I do not understand the this question, please help

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tvtt2010: Junior | Next Rank: 30 Posts; Posts: 19; Joined: Mon Mar 21, 2011 2:57 pm

by tvtt2010 » Mon Aug 08, 2011 5:24 pm

Please help to explain the question in the attachment. Thanks

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Source: Beat The GMAT — Problem Solving | Mon Aug 08, 2011 5:24 pm

smackmartine: Legendary Member; Posts: 516; Joined: Fri Jul 31, 2009 3:22 pmFollowed by: 13 members; Thanked: 112 times

by smackmartine » Mon Aug 08, 2011 5:59 pm

Posting the question:

The shaded region in the figure above represents a rectangular frame with length 18inches and width 15 inches. The frame encloses a rectangular picture that has the same area as the frame itself. If the length and width of the picture have the same ratio as the length and width of the frame, what is the length of the picture, in inches?
A. 9âˆš2
B. 3/2
C. 9/âˆš2
D. 15 (1-1/âˆš2)
E. 9/2

IMO A

Let the length of the picture be l and width be w.
The length and width of the picture have the same ratio as the length and width of the frame.

l/w = 18/15 = 6/5
we can easily assume l= 6x and w=5x

The question basically says that the area of picture = area of the frame itself (white part= blue part in the diagram)

6x * 5x =(18*15)-(6x * 5x)
2*(6x * 5x) = 18*15
x^2=18/4
x=3*rt(2)/2

So, length of the picture is 6x ---> [6*3*rt(2)]/2 = 9rt(2)

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tvtt2010: Junior | Next Rank: 30 Posts; Posts: 19; Joined: Mon Mar 21, 2011 2:57 pm

by tvtt2010 » Mon Aug 08, 2011 7:36 pm

Thanks, the reason i did not figure out the answer because I did not think about the area of the frame is just the area of the blue part. I forgot to minus the area of the white part.

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smackmartine: Legendary Member; Posts: 516; Joined: Fri Jul 31, 2009 3:22 pmFollowed by: 13 members; Thanked: 112 times

by smackmartine » Mon Aug 08, 2011 7:40 pm

OA is A, right?

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Tani: Legendary Member; Posts: 1255; Joined: Fri Nov 07, 2008 2:08 pmFollowed by: 90 members; Location: St. Louis; Thanked: 312 times

by Tani » Mon Aug 08, 2011 8:09 pm

Another way to look at this is: you know the area of the frame is 135 sq inches.

The ratio of length to width is 6:5 If we let the length = L, then the width = (5/6)L

L*W = 135 so L * (5/6)L = 135 L^2 = 162 L = sqrt 162 = 9 sqrt2

Tani Wolff

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tvtt2010: Junior | Next Rank: 30 Posts; Posts: 19; Joined: Mon Mar 21, 2011 2:57 pm

by tvtt2010 » Thu Aug 11, 2011 5:42 pm

Yes, the answer is A.

I have this question, my answer is C but the GMAT answer key is E. Can you explain to me

IF n and k are intergers whose product is 400, which of the following statements must be true?
A. n+k>0
B. n#k
C. Either n or k is a multiple of 10
D. if n is even then k is odd
E. if n is odd, then k is even

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Tani: Legendary Member; Posts: 1255; Joined: Fri Nov 07, 2008 2:08 pmFollowed by: 90 members; Location: St. Louis; Thanked: 312 times

by Tani » Thu Aug 11, 2011 6:32 pm

First look at E - If n is odd, K must be even. Since the product is even, one of the factors must be even. That makes E true.

Looking at C, consider the prime factors of 400: 2^4 * 5^2. To be divisible by 10, n or k would have to have one of each of the distinct prime factors (2 and 5). However, it is possible to split them so that n has all the 2s and k has all the 5s. That would give us 16 * 25 = 400. Neither factor divisible by 10.

Tani Wolff

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tvtt2010: Junior | Next Rank: 30 Posts; Posts: 19; Joined: Mon Mar 21, 2011 2:57 pm

by tvtt2010 » Thu Aug 11, 2011 7:37 pm

but it is can also be D, because we just need either n or k is even, why not D

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Tani: Legendary Member; Posts: 1255; Joined: Fri Nov 07, 2008 2:08 pmFollowed by: 90 members; Location: St. Louis; Thanked: 312 times

by Tani » Fri Aug 12, 2011 5:46 am

No, because they can both be even - for example 2 * 200. The key is that AT LEAST ONE must be even since 400 is even.

Tani Wolff

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tvtt2010: Junior | Next Rank: 30 Posts; Posts: 19; Joined: Mon Mar 21, 2011 2:57 pm

by tvtt2010 » Fri Aug 12, 2011 8:16 am

I still do not get it, because number D is just reverse of number E, and in D, either n or k is even.

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Tani: Legendary Member; Posts: 1255; Joined: Fri Nov 07, 2008 2:08 pmFollowed by: 90 members; Location: St. Louis; Thanked: 312 times

by Tani » Fri Aug 12, 2011 3:28 pm

Try picking numbers. You can have 10 and 40 - both even, or 5 and 16 - one odd and one even, but you do not have to have one of each.

E tells you that you have to have at least one even. D claims that you have to have one odd, which is not true.

Tani Wolff

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