"Silver" rectangle

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Goldfinger2001: Senior | Next Rank: 100 Posts; Posts: 50; Joined: Mon Jan 03, 2011 9:22 am

by Goldfinger2001 » Sun Feb 13, 2011 11:21 am

A rectangle is defined to be "silver" if and only if the ratio of its length to its width is 2 to 1. If rectangle S is silver, is rectangle R silver?

1) R has the same area as S
2) The ratio of one side of R to one side of S is 2 to 1.

OA: E

My answer would have been C since I think that if one side of R has a ratio of 2 to 1 to one side of S + the same area, it cannot be a silver rectangle....

Example

S: W=2, L=4

R: (Statement 2) lets take W.... W=4 so in order to have the same area, L has to be 2 -> no silver rectangle
R: (Statement2) lets take L...L=8 so in order to have the same area, W has to be 1 -> no silver rectangle....

Can someone help me out here?

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Source: Beat The GMAT — Data Sufficiency | Sun Feb 13, 2011 11:21 am

Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Sun Feb 13, 2011 5:22 pm

Goldfinger2001 wrote:A rectangle is defined to be "silver" if and only if the ratio of its length to its width is 2 to 1. If rectangle S is silver, is rectangle R silver?

1) R has the same area as S
2) The ratio of one side of R to one side of S is 2 to 1.

OA: E

l/w=2/1, 2w=l
st(1) R(l*w)=S, S cold be 2l and w Not Sufficient;
st(2) R(l)/S(l)=2/1 OR R(s)/S(l) no relative data about S(l and w) Not Sufficient
Combined st(1&2): Still not sufficient, as it doesn't precisely state which side is 2* the other side.

IOM E

My answer would have been C since I think that if one side of R has a ratio of 2 to 1 to one side of S + the same area, it cannot be a silver rectangle....

Example

S: W=2, L=4

R: (Statement 2) lets take W.... W=4 so in order to have the same area, L has to be 2 -> no silver rectangle
R: (Statement2) lets take L...L=8 so in order to have the same area, W has to be 1 -> no silver rectangle....

Can someone help me out here?

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David@VeritasPrep: GMAT Instructor; Posts: 2193; Joined: Mon Feb 22, 2010 6:30 pmFollowed by: 512 members; Location: Vermont and Boston, MA; Thanked: 1186 times; GMAT Score:770

by David@VeritasPrep » Sun Feb 13, 2011 6:06 pm

This question is in the Veritas Geometry book.

A rectangle is defined to be "silver" if and only if the ratio of its length to its width is 2 to 1. If rectangle S is silver, is rectangle R silver?

1) R has the same area as S
2) The ratio of one side of R to one side of S is 2 to 1.

OA: E

With statement 1 imagine two rectangles with the same area. Say the area is 8. This would mean that triangle S would have to have length 4 and width 2. This is silver. Now R could have the exact same measurements and also be silver or it could have length 8 and width 1 and not be silver. So this is not sufficient.

With statement 2 we can use exactly the same example. If both R and S have length of 4 and width of 2 then we could have length or R to width of S and satisfy statement 2. So R could be silver. It could also be that R has a length of 8 and width of 1 since no we could say the length of R (8) to the length of S (4) and this also satisfies statement 2. So this is also not sufficient.

Taking both statements together you can see that we have used the same example for each - always a clever strategy when possible - so that even together the statements still allow for R to be Length = 4 width =2 which is silver and length = 8 width = 1 which is not silver.

Therefore not sufficient and the answer is E.

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Goldfinger2001: Senior | Next Rank: 100 Posts; Posts: 50; Joined: Mon Jan 03, 2011 9:22 am

by Goldfinger2001 » Sun Feb 13, 2011 10:59 pm

Hi David,

I understand that statements 1 and 2 alone are insufficient.

If you combine both though you should not be able to get a silver rectangle:

Unless it doesn't matter if we switch W and L, the last figure is no silver rectangle since the ratio L to W is not 2:1, it is 1 to 2.

Do you see where I am stuck?

Please help me out there

Thanks!

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Fatehdeep Singh: Junior | Next Rank: 30 Posts; Posts: 10; Joined: Fri Feb 19, 2010 7:05 am; Thanked: 2 times

by Fatehdeep Singh » Mon Feb 14, 2011 4:05 am

Answer shud be C.

We can rule out both A and B

But if you consider C
Take II first:R has one side double than that of S.If dimensions of S are L=4,W=1.Area of S =4
Now for R.Dimensions can be L=8,W=x(say) or L=x(say),W=2.Area for R can be 8x or 2x
Take I: Area is same for both rectangles
So case 1: 8x=4; x=0.5.So,dimensions of rec R are L=8,W=0.5
Not a rectangle,because as per rec def,ratio of L to W needs to be 2:1
So,definite answer.
CaseII:2x=4;x=2.So,dimensions of rec R are L=2,W=2
Not a rectangle,because as per rec def,ratio of L to W needs to be 2:1
So,definite answer.

So,combining both statements,we can have definite answer.
So,answer is C.

Deg Teg Fateh!!!

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David@VeritasPrep: GMAT Instructor; Posts: 2193; Joined: Mon Feb 22, 2010 6:30 pmFollowed by: 512 members; Location: Vermont and Boston, MA; Thanked: 1186 times; GMAT Score:770

by David@VeritasPrep » Mon Feb 14, 2011 6:25 am

Sorry to have to tell you guys but you made a small assumption here...

Look at statement 2. it says 2) The ratio of one side of R to one side of S is 2 to 1.

This does not say that it needs to be the length of R compared to the length of S, just one side to one side.

So Goldfinger you are correct when you say "unless it does not matter if we compare length to width" instead of length to length or width to width.

So that is the reason why the drawing that you have for S would also work as a drawing for R since it would satisfy statement 1 to have the same area if R and S have the same exact dimensions but it would also satisfy statement 2 as we compare the length of R at 4 to the width of S at 2.

This may seem even a little unfair but remember that the one thing that official GMAT questions (and this is not one but very similar) are best at is getting you to make a small assumption or answer the wrong question or in some way trick you!

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Goldfinger2001: Senior | Next Rank: 100 Posts; Posts: 50; Joined: Mon Jan 03, 2011 9:22 am

by Goldfinger2001 » Mon Feb 14, 2011 6:46 am

Thanks David,

you guys tricked me here.

One question about the veritas books (I took the 7 days accelerated course 3 weeks ago):

How to you feel about the questions at the end of each book? Do you think if you do all questions and understand them, that you can get a 720+?

I m asking, because I also have the OG12 and did some 700-800 math questions at the end, they seem quite easy for me?! Can that be?

I hope to get an unbiased answer

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Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Mon Feb 14, 2011 7:09 am

@Fatehdeep Singh

Night reader wrote:
Goldfinger2001 wrote:A rectangle is defined to be "silver" if and only if the ratio of its length to its width is 2 to 1. If rectangle S is silver, is rectangle R silver?

1) R has the same area as S
2) The ratio of one side of R to one side of S is 2 to 1.

OA: E

l/w=2/1, 2w=l
st(1) R(l*w)=S, S cold be 2l and w Not Sufficient;
st(2) R(l)/S(l)=2/1 OR R(s)/S(l) no relative data about S(l and w) Not Sufficient
Combined st(1&2): Still not sufficient, as it doesn't precisely state which side is 2* the other side.

IOM E

My answer would have been C since I think that if one side of R has a ratio of 2 to 1 to one side of S + the same area, it cannot be a silver rectangle....

Example

S: W=2, L=4

R: (Statement 2) lets take W.... W=4 so in order to have the same area, L has to be 2 -> no silver rectangle
R: (Statement2) lets take L...L=8 so in order to have the same area, W has to be 1 -> no silver rectangle....

Can someone help me out here?

Like this post

David@VeritasPrep: GMAT Instructor; Posts: 2193; Joined: Mon Feb 22, 2010 6:30 pmFollowed by: 512 members; Location: Vermont and Boston, MA; Thanked: 1186 times; GMAT Score:770

by David@VeritasPrep » Mon Feb 14, 2011 7:12 am

2 kinds of difficulty on the GMAT quantitative conceptual difficulty and sneaky difficulty. The question above is definitely in the sneaky difficulty line.

The OG has some nice sneaky questions but nearly all will agree that they do not give you enough of the conceptual difficulty. So maybe that is what you mean by they seemed easy to you. It is interesting that you call them 700-800 questions? How do you know what level they are? The consensus is that these are not very many so-called 700 level questions in the OG...

The Veritas books have much more conceptual difficulty than the OG. The Veritas books also have sneaky difficulty, although sneaky is what the OG does best.

To your question, if you answer and understand all of the Veritas questions in the books you will have the tools to get over a 720 but there are so many variables, such as timing, your strategy, how you perform on test day, etc. So I cannot say what you will score but you will have the tools to get there.

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Goldfinger2001: Senior | Next Rank: 100 Posts; Posts: 50; Joined: Mon Jan 03, 2011 9:22 am

by Goldfinger2001 » Mon Feb 14, 2011 7:17 am

The MGMAT OG guide offers these "difficulty levels" for quant OG questions.

Thanks again for the explanaiton!

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