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HELP AGAIN.....Need the concept....

Expert replies
Source: — Data Sufficiency |

Re: HELP AGAIN.....Need the concept....

by lunarpower » Thu Jul 10, 2008 11:04 pm
smclean23 wrote:Is x2 equal to xy?
(1) x2 – y2 = (x + 5)(y - 5)
(2) x = y

Answer is B.....HUH?
let's hit statement (2) first, since that's easier.
if x and y are equal, then you can substitute x for y anywhere you see either one.
therefore, xy = xx (because you can take out y and substitute x) = x^2.
sufficient.

before we hit the other statement, let's rephrase:
x^2 = xy would mean x^2 - xy = 0, which would mean x(x - y) = 0.
therefore, here's a rephrase: is x = 0 or x = y?
(note that statement (2) becomes absolutely trivial to answer with this rephrase)
also, another rephrase, which is weird-looking but, as it turns out, supremely convenient for statement 1: is x(x - y) = 0?

statement (1):
number picking may be the best way to go here. it's a bit tricky, but we can cherry-pick a couple of choices that make both sides 0.
x = y = 5: answer to question = 'yes'
x = -5, y = 5: answer to question = 'no'
insufficient.
if we don't want to pick numbers, let's play with the algebra and see where it goes.
we could factor x^2 - y^2 into (x - y)(x + y), as we are wont to do in the vast majority of cases involving that expression, but that's a dead end here because there aren't any common factors with the expression on the right side.
therefore, try expanding the right side:
x^2 - y^2 = xy - 5x + 5y - 25
when we rephrased the question, we ran into the expression x^2 - xy along the way. since both of those terms are in the above equation, let's isolate the same combo: add y^2 to both sides and subtract xy from both sides.
x^2 - xy = y^2 - 5x + 5y - 25
x(x - y) = y^2 - 5x + 5y - 25
if the right side is 0, then the answer to the question is 'yes'; if it's not 0, the answer is 'no'.
the right side can be 0 (for instance, if x = y = 5). also, it's possible for the right side not to be 0 (for instance, if x = -5 and y = 5).
insufficient.

answer = a
Ron has been teaching various standardized tests for 20 years.

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by amitdgr » Fri Jul 11, 2008 3:58 am
Hi Ron,

Thanks for patiently typing in such a big explanation. I generally understand your explanations. This one went above my head :(

Amit
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by target790 » Fri Jul 11, 2008 8:57 am
(1)
x2 - y2= (x+5)(x-5)

=>x2 - y2 = x2 - 25
=>y2 = 25

so this option doesn't lead to anywhere by which we can prove x2=xy.
It can only get values for y(5,-5)

(2)x=y

Considering x2=x*x=x*y(replacing one x by y)
so this is sufficient.


So option B is correct
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by target790 » Fri Jul 11, 2008 9:02 am
Please ignore my previous post.I completely missed the 1st expression.
My mistake :oops:
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by lunarpower » Fri Jul 11, 2008 11:24 am
target790 wrote:(1)
x2 - y2= (x+5)(x-5)

=>x2 - y2 = x2 - 25
=>y2 = 25

so this option doesn't lead to anywhere by which we can prove x2=xy.
It can only get values for y(5,-5)

(2)x=y

Considering x2=x*x=x*y(replacing one x by y)
so this is sufficient.


So option B is correct
whoa whoa, wait juuuust a minute there.
the original statement 1 is (x + 5)(x - 5)?
oh man, that makes things so much easier (see previous post).
i was going with the original post, which contained (x + 5)(y - 5).
which one is it?
Ron has been teaching various standardized tests for 20 years.

--

Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi

--

Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.

Yves Saint-Laurent

--

Learn more about ron
Join the discussion