If a^1/2 > b > c^2, which of the following could be true?

let a = 9
b = 2
c = 1

Then 9 > 2 > 1 and this maintains the question stem condition.

3 > 2 > 1

c = 1/2
b = 1/3
a = 1/4

1/2 > 1/3 > 1/4 this the second case.

Sqrt of 1/4 = 1/2

1/2 > 1/3 > 1/4 is possible.

Now last case

a > c > b

a = 1/4
c = 1/6
b = 1/8

1/2 > 1/8 > 1/36 possible.

All the cases are possible.

Answer choice E

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@experts is there a shorter way to do such questions? it takes a lot of time in trial and error.

Bumping it again. Is there any shorter way to do it apart from hit and trial?

Kritisood ajaygaur319

Whenever you see \sqrt{a}, you can consider the value as .404
for .404 \sqrt{a}>a>a^2.

For each option consider the values of a, b, c as .404, . 405 and .406 or .404, .403, .402 depending on the option because \sqrt{.403}>.404.
That's the fastest way to tackle the Square root questions
Please give Kudos if you found this Helpful.

\(\sqrt{a}\) > b > \(c^2\)

First thing we can infer is a>0 ,b>=0, but c can be negative.

Hence I. a > b > c and III. a > c > b can be true without even checking. ( Take 'a' very large number(10000000). Now if you draw \(b=c^2\), it get cuts twice in that range by b=c. Hence b>c or c>b can be possible in range (0<b<1)

If this can be true c > b > a , we must take +ve value of c and b as a can't be negative. Since \(c>b>c^2\), c<1.

c=0.5; b=0.4; a=0.36

\(\sqrt{a} =0.6; b=0.4; c^2=0.25\)

E

Im sorry, couldn't understand this approach. chetan2u GMATinsight any shortcuts for such questions? Would appreciate the help.

Kritisood

The language says "which of the following could be true?"

In such question we have seek one example for which given expression is/are true

What we have is \(\sqrt{a}\) > b > \(c^2\)

So first example, c = 1, b = 2, a = 9 i.e. a > b > c Also \(√9 > 2 > (1)^2\) I part true

So Second example, keeping in mind II. c > b > a , Since a myst be smaller than b while √a should be greater so it's possible only for values between 0 and 1

we can take, c = 1/2, b = 1/3, a = 1/4 i.e. c > b > a because \(√1/4 > 1/3 > (1/2)^2\) II part true

III. a > c > b, This is easy, we can take a = 1, c = 1/2 and b= 1/3 because \(√1 > 1/3 > (1/2)^2\) III part true

is there any easy solution , i can not understand

If \(\sqrt{a} > b^2 > c^4\), which of the following could 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 only

I dont think you need to have trial and error for this solution. Under the assumption that the a, b, and c are Real Numbers, it is quite obvious that all of these three could be true. In other words a, b, c do not have any contraint and you can imagine any number you want to fit the 3 scenarios. If one notices it, then it is 30 second question.

Hi IanStewart I would like to know your take on problems like this. How do we approach similar problems? For me plugging in takes so much time, its not even worth it. Please help.

I'd never just plug arbitrary numbers into a question like this. The first thing I'd want to work out is which types of numbers might matter. Often in questions like this, you can make things easy by thinking about positives and negatives. So, in a different question, if you knew x^2 > y^2, and you're wondering if y > x can be true, it's very easy to get a yes answer just by making x negative and y positive.

Unfortunately thinking about negatives doesn't help on this question, because we learn here that a is positive (we're taking its square root) and so is b (it's larger than c^2, a square). Instead what matters here are numbers larger than 1, because those increase when you square them and decrease when you square root them, and the 'fractions' between 0 and 1, because those do the opposite; they decrease when you square them and increase when you square root them:



We know √a > b > c^2. If a, b and c are larger than 1, then a > b > c is going to be true for sure (because in that case, a > √a, and c^2 > c). So I can be true.

So the only way we'll find a situation where c > b, which is what we need in II and III, is if our letters are between 0 and 1. So I'd just take the simplest fraction possible, 1/2, and see what happens if √a = 1/2, b = 1/2, and c^2 = 1/2. Then a = 1/4, b = 1/2, and c = 1/√2 = √2/2 ~ 0.7. So in this case we find c > b > a is clearly true. Now, it can't technically be true that √a = b = c^2 = 1/2, but if √a is just negligibly larger than 1/2, and c^2 is negligibly smaller than 1/2, then √a > b > c^2 will be true, and c > b > a will be true. So II can be true, and so can III, by the same reasoning (a can be large, and b and c can be roughly 0.5).

But there are definitely other ways to look at this, and other good sets of numbers you could plug in. The important thing though is to identify that it's only the numbers between 0 and 1 that matter when you're looking at II and III.

Hi Kritisood
Missed out earlier.

\(\sqrt{a}\) > b > \(c^2\geq{0}\)
Certain things we can say for sure a>0 and b>0
1) When a, b and c are positive integers and \(\sqrt{a}\) > b > \(c^2\), a>b>c for all values.
2) When a, b and c are fractions and \(\sqrt{a}\) > b > \(c^2\), it is possible that c>b>a for certain values.
a=1/9, b=1/4 and c=1/3
3) This is a mix of above two cases. Surely we can find a way that a is the greatest and the other two maintain the order as in (2) above. That is a positive integer and b and c are fraction. a=1, b=1/4 and c=1/3.

E
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