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Couldnt locate a discussion session for the ques below...if

DeeptiM

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Couldnt locate a discussion session for the ques below...if u knw it, pls direct me to d forum..

If A & B are non-negative, is A^5>B^2?

1) A^1/3>B^2
2) A>B^2

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fluke

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28 Aug 2011, 14:29

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DeeptiM

Couldnt locate a discussion session for the ques below...if u knw it, pls direct me to d forum..

If A & B are non-negative, is A^5>B^2?

1) A^1/3>B^2
2) A>B^2

Please refer this:
r-and-s-38046.html#p967306

A=0.5;B=0.5; is A^5>B^2? NO
A=2;B=1; is A^5>B^2? Yes
Not Sufficient.

Ans: "E"

DeeptiM

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28 Aug 2011, 21:00

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Thanks a lot fluke..i didnt include decimals hence, was getting a diff answer..

also, would really appreciate if can u help me with the workings of statement 1.
Thnx

ankit1234suhane

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28 Aug 2011, 23:34

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i think ans is D. Why see no matter what kind of number you are decimal or integer, if A^1/3 >b^2 then a must be greater than B

subhajeet

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29 Aug 2011, 04:08

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I think the answer should be E

Here using the conditions we have to prove that A^5> B^2

Condition1: A^1/3>B^2 -> If I put A=B=2 or A=1 and B=2 then the condition becomes false.

Condition2: A> B^2 -> Again here also if I put A=B=2 or A=1 and B=2 then the condition becomes false.

If there is some other way to solve this question then please let me know.

pike

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29 Aug 2011, 05:50

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subhajeet

This logic is not sound. First, you are not trying to prove the equation, you are trying to prove can you definitively say yes or no.

But second, the numbers you have chosen are not sensible. For instance, take for 2) your selection of a=1 b=2. These do not fit the criteria, hence are not relevant. E.g:

Is x > 10?

1) x > 8

It makes no sense to test x=7, as we have just been told that x>8. On the other hand, x=9 fits within 1), but is not >10. x=11 fits within 1), but is >10.

That is the type of thought process and numbers we should be testing.

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subhajeet

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29 Aug 2011, 07:47

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pike

subhajeet

I totally agree with the example you have mentioned above, but in your case it is definite that we will start with values of x that are greater than 8, it can also be 8.1.

In the question which we are discussing, it has only been specified that A and B are non negetive integers. Nothing has been specified whether A>B or vice versa, so I took those values using which I could solve the problem.

If there is any other please please mention it here. It will help us.

pike

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29 Aug 2011, 19:36

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subhajeet

Here is why I think your logic is wrong.

subhajeet

Here using the conditions we have to prove that A^5> B^2

As you mentioned, we are given the following: A > 0 and B > 0
And asked: is A^5> B^2?

At this point, we do not have sufficient information to answer this question (otherwise it wouldn't be a GMAT DS question!). Lets delve into the conditions to see what further information we can glean that might help us answer the question.

subhajeet

Condition1: A^1/3>B^2 -> If I put A=B=2 or A=1 and B=2 then the condition becomes false.

This is where I believe things go wrong in your logic. We are told that A^(1/3) > B^(1/2). This is now an immutable truth, no matter what the true values of A and B are, they must fit the above equation.

At this point you have suggested testing A=B=2; A=1 and B=2
2^(1/3) > 2^(1/2)
1.25 > 1.4

1^(1/3) > 2^(1/2)
1.25 > 1.4

You are correct that these do not hold under 1), but that's the exact information that 1) offers us! Do you see how this is akin to:
Is x > 10

1) x > 8
Let's assume x = 7...

We can't assume that A=B=2 because from 1) this is clearly not true.

The numbers you should be picking are those that are true under 1). Then we test those against the question. If we can find numbers that are true under 1), but give contradictory answers to the yes/no question, then we can rule out 1) as offering useful information.

So for instance, some pairs that we might test are the ones that fluke suggested:
A=0.5;B=0.5
A=2;B=1

Fluke hasn't just chosen these randomly, he has chosen them because they hold under 1), yet give us contradictory answers when put back into our yes/no question.

.5^(1/3) > .5^(1/2)
.8 > . <--- is true!

2^(1/3) > 1^(1/2)
1.25 > 1 <--- is true!

OK, so now we have some pairs of numbers that are true under 1), but we still can't answer the question just from this alone. We need to go back to the question, and see if these numbers give a contradiction. Let's have a look:

Is A^5> B^2?

A=0.5; B=0.5
.03 > .25. No.

A=2; B=1
32 > 1 Yes.

Now we see why 1) is not helpful to us. We have picked two sets of numbers that are true under 1), but give contradictory answers to the yes/no question. We can now stop testing numbers, and move onto 2).

The numbers you used to rule out question 2) were not helpful for the same reasons as above.

I highly recommend you focus on the numbers fluke suggested and compare them to the ones you used. It's really important to get this approach locked down for DS.

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30 Aug 2011, 08:00

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Pike
Thanks you very much for the explanation. My approach was flawed. Thanks for rectifying me.

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30 Aug 2011, 18:44

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A^5>B^2?

1. Not sufficient

when A= 1/2, B =1/2 ,answer is No

when A=2, B=1 , answer is Yes

2. Not sufficient

same examples as in 1 applies here.

Together, Not sufficient

same examples as in 1 holds good.

Answer is E.

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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.

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