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M10-05

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New post 16 Sep 2014, 00:41
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What is the ratio of the area of a rectangular TV screen with a diagonal of 18'' to that of a rectangular screen with a diagonal of 15''?


(1) The ratio of width to length is the same for both screens.

(2) The width of the 18''-screen is 20% greater than that of the 15''-screen.

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New post 16 Sep 2014, 00:41
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Official Solution:


In two similar triangles (or rectangles), the ratio of their areas is the square of the ratio of their sides.

Given: ratio of diagonals \(= \frac{18}{15}=1.2\) (note that diagonals are the hypotenuses in right triangles made by width and length).

(1) The ratio of width to length is the same for both screens. This implies that the rectangles are similar, hence right triangles made by diagonals are similar as well, which means that the ratio of areas of triangles = ratio of areas of rectangles \(= ( \frac{18}{15} )^2 = 1.44\). Sufficient.

(2) The width of the 18''-screen is 20% greater than that of the 15''-screen. Given: the ratio of widths = 1.2 = ratio of diagonals (hypotenuse), so right triangles made by width and length are similar, (in right triangles if 2 corresponding sides are in the same ratio, in our case \(\frac{W}{w}=\frac{D}{d}=1.2\), then these right triangles are similar). Therefore the rectangles are similar too, which means that the ratio of areas of triangles = the ratio of areas of rectangles \(= ( \frac{18}{15} )^2=1.44\). Sufficient.


Answer: D
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New post 09 Oct 2014, 15:07
Bunuel wrote:
Official Solution:


In two similar triangles (or rectangles), the ratio of their areas is the square of the ratio of their sides.

Given: ratio of diagonals \(= \frac{18}{15}=1.2\) (note that diagonals are the hypotenuses in right triangles made by width and length).

(1) The ratio of width to length is the same for both screens. This implies that the rectangles are similar, hence right triangles made by diagonals are similar as well, which means that the ratio of areas of triangles = ratio of areas of rectangles \(= ( \frac{18}{15} )^2 = 1.44\). Sufficient.

(2) The width of the 18''-screen is 20% greater than that of the 15''-screen. Given: the ratio of widths = 1.2 = ratio of diagonals (hypotenuse), so right triangles made by width and length are similar, (in right triangles if 2 corresponding sides are in the same ratio, in our case \(\frac{W}{w}=\frac{D}{d}=1.2\), then these right triangles are similar). Therefore the rectangles are similar too, which means that the ratio of areas of triangles = the ratio of areas of rectangles \(= ( \frac{18}{15} )^2=1.44\). Sufficient.


Answer: D



I did not follow the highlighted part. Ratio of widths is same as ratio of hypotenuse.
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New post 10 Oct 2014, 02:41
earnit wrote:
Bunuel wrote:
Official Solution:


In two similar triangles (or rectangles), the ratio of their areas is the square of the ratio of their sides.

Given: ratio of diagonals \(= \frac{18}{15}=1.2\) (note that diagonals are the hypotenuses in right triangles made by width and length).

(1) The ratio of width to length is the same for both screens. This implies that the rectangles are similar, hence right triangles made by diagonals are similar as well, which means that the ratio of areas of triangles = ratio of areas of rectangles \(= ( \frac{18}{15} )^2 = 1.44\). Sufficient.

(2) The width of the 18''-screen is 20% greater than that of the 15''-screen. Given: the ratio of widths = 1.2 = ratio of diagonals (hypotenuse), so right triangles made by width and length are similar, (in right triangles if 2 corresponding sides are in the same ratio, in our case \(\frac{W}{w}=\frac{D}{d}=1.2\), then these right triangles are similar). Therefore the rectangles are similar too, which means that the ratio of areas of triangles = the ratio of areas of rectangles \(= ( \frac{18}{15} )^2=1.44\). Sufficient.


Answer: D



I did not follow the highlighted part. Ratio of widths is same as ratio of hypotenuse.


Stem says that the ratio of the diagonals is 18/15 = 1.2 and (2) says that the ration of widths is also 1.2. So, the ratio of widths = 1.2 = ratio of diagonals (hypotenuse).
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New post 01 Dec 2014, 15:49
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are we assuming that the screen is a square or a rectangle? The question doesn't specify and that is why I chose E.
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New post 16 Jan 2015, 03:03
I think this question is good and helpful.
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New post 27 Apr 2015, 17:28
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I think the questions needs to explicitly state that the screen is "rectangle" . I marked E because these days we have curved TV's as well.
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New post 30 Jun 2015, 11:33
Hi,

I got this question right doing educated guess... I was sure about (a) BUT was short on time and I decided to guess on (d) because I identified that 20 % more combined with the known lenghts of the diagonal (15 and 18) was enough to know de ratio of the areas; but in the explanation it says that this is concluded because 18/15 = 1.2 and 20 % more is also 1.2... then I doubt If my thought was roght, for instance if instead of 20 % more would be 30 % more then what would be the answer? The same or it would change something?

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New post 01 Jul 2015, 01:50
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luisnavarro wrote:
Hi,

I got this question right doing educated guess... I was sure about (a) BUT was short on time and I decided to guess on (d) because I identified that 20 % more combined with the known lenghts of the diagonal (15 and 18) was enough to know de ratio of the areas; but in the explanation it says that this is concluded because 18/15 = 1.2 and 20 % more is also 1.2... then I doubt If my thought was roght, for instance if instead of 20 % more would be 30 % more then what would be the answer? The same or it would change something?

Thanks

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No, in this case (30%) the statement wouldn't be sufficient.

Ratio of the areas of the triangles with sides \(18, \ 13, \ \sqrt{18^2-13^2}\) and \(15, \ 10, \ \sqrt{15^2-10^2}\) is NOT the same as the ratio of the areas of the triangles with sides \(18, \ 6.5, \ \sqrt{18^2-6.5^2}\) and \(15, \ 5, \ \sqrt{15^2-5^2}\).
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New post 01 Jul 2015, 05:43
Bunuel wrote:
luisnavarro wrote:
Hi,

I got this question right doing educated guess... I was sure about (a) BUT was short on time and I decided to guess on (d) because I identified that 20 % more combined with the known lenghts of the diagonal (15 and 18) was enough to know de ratio of the areas; but in the explanation it says that this is concluded because 18/15 = 1.2 and 20 % more is also 1.2... then I doubt If my thought was roght, for instance if instead of 20 % more would be 30 % more then what would be the answer? The same or it would change something?

Thanks

Best regards

Luis Navarro
Looking for 700


No, in this case (30%) the statement wouldn't be sufficient.

Ratio of the areas of the triangles with sides \(18, \ 13, \ \sqrt{18^2-13^2}\) and \(15, \ 10, \ \sqrt{15^2-10^2}\) is NOT the same as the ratio of the areas of the triangles with sides \(18, \ 6.5, \ \sqrt{18^2-6.5^2}\) and \(15, \ 5, \ \sqrt{15^2-5^2}\).



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New post 23 Aug 2015, 07:52
I think this is a high-quality question and I agree with explanation. Can we solve this prob by doing this :

Let's assume 2x=18, then we know that in a rectangle triangle we have the legs given by x and xsqrt(3)
same with 2x=15
We found that the stem is sufficient to solve the problem.
the ratio is
(9*9*sqrt(3))/(15/2*15/2*sqrt(3))=1.44

What is wrong with this way ?
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New post 22 Nov 2015, 09:22
if the screens were squares then the question would become a problem solving type of question, right?
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New post 17 Dec 2016, 00:42
(2) The width of the 18''-screen is 20% greater than that of the 15''-screen. Given: the ratio of widths = 1.2 = ratio of diagonals (hypotenuse), so right triangles made by width and length are similar


How can u say the ratio of widths= ratio of diagonals?

Ratio of lengths should also be given right!!
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New post 17 Dec 2016, 03:48
Sumanth8492 wrote:
(2) The width of the 18''-screen is 20% greater than that of the 15''-screen. Given: the ratio of widths = 1.2 = ratio of diagonals (hypotenuse), so right triangles made by width and length are similar


How can u say the ratio of widths= ratio of diagonals?

Ratio of lengths should also be given right!!


Stem says that the ratio of the diagonals is 18/15 = 1.2 and (2) says that the ration of widths is also 1.2. So, the ratio of widths = 1.2 = ratio of diagonals (hypotenuse).
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New post 25 Aug 2017, 22:44
Hi Bunuel,

explanation is clear, however although concluded it as sufficient as well, I wasn't thinking of the logic you presented for S2 (pressed for time). So I want to clear up whether, in this case, doesn't the very fact that S1 proves sufficient (i.e. similar rectangles thus similar ratio of areas) necessarily mean that if S2 provides measurements of the rectangles' sides, such measurements must also be in a 1.2 ratio? For instance, could S2 have been 30% instead of 20% or wouldn't that be a flawed question where the statements provide incongruent answers (no pun intended)? Thx
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New post 26 Aug 2017, 02:11
ulysses02 wrote:
Hi Bunuel,

explanation is clear, however although concluded it as sufficient as well, I wasn't thinking of the logic you presented for S2 (pressed for time). So I want to clear up whether, in this case, doesn't the very fact that S1 proves sufficient (i.e. similar rectangles thus similar ratio of areas) necessarily mean that if S2 provides measurements of the rectangles' sides, such measurements must also be in a 1.2 ratio? For instance, could S2 have been 30% instead of 20% or wouldn't that be a flawed question where the statements provide incongruent answers (no pun intended)? Thx


On the GMAT, two data sufficiency statements always provide TRUE information and these statements NEVER contradict each other or the stem.
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New post 26 Aug 2017, 09:55
Bunuel wrote:
ulysses02 wrote:
Hi Bunuel,

explanation is clear, however although concluded it as sufficient as well, I wasn't thinking of the logic you presented for S2 (pressed for time). So I want to clear up whether, in this case, doesn't the very fact that S1 proves sufficient (i.e. similar rectangles thus similar ratio of areas) necessarily mean that if S2 provides measurements of the rectangles' sides, such measurements must also be in a 1.2 ratio? For instance, could S2 have been 30% instead of 20% or wouldn't that be a flawed question where the statements provide incongruent answers (no pun intended)? Thx


On the GMAT, two data sufficiency statements always provide TRUE information and these statements NEVER contradict each other or the stem.


Thus, S2 could not possibly have been 30%, correct? (I'm implicitly referring to an earlier reply you made to luisnavarro which said if S2 was 30%, then it would be insufficient)
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New post 27 Aug 2017, 03:52
ulysses02 wrote:
Bunuel wrote:
ulysses02 wrote:
Hi Bunuel,

explanation is clear, however although concluded it as sufficient as well, I wasn't thinking of the logic you presented for S2 (pressed for time). So I want to clear up whether, in this case, doesn't the very fact that S1 proves sufficient (i.e. similar rectangles thus similar ratio of areas) necessarily mean that if S2 provides measurements of the rectangles' sides, such measurements must also be in a 1.2 ratio? For instance, could S2 have been 30% instead of 20% or wouldn't that be a flawed question where the statements provide incongruent answers (no pun intended)? Thx


On the GMAT, two data sufficiency statements always provide TRUE information and these statements NEVER contradict each other or the stem.


Thus, S2 could not possibly have been 30%, correct? (I'm implicitly referring to an earlier reply you made to luisnavarro which said if S2 was 30%, then it would be insufficient)

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Re: M10-05  [#permalink]

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New post 27 Dec 2017, 14:44
Bunuel wrote:
Official Solution:


In two similar triangles (or rectangles), the ratio of their areas is the square of the ratio of their sides.

Given: ratio of diagonals \(= \frac{18}{15}=1.2\) (note that diagonals are the hypotenuses in right triangles made by width and length).

(1) The ratio of width to length is the same for both screens. This implies that the rectangles are similar, hence right triangles made by diagonals are similar as well, which means that the ratio of areas of triangles = ratio of areas of rectangles \(= ( \frac{18}{15} )^2 = 1.44\). Sufficient.

(2) The width of the 18''-screen is 20% greater than that of the 15''-screen. Given: the ratio of widths = 1.2 = ratio of diagonals (hypotenuse), so right triangles made by width and length are similar, (in right triangles if 2 corresponding sides are in the same ratio, in our case \(\frac{W}{w}=\frac{D}{d}=1.2\), then these right triangles are similar). Therefore the rectangles are similar too, which means that the ratio of areas of triangles = the ratio of areas of rectangles \(= ( \frac{18}{15} )^2=1.44\). Sufficient.


Answer: D



Hi Bunuel,
I have encountered a genuine doubt while solving this question. Are not the diagonals of a rectangle equal in length? If it's so, then in any case the area of each rectangle can be determined using the formula 1/2*(diagonal)^2 and the ratio of their areas can be determined. This would make the 2 statements not required to answer the question itself.
Please, clarify this doubt. I am really confused.
Thanks.
Re: M10-05 &nbs [#permalink] 27 Dec 2017, 14:44

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