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Which is greater?

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Bunuel
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Which is greater? [#permalink] New post   12 Nov 2017, 01:52
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A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given

Kudos for correct solution.


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B
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Peter
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Re: Which is greater? [#permalink] New post   12 Nov 2017, 04:34
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Bunuel wrote:
Image

Image




A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given

Kudos for correct solution.


Show: ::
Attachment:
gqgpp_img11.png

Attachment:
gqgpp_img12.png


Don't know how it works; can anyone give explanation?
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IlCreatore
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Re: Which is greater? [#permalink] New post   12 Nov 2017, 10:15
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Let's start from the fact that \(\frac{AB}{AC}=\frac{AC}{BC}\), then given AB = AC+BC, we can rewrite the equality as \(\frac{AC+BC}{AC}=1+\frac{BC}{AC}=\frac{AC}{BC}\). Then, given that BC is smaller than AC, BC/AC is smaller than 1 so that summed to 1 is surely less than 3. Quantity B is greater.

Answer B
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Peter
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Re: Which is greater? [#permalink] New post   12 Nov 2017, 18:16
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IlCreatore wrote:
Let's start from the fact that \(\frac{AB}{AC}=\frac{AC}{BC}\), then given AB = AC+BC, we can rewrite the equality as \(\frac{AC+BC}{AC}=1+\frac{BC}{AC}=\frac{AC}{BC}\). Then, given that BC is smaller than AC, BC/AC is smaller than 1 so that summed to 1 is surely less than 3. Quantity B is greater.

Answer B


thank you for the info: but how do you make sure that BC is smaller than AC given that the picture might not be drawn to scale?
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wongpcla
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Re: Which is greater? [#permalink] New post   07 Dec 2017, 19:47
IlCreatore wrote:
Let's start from the fact that \(\frac{AB}{AC}=\frac{AC}{BC}\), then given AB = AC+BC, we can rewrite the equality as \(\frac{AC+BC}{AC}=1+\frac{BC}{AC}=\frac{AC}{BC}\). Then, given that BC is smaller than AC, BC/AC is smaller than 1 so that summed to 1 is surely less than 3. Quantity B is greater.

Answer B


how do you get the part 1+\frac{BC}{AC} ?
thank you, please help
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Re: Which is greater? [#permalink] New post   07 Dec 2017, 23:03
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Peter wrote:
IlCreatore wrote:
Let's start from the fact that \(\frac{AB}{AC}=\frac{AC}{BC}\), then given AB = AC+BC, we can rewrite the equality as \(\frac{AC+BC}{AC}=1+\frac{BC}{AC}=\frac{AC}{BC}\). Then, given that BC is smaller than AC, BC/AC is smaller than 1 so that summed to 1 is surely less than 3. Quantity B is greater.

Answer B


thank you for the info: but how do you make sure that BC is smaller than AC given that the picture might not be drawn to scale?


Good question,

But you can try putting the some values,

But remember it should satisfy the equation \(\frac{AB}{AC}=\frac{AC}{BC}\)

Let AB =12
AC= 6
BC = 3

These values satisfy the above equation , and \(\frac{AC}{BC}= 2\) which is less than 3
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Re: Which is greater? [#permalink] New post   07 Dec 2017, 23:07
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wongpcla wrote:
IlCreatore wrote:
Let's start from the fact that \(\frac{AB}{AC}=\frac{AC}{BC}\), then given AB = AC+BC, we can rewrite the equality as \(\frac{AC+BC}{AC}=1+\frac{BC}{AC}=\frac{AC}{BC}\). Then, given that BC is smaller than AC, BC/AC is smaller than 1 so that summed to 1 is surely less than 3. Quantity B is greater.

Answer B


how do you get the part 1+\frac{BC}{AC} ?
thank you, please help


Because AB = AC + BC

Now
\(\frac{AB}{AC}\) can be written as

\(\frac{(AC +BC)}{AC} = \frac{AC}{AC} + \frac{BC}{AC}\)

= \(1 + \frac{BC}{AC}\)
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Re: Which is greater? [#permalink] New post   24 Jan 2018, 07:00
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pranab01 wrote:
Peter wrote:
IlCreatore wrote:
Let's start from the fact that \(\frac{AB}{AC}=\frac{AC}{BC}\), then given AB = AC+BC, we can rewrite the equality as \(\frac{AC+BC}{AC}=1+\frac{BC}{AC}=\frac{AC}{BC}\). Then, given that BC is smaller than AC, BC/AC is smaller than 1 so that summed to 1 is surely less than 3. Quantity B is greater.

Answer B


thank you for the info: but how do you make sure that BC is smaller than AC given that the picture might not be drawn to scale?


Good question,

But you can try putting the some values,

But remember it should satisfy the equation \(\frac{AB}{AC}=\frac{AC}{BC}\)

Let AB =12
AC= 6
BC = 3

These values satisfy the above equation , and \(\frac{AC}{BC}= 2\) which is less than 3


I think if AB=12 and AC = 6 then BC must equal to 6 and not 3 as BC = AB-AC
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Matteo91
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Re: Which is greater? [#permalink] New post   24 Jan 2018, 10:44
Peter wrote:
IlCreatore wrote:
Let's start from the fact that \(\frac{AB}{AC}=\frac{AC}{BC}\), then given AB = AC+BC, we can rewrite the equality as \(\frac{AC+BC}{AC}=1+\frac{BC}{AC}=\frac{AC}{BC}\). Then, given that BC is smaller than AC, BC/AC is smaller than 1 so that summed to 1 is surely less than 3. Quantity B is greater.

Answer B


thank you for the info: but how do you make sure that BC is smaller than AC given that the picture might not be drawn to scale?


BC must be smaller, otherwise the equation in the question cannot hold.
Try to plug in numbers (with BC > AC) and you'll see :-)
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Re: Which is greater? [#permalink] New post   18 Jan 2021, 12:31
In case you're familiar with the golden ratio, that's what this question is. Short length over medium length = medium length over whole. The golden ratio is ~1.6
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Devesh26
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Which is greater? [#permalink] New post   07 Oct 2023, 21:59
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IlCreatore wrote:
Let's start from the fact that \(\frac{AB}{AC}=\frac{AC}{BC}\), then given AB = AC+BC, we can rewrite the equality as \(\frac{AC+BC}{AC}=1+\frac{BC}{AC}=\frac{AC}{BC}\). Then, given that BC is smaller than AC, BC/AC is smaller than 1 so that summed to 1 is surely less than 3. Quantity B is greater.

Answer B




Okay here since 1+ BC/AC = AC/BC. for AC/BC>1 from previous equation AC>BC and from that we can say BC/AC <1 and 1+(<1)<2

therefore AC/BC<2

and therefore quantity B is greater than quantity A
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