If a^3 b^2 c, which of the following may be true?

Question

If a^3 > b^2 > c, 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

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Adit_ · Answer

All of them seems true.
That is because we have no cinstraints and the value of a^3 and b^2 can always grow faster than a single term, so one can rearrange the original values of a,b,c in any order.

Answer:  Option E

Aman2487 · Answer

But it's given a>b>c so how can c be greater than a

Adit_ · Answer

a^3>b^2>c is what is given.

Take the example of a=5, b=6, c=7.
Here a^3 = 125, b^2 = 36, c=7.

125>36>7
Also 7>6>5 which is c>b>a thus satisfied.

Hope it helps.

thisisakr · Answer

Key insight: b2 ≥ 0 always, so c < b2 means c could be negative. Also a3 > b2 ≥ 0 means a must be positive.

Statement 1:
A>b>c
A=4 b=3 c=2 will solve this as it satisfies a^3>b^2>c

Statement 2:

C>b>a here a is least

But remember a has to be greater than 0 so 
a =1.9 b =2 c=3 
Will satisfy both conditions

Statement 3 a>c>b

Here 
C=0 b=-1 a= 2 satisfied both equation

So by trail and error and using the logic of a> 0 we can see  answer is E

ExpertsGlobal5 · Answer

E is the correct answer choice.

Video explanation:

https://www.youtube.com/watch?v=5xmuqn72qj8

Edskore · Answer

Aman2487's confusion here is actually super common and worth addressing clearly.

The given condition is a^3 > b^2 > c. That's it. We are NOT told anything about the relative ordering of a, b, and c themselves. The question is asking which orderings of a, b, c are POSSIBLE -- not which ones must always be true.

Here's the key insight this question is testing in Problem Solving: a^3 can grow much faster or slower than a depending on whether a is greater or less than 1, and b^2 is always non-negative. So the values of a^3, b^2, and c tell you relatively little about the actual values of a, b, and c.

For Statement I (a > b > c): try a = 10, b = 3, c = 2. Then a^3 = 1000, b^2 = 9. Check: 1000 > 9 > 2. And a > b > c: 10 > 3 > 2. Works.

For Statement II (c > b > a): try a = 2, b = 3, c = 7. Then a^3 = 8, b^2 = 9. Check: 9 > 9? No. Adjust: a = 1.9, b = 2, c = 3. a^3 = 6.86, b^2 = 4. Check: 6.86 > 4 > 3. And c > b > a: 3 > 2 > 1.9. Works.

For Statement III (a > c > b): try a = 5, b = 0, c = 2. a^3 = 125, b^2 = 0. Check: 125 > 0 -- wait c needs to be less than b^2. Try b = -1, c = 0.5, a = 2. b^2 = 1, a^3 = 8. Check: 8 > 1 > 0.5. And a > c > b: 2 > 0.5 > -1. Works.

So all three can be true. Answer: E.

The trap is thinking a^3 > b^2 tells you a > b. It doesn't, because squaring and cubing can flip orderings depending on the sign and magnitude of the base.

If a^3 > b^2 > c, which of the following may be true?

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