Problem Solving

For (x=/= 0 ,+-1,) the expression [x^1/2007 - x^1/2009] / [x^1/2008 - x^1/2010] is equivalent to :


A> x ,
B> x - 1 ,
C> x^2 - 1,
D>1/x .

It's sad to see you lost the fifth choice on this occasion, but let's see what's in store

One thing is here to mention before we go for the elucidation, that on GMAT, whenever we find the phrases like x ≠ 0, ±1, etc, we should make sure that we're going to deal with some rational expression(s) that could contain factors like (x - 0), (x ± 1), etc in its denominators. In order to steer clear of an indeterminate state, they already make us observant with the precincts like that.

From the distance, it could be easily seen that the given rational expression has x^1/2009 as a common factor in the numerator with (x^2 - 1) to spare, and that it has x^1/2010 as a common factor in the denominator, once again with (x^2 - 1) to spare. Hence, (x^2 - 1) is the most visible common factor in the given rational expression, canceled and we are left with x^1/2009/ x^1/2010 = x (x^1/2010)/ x^1/2010 = [spoiler]x[/spoiler].

[spoiler]A[/spoiler]

sanju09 wrote:It's sad to see you lost the fifth choice on this occasion, but let's see what's in store

One thing is here to mention before we go for the elucidation, that on GMAT, whenever we find the phrases like x ≠ 0, ±1, etc, we should make sure that we're going to deal with some rational expression(s) that could contain factors like (x - 0), (x ± 1), etc in its denominators. In order to steer clear of an indeterminate state, they already make us observant with the precincts like that.

From the distance, it could be easily seen that the given rational expression has x^1/2009 as a common factor in the numerator with (x^2 - 1) to spare, and that it has x^1/2010 as a common factor in the denominator, once again with (x^2 - 1) to spare. Hence, (x^2 - 1) is the most visible common factor in the given rational expression, canceled and we are left with x^1/2009/ x^1/2010 = x (x^1/2010)/ x^1/2010 = [spoiler]x[/spoiler].

[spoiler]A[/spoiler]

I do not see it that way Sanju

[x^1/2007 - x^1/2009] if we take x^1/2009 as the common factor

x^1/2009 [x^(1/2007-1/2009) -1) which is NOT equal to x^1/2009 (x^2 -1)

Similar is the case with denominator

on top of that x^1/2009/x^1/2010 is not equal to x

ajith wrote:

sanju09 wrote:It's sad to see you lost the fifth choice on this occasion, but let's see what's in store

One thing is here to mention before we go for the elucidation, that on GMAT, whenever we find the phrases like x ≠ 0, ±1, etc, we should make sure that we're going to deal with some rational expression(s) that could contain factors like (x - 0), (x ± 1), etc in its denominators. In order to steer clear of an indeterminate state, they already make us observant with the precincts like that.

From the distance, it could be easily seen that the given rational expression has x^1/2009 as a common factor in the numerator with (x^2 - 1) to spare, and that it has x^1/2010 as a common factor in the denominator, once again with (x^2 - 1) to spare. Hence, (x^2 - 1) is the most visible common factor in the given rational expression, canceled and we are left with x^1/2009/ x^1/2010 = x (x^1/2010)/ x^1/2010 = [spoiler]x[/spoiler].

[spoiler]A[/spoiler]

I do not see it that way Sanju

[x^1/2007 - x^1/2009] if we take x^1/2009 as the common factor

x^1/2009 [x^(1/2007-1/2009) -1) which is NOT equal to x^1/2009 (x^2 -1)

Similar is the case with denominator

on top of that x^1/2009/x^1/2010 is not equal to x

Oh really! Let each of us give it another and closer look, see you soon, ajith

Ajith is absolutely correct, Sanju please prove your point.

Since if you take x^1/2009 commom you get x^1/2009[x^(1/2007-1/2009) - 1]

Ajith can you solve the problem, as it would be a great deal of help for me.

Anchit 777 wrote:Ajith is absolutely correct, Sanju please prove your point.

Since if you take x^1/2009 commom you get x^1/2009[x^(1/2007-1/2009) - 1]

Ajith can you solve the problem, as it would be a great deal of help for me.

I do not think any of the answer choices are right

Because I tried with x=2 and x=3 and I ended up getting values close to 1

I am sorry, I cannot think of anything

The answer is option A> x.

I definitely agree with the fact that no matter what number you put in place of x, you would always end up getting 1.00..something , but the answer is x, and i really wanted to know how.

I got solutions from a few people on Yahoo answers, but all the solutions seemed to be wrong like sanju's method of solving the problem.
I even tried goiit.com but no one has been able to answer me yet.

Anchit 777 wrote:The answer is option A> x.

I definitely agree with the fact that no matter what number you put in place of x, you would always end up getting 1.00..something , but the answer is x, and i really wanted to know how.

I got solutions from a few people on Yahoo answers, but all the solutions seemed to be wrong like sanju's method of solving the problem.
I even tried goiit.com but no one has been able to answer me yet.

I think the question is wrong, I honestly do.

Because if it is indeed x, it should work well with 2 and 3 also.

the question above doesn't give any bounds to x aside from saying that x <> -1,0,1

let's park that for now and simplify

(x^1/2007 - x^1/2009)/(x^1/2008 - x^1/2010)

[x^1/2009 * (x^2009/2007 - 1)] / [x^1/2010 * (x^2010/2008 - 1)]

x^1/2009 / x^1/2010 * (x^2009/2007 - 1)/(x^2010/2008 - 1)

x^(1/2009 - 1/2010) * (x^2009/2007 - 1)/(x^2010/2008 - 1)

x^[1/(2010*2009)] * (x^2009/2007 - 1)/(x^2010/2008 - 1)

if the question said |x| < 1,000,000,000 and the question asked what is the "approximate" value then we could assume


(x^2009/2007 - 1) ~ (x^2010/2008 - 1) AND
(x^2009/2007 - 1) / (x^2010/2008 - 1) ~ 1 AND THEREFORE

above equals

x^[1/(2010*2009)]

Turns out that this is a relatively good assumptions as long as x is small - when x gets HUGE (in the order of 10^100) this falls apart, see img below:






As posted, I don't think GMAC would ask this question - my 2 cents

sanju09 wrote:

ajith wrote:

sanju09 wrote:It's sad to see you lost the fifth choice on this occasion, but let's see what's in store

One thing is here to mention before we go for the elucidation, that on GMAT, whenever we find the phrases like x ≠ 0, ±1, etc, we should make sure that we're going to deal with some rational expression(s) that could contain factors like (x - 0), (x ± 1), etc in its denominators. In order to steer clear of an indeterminate state, they already make us observant with the precincts like that.

From the distance, it could be easily seen that the given rational expression has x^1/2009 as a common factor in the numerator with (x^2 - 1) to spare, and that it has x^1/2010 as a common factor in the denominator, once again with (x^2 - 1) to spare. Hence, (x^2 - 1) is the most visible common factor in the given rational expression, canceled and we are left with x^1/2009/ x^1/2010 = x (x^1/2010)/ x^1/2010 = [spoiler]x[/spoiler].

[spoiler]A[/spoiler]

I do not see it that way Sanju

[x^1/2007 - x^1/2009] if we take x^1/2009 as the common factor

x^1/2009 [x^(1/2007-1/2009) -1) which is NOT equal to x^1/2009 (x^2 -1)

Similar is the case with denominator

on top of that x^1/2009/x^1/2010 is not equal to x

Oh really! Let each of us give it another and closer look, see you soon, ajith

Sorry cronies for reappearing so late, thanks to a lingering server problem in this part of land since then. Two things to submit, first is that my distance vision didn't work well on this occasion and resulted in assuming rope in a snake sort of thing for me, that's why I suggested my/our self to have a closer look now. I knew the folly in my elucidation through ajith, so I started trying it again on paper. Secondly, I didn't find it to be a correct problem, as long as the choices are concerned. Just useless to say that it's NOT a GMAT query. Following are few efforts made by me, but to no avail:

Multiplying numerator and denominator by x^2010, we get

[x^ (2010/2007) - x^ (2010/2009)]
[x^ (2010/2008) - x^ (2010/2010)]

=

[x^ (1 + 3/2007) - x^ (1 + 1/2009)]
[x^ (1 + 2/2008) - x]

=

x [x^ (3/2007) - x^ (1/2009)]
x [x^ (2/2008) - 1]

And then a weak assumption ([spoiler]following words are not to be taken as valid[/spoiler])

Owing to the fact that the roots of the order of 2000 applied to a non-zero, non-unity number, x, is so small to assume that we can safely assume

That x^1/2007 = x^1/2008 = x^1/2009 = t, where t tends to 0, but it's never 0 either way.

Hence,

[x^ (1/2007)]^3 - [x^ (1/2009)]
[x^ (1/2008)]^2 - 1

=

t^3 - t
t^2 - 1

=

t(t^2 - 1 )
t^2 - 1

= t

???????????????????

There were few more efforts made by me, but I don't feel like typing so much for nothing. If Anchit is sure that the question is correct from all corners, and also, if he can provide us the source, then I can try it beyond GMAT specifications before saying died.

Hey m&m and sanju, I really appreciate the effort of you guys. Thanks.

I firmly agree with Ajith with the fact that this question is incorrect, since almost all answers are close to one.

Hence if any of the options would have one, that would be it.

Thanks again!

Anchit 777 wrote:For (x=/= 0 ,+-1,) the expression [x^1/2007 - x^1/2009] / [x^1/2008 - x^1/2010] is equivalent to :


A> x ,
B> x - 1 ,
C> x^2 - 1,
D>1/x .

x^1/2007 -x^1/2009 = x^1/2009 (x^2009/2007-1)
x^1/2008 - x^1/2010 = x^1/2010 (x^2010/2008 -1)
approximating, 2009/2007 =2010/2008 = 1
so exp = x^1/2009/x^1/2010 = x^ (1/2009 - 1/2010)

I think something is fishy in this question
ok archit ur soln. seems to be quite correct...