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One PR DS question !!

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One PR DS question !!

by amitansu » Sun Mar 30, 2008 9:16 pm
Yesterday i came across this DS question in PR...

The q says :

ABCD is a rectangle with sides of length as 'x' cm and width as 'y' cm.It has a diagonal of length 'z' cm (connection between A and C).What is the perimeter of rectangle in cm ?

1. x-y=z
2. z=13

Consider it as a data suffiency prob and suggest accordingly.

My pick was B i.e. 2 alone is only sufficient but answer says someting else !!Though i have gone through the solution still i am not convinced !!

Any suggestion ??
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by codesnooker » Sun Mar 30, 2008 11:51 pm
Hi is answer is (E)?

It is of course not (B) as follows:

By the property of triangle:- X^2 + Y^2 = Z^2

now if z = 13. then X^2 + Y^2 = 169.

I guess, here you have simply put X = 12 and Y=5 and deduce the perimeter as 34cms. Am I correct?

However, there are many other possible values for X and Y. For example, X = 11 and Y = 6.92 => Perimeter = 35.84cms.

So, perimeter is changed with the values of X and Y. So, statement 2nd is not alone sufficient to get constant perimeter.

Now. on including statement 1 also, => X - Y = Z

squaring both sides => X^2 + Y^2 -2XY = Z^2
=> X^2 + Y^2 -2XY = X^2 + Y^2
=> 2XY = 0

XY could only be zero when either of X or Y is zero, which hypothetically impossible for a property of rectangle or triangle.

So statement 1 is also not alone sufficient.

Now, consider both statement as together.
X - Y = Z & Z = 13
=> X - Y = 13

Again this is impossible, as by property of triangle X + Y > Z and X^2 + Y^2 = Z^2

X - Y can only be equal to 13 when X = 13 and Y = 0 or X > 13 and Y < 0, which again not possible.

So answer should be (E), none of the statement is sufficient to deduce the answer.

Let me know the correct answer and the explanation mentioned in the book.
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by amitansu » Mon Mar 31, 2008 12:15 am
Normally when i get some question wrong, i give it one try again without seeing the answer.Like you picked here, i also picked E as my second choice.

But the answer is C here as is explained by Princeton Review.
It's a lengthy explanation.I would try to put it in forum in evening as i am in office now :-)

thanks
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Re: One PR DS question !!

by camitava » Mon Mar 31, 2008 3:00 am
amitansu wrote:Yesterday i came across this DS question in PR...

The q says :

ABCD is a rectangle with sides of length as 'x' cm and width as 'y' cm.It has a diagonal of length 'z' cm (connection between A and C).What is the perimeter of rectangle in cm ?

1. x-y=z
2. z=13

Consider it as a data suffiency prob and suggest accordingly.

My pick was B i.e. 2 alone is only sufficient but answer says someting else !!Though i have gone through the solution still i am not convinced !!

Any suggestion ??
amitansu,
U faced this Qs in PR? Strange! This Qs seems to be little awkward. Dnt mind!
Now if u go with the stmt-1,
we know x^2 + y^2 = z^2
or (x - y)^2 + 2xy = z^2
or z^2 + 2xy = z^2
or xy = 0 -> x = 0 or y = 0
Now with stmt-2, we can not determine the value of x and y. Because x and y can be any value except 12 and 5 where z comes equal to 13.
Am I wrong here?
So the only way left is E to chose. IMO E. What's the OA?
Correct me If I am wrong


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Re: One PR DS question !!

by codesnooker » Mon Mar 31, 2008 3:50 am
camitava wrote: Now with stmt-2, we can not determine the value of x and y. Because x and y can be any value except 12 and 5 where z comes equal to 13.
Am I wrong here?
Yes you are correct. There are many more values which can satisfy X^2 + Y^2 = 13^2

As I mentioned above, X = 11 and Y = 6.92 and so on. So 12 and 5 are not only the values over here. Also it can be satisfied at X = 13 and Y = 0. However, then it wouldn't be a triangle or later on rectangle. :D
camitava wrote: So the only way left is E to chose. IMO E. What's the OA?
Answer is 'C' as mentioned by Amit
Last edited by codesnooker on Mon Mar 31, 2008 4:41 am, edited 1 time in total.
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by amitansu » Mon Mar 31, 2008 4:23 am
Strange !! the fact that i got this DS question on PR.
This is a part of BIN 4 (Math) which has only the hard category probs.

Few of the questions here i found are really mind benders like this one !!
In fact Kaplan 800 math to me came as easier than few questions right here in PR !!

Amit
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by codesnooker » Mon Mar 31, 2008 4:31 am
I am waiting for the solution.
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by amitansu » Mon Mar 31, 2008 4:38 am
Yes..once i reach home i would just copy-paste the answer... :-) though i am still not convinced with the answer as i said earlier !!
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Re: One PR DS question !!

by camitava » Mon Mar 31, 2008 4:45 am
codesnooker wrote:
camitava wrote: Now with stmt-2, we can not determine the value of x and y. Because x and y can be any value except 12 and 5 where z comes equal to 13.
Am I wrong here?
Yes you are wrong. There are many more values which can satisfy X^2 + Y^2 = 13^2

As I mentioned above, X = 11 and Y = 6.92 and so on. So 12 and 5 are not only the values over here. Also it can be satisfied at X = 13 and Y = 0. However, then it wouldn't be a triangle or later on rectangle. :D
camitava wrote: So the only way left is E to chose. IMO E. What's the OA?
Answer is 'C' as mentioned by Amit
Hey Codesnooker, Just look what I said. I said - Because x and y can be any value except 12 and 5 where z comes equal to 13. By this I wanted to mean that x and y can be any value. You also agreed to the point; but ...
So hence proved - my expression is not easily understandable! :cry: :wink: 8) :lol:
Ha ha ha!!! Just kidding... However, I am also curious to know about the explanation? Otherwise, after returning home from office, I have to look into PR! Hope for the best ...
Correct me If I am wrong


Regards,

Amitava
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Re: One PR DS question !!

by codesnooker » Mon Mar 31, 2008 4:50 am
Dude, I have corrected my post above. I am having headache from morning and mis-interpreted your statement, however when amit replied then I analysed my post and make it correct. Sorry to trouble you.
camitava wrote:
codesnooker wrote:
camitava wrote: Now with stmt-2, we can not determine the value of x and y. Because x and y can be any value except 12 and 5 where z comes equal to 13.
Am I wrong here?
Yes you are wrong. There are many more values which can satisfy X^2 + Y^2 = 13^2

As I mentioned above, X = 11 and Y = 6.92 and so on. So 12 and 5 are not only the values over here. Also it can be satisfied at X = 13 and Y = 0. However, then it wouldn't be a triangle or later on rectangle. :D
camitava wrote: So the only way left is E to chose. IMO E. What's the OA?
Answer is 'C' as mentioned by Amit
Hey Codesnooker, Just look what I said. I said - Because x and y can be any value except 12 and 5 where z comes equal to 13. By this I wanted to mean that x and y can be any value. You also agreed to the point; but ...
So hence proved - my expression is not easily understandable! :cry: :wink: 8) :lol:
Ha ha ha!!! Just kidding... However, I am also curious to know about the explanation? Otherwise, after returning home from office, I have to look into PR! Hope for the best ...
:D
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by camitava » Mon Mar 31, 2008 4:56 am
Hey codesnooker! Are u mad? Hey dnt take me in other sense! No need to be so formal. It was not ur mistake; may be I failed to explain it correctly. Ohhh! I would like to mention one thing -
When I started to post my response, there was none who placed his/ her responses. But after when I submitted, I found so many replies... Actually I really forgot to send my responses and became busy in office works. Ha ha ha! So sad with me ...
Hey this is just my experience I am sharing with u guys... Pls dnt take it in other sense ...
Correct me If I am wrong


Regards,

Amitava
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by amitansu » Mon Mar 31, 2008 7:20 am
Ok !! here is the expanation that PR gives...

Statement 1 tells us that x-y=z.An infinite no. of combination satisfy this eq., yielding an infinite no. of perimeters.NOT SUFFICIENT.

FromsStatement 2: there are many different ways to draw a rectangle with diagonal 13.Not all these rectangle have the same perimeter.So answer B is out of the mix.So answer must be C or E.At first glance the two statements together do not appear to offer much help.Start by squaring the statement 1 i.e. x^2-2xy+y^2=49 (Where from this 49 came into pic. don't know !!)

From statement 2 : Using Pythgoras theorem (x^2+y^2=z^2) we can rewrite as 169-2xy=49.
So xy=60
Thus we know x^2-2xy+y^2=289.
The sq. root of above eq.is (x+y) ?? (How ?) :?:
So we also know that x+y=17 (it can't equal -17 because x n yh are cm length).This eq. yielded by info. from both statements, can be used in conjunction with statement 1 to solve simoulteneously for x n y, thus allowing you to calculate the perimeter of rectangle.So correct answer is "C".

Somehow this explanation drives me insane !! I don't get convinced...
Can any of you explain me after reading this explanation please....


Amit
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by Stuart@KaplanGMAT » Mon Mar 31, 2008 9:21 am
There is NO solution to this problem. Either you mistranscribed it or PR messed it up to start with.

Let's try to simplify things. Remember, DS isn't about actually solving problems, it's about deciding if problems are solvable.

To find the perimiter, we need the length and the width. If we can find the legs (i.e. the non-hypotenuse sides) of our x/y/z triangle, we have length and width. So, we're trying to solve for either x and y individually OR x + y collectively (since 2(l+w) = perimiter).

(1) x - y = z

No numbers at all, no way to solve for perimiter: insufficient

(2) z = 13

We can determine that:

x^2 + y^2 = 169,

but from this there's no way to solve for either (x + y) or the individual values of x and y: insufficient.

So, neither statement is sufficient on its own - combination time!

We know that:

x - y = z

Squaring both sides, we get:

(x - y)^2 = z^2

and subbing in for z we get:

(x - y)^2 = 169

So:

x^2 - 2xy + y^2 = 169

or

x^2 + y^2 = 169 + 2xy

Now, Pythagoras also tells us that:

x^2 + y^2 = 169

So, we end up with:

169 + 2xy = 169

or

2xy = 0

which means that either x=0 or y=0, a geometrical impossibility.

In DS, there is ALWAYS at least one possible solution to the problem. Therefore, this question is IMPOSSIBLE.

Let's see why - and we just have to think about statement (1) for a second.

(1) x - y = z

or

x = y + z

Err... it's geometry, so x, y and z all have to be positive.

Umm... that means that x is GREATER THAN z.

Doh! If z is the hypotenuse of the triangle, then isn't Z automatically the BIGGEST of the 3 sides?

So, if z is the hypotenuse, it's IMPOSSIBLE for statement (1) to be true. We will NEVER see an impossible statement in a GMAT DS question.
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by mba.dude » Mon Mar 31, 2008 7:08 pm
I guess there must have been many typo's in the edition of PR that you are using as I just verified in my Book everything seems fine with this question.

The first statement is x-y = 7, and that is how x^2-2xy+y^2=49 .

z = 13

xy = 60.

And x^2+y^2 = z^2 = 169

So, (x+y)^2 = x^2+y^2+2xy = 169+120 = 289.
All the Best!
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by amitansu » Mon Mar 31, 2008 8:08 pm
Ok.Then there must be a typo error....because z could be 7 actually as mba-dude says... :)

But looking at the given solution as it is , i surely agree with Stuart.

Thanks all for the views.

Amit
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