it's true for at least one case.
Third option: a>c>b
In this case we can just think about b and c (since for a we can take whatever bigger value after we choose b and c and both inequalities will be verified). We already seen that there is a case in which \(b^2\)>\(c^3\) while c>b so we already know that even this option is verified for at least one case.
Answer: E
