I do not understand the this question, please help
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Please help to explain the question in the attachment. Thanks
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- smackmartine
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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. 9√2
B. 3/2
C. 9/√2
D. 15 (1-1/√2)
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)
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. 9√2
B. 3/2
C. 9/√2
D. 15 (1-1/√2)
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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- smackmartine
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OA is A, right?
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- Tani
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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
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
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
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
- Tani
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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.
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
- Tani
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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.
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