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12 Days of Christmas GMAT Competition - Day 3: If a > b^2 > c^3

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12 Days of Christmas GMAT Competition - Day 3: If a > b^2 > c^3 [#permalink] 12 Days of Christmas GMAT Competition - Day 3: If a > b^2 > c^3   16 Dec 2022, 07:00
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12 Days of Christmas GMAT Competition with Lots of Fun

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


 


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Re: 12 Days of Christmas GMAT Competition - Day 3: If a > b^2 > c^3 [#permalink] 12 Days of Christmas GMAT Competition - Day 3: If a > b^2 > c^3   16 Dec 2022, 08:20
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a > b^2 > c^3

Think about the graph to solve it easily.

In may be true, we just need to find ONE example where each of the following hold:
I. a > b > c
II. c > b > a
III. a > c > b

If a > b^2 > c^3, only values between 0 to 1 will hold. See graph.
You will see that I, II, and III - all can hold.

I. Clearly true for the region (0,1).
II. c=0.6, b=0.5, a=0.4 holds. Take difference between a,b,c smaller and you will see this hold even without a precise graph. The attached graph is only for understanding purposes.
III. If II holds (where c>b), and a is greater, then III definitely holds.

Answer: E
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Re: 12 Days of Christmas GMAT Competition - Day 3: If a > b^2 > c^3 [#permalink] 12 Days of Christmas GMAT Competition - Day 3: If a > b^2 > c^3   16 Dec 2022, 10:59
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E is the answer. It is possible to create a scenario that all of the above statements are true given the original statement.

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

For questions like this, always consider "Special Numbers." These numbers are numbers that behave differently in different operations in math. Special numbers can include: {negatives, 0, fractions less than 1, 1, 2, and extremes.}

Too many possible scenarios exist with the original statement to either eliminate possibilities that won't work or to include only certain groups of numbers. For this reason, let's jump in and consider the options I, II, and III.

I. Is it possible that √a > b^2 > c^3 AND a>b>c?
This can be possible if a is a positive number (adequately) greater than b (they can even be fractions less than 1) and c is negative
Example: a=100
b=2 and
c=-1

II. Is it possible that √a > b^2 > c^3 AND c>b>a?
This can be possible if a , b and c are appropriately chosen fractions less than 1
Example: a=1/100
b=1/4 and
c=1/3

III. Is it possible that √a > b^2 > c^3 AND a>c>b?
This can be possible if a and c are appropriately chosen positive numbers and b is a negative number
Example:
a=100
b=-1 and
c=1/2

The answer is that I, II and III are ALL possible scenarios based on the properties of positive numbers, fractions less than one and negative numbers.

E is the correct answer!
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Re: 12 Days of Christmas GMAT Competition - Day 3: If a > b^2 > c^3 [#permalink] 12 Days of Christmas GMAT Competition - Day 3: If a > b^2 > c^3   16 Dec 2022, 11:42
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I. a > b > c


Let a = 25, b= 2 and c=1

Applying to the condition √a > b^2 > c^3, we get,

5 > 4 > 1 which is true.

It also satisfies given condition I, 25 > 2 > 1.

So keep the choices that contain I. Eliminate B.


II. c > b > a

In this case, we want an example, where c to be the largest number at the same time c^3 is the smallest.
Also note, since GMAT covers only real numbers a cannot be negative :)
That means a,b and c all must be positive.

Consider c to be between 0< c < 1, then a and b should be less than 1 .

Let c = 1/3, b= 1/4 and a=1/64
Applying to the condition √a > b^2 > c^3, we get,

1/8 > 1/16 > 1/27 is true.

Also, here c > b > a is true.

So keep the choices that contain II.


Eliminate A and D. So the remaining options are C and E.


III. a > c > b

Let us consider b = -2, c= -1 and a = 25.


Applying to the condition √a > b^2 > c^3, we get,

5 > 4 > -1 which is true.

Also here a > c > b is true.

So keep the choice that contains III.

Eliminate C.

So the correct answer choice is E.
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Re: 12 Days of Christmas GMAT Competition - Day 3: If a > b^2 > c^3 [#permalink] 12 Days of Christmas GMAT Competition - Day 3: If a > b^2 > c^3   11 Aug 2024, 21:08
First we see the question.
a^1/2 > b^2 > c^3
Which of the folowing can be true.

1. a>b>c

So both question and answer is in same direction. I would pick numbers which has sufficiently large difference between them and any operation on them are inconsequential.

eg: a^1/2 > b^2 > c^3 -> 1000,100,1
So a>b>c -> 10^6,10,1

2. a<b<c
Here root(a) should be largest then a becomes smallest
and c^3 should be smallest then c becomes largest.

These are usually the property of 0<x<1
For easier manipulation lets take them as decimal numbers
(0.1)^2 = 0.01
(0.008)^1/3 = .2

So we can see that its in right direction.
So first I pick 0.1,X,0.008

I like to keep things simple. So I would go to -ve if things doesnt work out only and always keep decimals in easily rootable format.
So by this standard, I would have picked b^2=0.04 then b=0.2 (Thats same as our C). Since there is less room in the middle. I would try move C further. I will pick c^3 = 0.027 then c=.3

eg: a^1/2 > b^2 > c^3 -> 0.1,0.04,0.027
So a<b<c -> 0.01<0.2<0.3

3. Its almost same as first except now square root is smaller than cube root.
Easiest way would be to take a negative root of the one you want to make smaller.
eg: a^1/2 > b^2 > c^3 -> 1000,100,1
So a>c>b -> 10^6,1,-10
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Re: 12 Days of Christmas GMAT Competition - Day 3: If a > b^2 > c^3 [#permalink]
11 Aug 2024, 21:08
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