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Is the eqn > 2

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Source: — Data Sufficiency |
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by vittalgmat » Fri Apr 03, 2009 12:12 am
Problem is
is x/3 + 3/x > 2 ?

Stmt 1: x < 3

This is insufficient coz of the following cases:
let x = 2
2/3 + 3/2
> 2

let x = 0
value is indeterminate
So insufficient


Stmt 2. x > 1

For all values of x > 1, the value of 3/x + x/3 is > 2.
Sufficient.


So ans is B.

What is the OA ?

thanks
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by bluementor » Fri Apr 03, 2009 1:26 am
vittalgmat wrote:Stmt 2. x > 1

For all values of x > 1, the value of 3/x + x/3 is > 2.
Sufficient.


So ans is B.
Wrong.

If x = 3, then 3/x + x/3 = 2, which disqualifies the inequality. Hence statement 2 is insufficient on its own.

Both statements together specify the lower and upper boundaries for x to satisfy the inequality. Choose C.

-BM-
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by GID09 » Fri Apr 03, 2009 1:55 am
IMO A. Is x/3 + 3/x >2 ? could be simplified to x^2 - 6x+9 >0?

which could be further simplified as x-3 >0 => so the question is x>3?

St 1 tells is x<3 ...Hence sufficient
St 2 x>1 doesn't help much.

Hence A.
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by bluementor » Fri Apr 03, 2009 2:28 am
GID09 wrote:IMO A. Is x/3 + 3/x >2 ? could be simplified to x^2 - 6x+9 >0?

which could be further simplified as x-3 >0 => so the question is x>3?
I don't think you can treat the question stem as a known fact. Try plugging a negative value into the question stem and you will see that the inequality does not hold.

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by GID09 » Fri Apr 03, 2009 3:03 am
BM, Could you explain what you mean by "Try plugging a negative value into the question stem and you will see that the inequality does not hold" ?

I was just rearranging the variables without disturbing the inequality right?

Thanks,
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by bluementor » Fri Apr 03, 2009 3:40 am
GID09 wrote:BM, Could you explain what you mean by "Try plugging a negative value into the question stem and you will see that the inequality does not hold" ?

I was just rearranging the variables without disturbing the inequality right?
hi GID09,

in your solution, you did the following:

"is x/3 + 3/x > 2 ?" was changed to -> "is x^2 - 6x + 9 > 2?".

when you manupulated the inequality, you assumed x to be positive. thats where the problem is. we don't know whether x is positive or negative. if x is negative, you will have to flip the sign when you multipled the inequality with x, and you will arrive at a different inequality.

the point of the question is to prove whether (x/3 + 3/x) is always greater than 2. Without any boundaries for x, (x/3 + 3/x) can take values less than 2 or greater than 2.

All we know in statement 1 is that x<3. But is this sufficient? For eg. say x = -1, then (x/3 + 3/x) will be a negative value, i.e. less than 2. but if x = 2, then (x/3 + 3/x) is greater than 2. so statement 1 is insufficient on its own.

hope this explanation helps.

-BM-
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by GID09 » Fri Apr 03, 2009 5:33 am
BM, I agree. Thanks!
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Re: Is the eqn > 2

by Ian Stewart » Fri Apr 03, 2009 6:28 am
hk wrote:Here is one for you all:

Is (x/3) + (3/x) > 2

(1) x<3
(2) x>1
You might notice first that if x is negative, then (x/3) + (3/x) will be negative, so the answer to the question will be 'no'. On the other hand, if x is, say, 2, the answer will be 'yes'. So Statement 1 is not sufficient, since x could be 2, or x could be negative.

If we use Statement 2, we know that x is positive, so we can safely rewrite the inequality in the question by multiplying by 3x on both sides:

(x/3) + (3/x) > 2
x^2 + 9 > 6x
x^2 - 6x + 9 > 0
(x-3)^2 > 0

We've just rephrased the question here, and to rewrite the question in this way, we are using Statement 2 -- we needed to know that x was positive in order to multiply by x on both sides. Now, (x-3)^2 will always be greater than zero unless x = 3. Since it is possible that x=3 from Statement 2 alone, we also need Statement 1 to rule out that possibility, and the answer is C.

The question is from GMATFocus, and the solution they provide isn't a solution at all; it isn't worth reading.
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by hk » Fri Apr 03, 2009 7:09 am
bluementor wrote: Wrong.

If x = 3, then 3/x + x/3 = 2, which disqualifies the inequality. Hence statement 2 is insufficient on its own.

Both statements together specify the lower and upper boundaries for x to satisfy the inequality. Choose C.

-BM-
The OA is C.

But i still dont understand one thing.

Statement 1. This is insufficient which can be determined by plugging in values 2 and 0 for x which leads to different results.

Statement 2. This is where i am lost. If i plug in 3, it would yield that x/3 + 3/x = 2. So 2=2 and would lead to an answer of NO. But when we plug in 2 for x, this would yield that x/3 + 3/x = 0.66 + 1.5 = 2.16> 2 and would lead to an answer of YES. So this statement is insufficient.

Now taking both statements together, we get 1<x<3, plug in 2 we get an answer of YES the eqn is > 2. Now substitute x = 1.5, we get 1.5/3 + 3/1.5 = 0.5 + 2 = 2.5 > 2. Similarly if we plug in x = 2.5 we get 2.5/3 + 3/2.5 = 0.83 + 1.2 = 2.03 > 2. So yes. Hence both together are sufficient.

Although i got this right, the explanation said "Using statement 2, if we substitute x=3, then it would result in the eqn being 2=2, so x cannot be 3 hence this statement is insufficient because it violates the inqeuality." --> I also see that bluementor also gave the same explanation. But i dont understand this. The question asks if the eqn is > 2 and if the eqn is = 2, then the question is answered as NO its is not greater than 2. Whats with this "disqualify the inequality." Please explain this to me.

I hope i didn't make it too complicated to understand.

Thanks in advance..
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by Ian Stewart » Fri Apr 03, 2009 7:20 am
hk wrote: Statement 2. This is where i am lost. If i plug in 3, it would yield that x/3 + 3/x = 2. So 2=2 and would lead to an answer of NO. But when we plug in 2 for x, this would yield that x/3 + 3/x = 0.66 + 1.5 = 2.16> 2 and would lead to an answer of YES. So this statement is insufficient.

Although i got this right, the explanation said "Using statement 2, if we substitute x=3, then it would result in the eqn being 2=2, so x cannot be 3 hence this statement is insufficient because it violates the inqeuality." --> I also see that bluementor also gave the same explanation. But i dont understand this. The question asks if the eqn is > 2 and if the eqn is = 2, then the question is answered as NO its is not greater than 2. Whats with this "disqualify the inequality." Please explain this to me.
You've understood the logic of the question perfectly, and it may just be the terminology used in one of the solutions above that you're finding confusing. If x = 3, we find that the inequality is false; 3/x + x/3 is not greater than 2, and the answer to the question is 'no'. That's what BM meant by 'disqualifies the inequality'; he meant that the inequality is not true when x = 3. If, on the other hand, 1 < x < 3, then 3/x + x/3 is always greater than 2, and the inequality is true.
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by vittalgmat » Fri Apr 03, 2009 9:56 am
bluementor wrote:
vittalgmat wrote:Stmt 2. x > 1

For all values of x > 1, the value of 3/x + x/3 is > 2.
Sufficient.


So ans is B.
Wrong.

If x = 3, then 3/x + x/3 = 2, which disqualifies the inequality. Hence statement 2 is insufficient on its own.

Both statements together specify the lower and upper boundaries for x to satisfy the inequality. Choose C.

-BM-
Damn!! so silly of me!!! .. careless error!!.
Thanks BM for pointing it out.
Will be more careful next time.

rgds
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by bluementor » Mon Apr 06, 2009 12:39 am
hk wrote:
Although i got this right, the explanation said "Using statement 2, if we substitute x=3, then it would result in the eqn being 2=2, so x cannot be 3 hence this statement is insufficient because it violates the inqeuality." --> I also see that bluementor also gave the same explanation. But i dont understand this. The question asks if the eqn is > 2 and if the eqn is = 2, then the question is answered as NO its is not greater than 2. Whats with this "disqualify the inequality." Please explain this to me.

I hope i didn't make it too complicated to understand.

Thanks in advance..
hk, sorry for replying so late. I only saw your message this morning.

Anyway, I guess Ian has already clearified your doubt. Post back if something still bothers you.

-BM-
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