Question: Rectangle’s height

Comment on Rectangle’s height

Hi,
As per the ques https://gre.myprepclub.com/forum/a-project-requires-a-rectangular-sheet-of-cardboard-satisfyi-12275.html

The resulting rectangles has the same ratio of length to width as the original sheet : what is exactly meant by that
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/a-project-requires-a-rectangular-sheet-...

So, for example, let's we have a rectangle such that length : width = x : y
If we were to cut that rectangle in half, we want EACH of resulting rectangles to have a length to width ratio of x : y

In the following solution, I provide a diagram that may help: https://gre.myprepclub.com/forum/a-project-requires-a-rectangular-sheet-...

Please let me know if you'd like me to elaborate on my solution.

Cheers,
Brent

Hi,

I got that but Iam not able understand

as you wrote "we want x/y = y/(x/2)" . I have checked by plotting some values and your statement is correct

but on the other hand, as per the ques should I interpret that x : y = x : 2y ( taking half of length) and not
x/y = 2y/x .

Plz let me know what am I missing
Thanks
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/a-project-requires-a-rectangular-sheet-...

The question isn't worded very well.

However, for this question, the equation x : y = x : 2y is unsolvable (unless x = y = 0)

We're comparing the dimensions of the original rectangle to the dimensions of one of the newly-formed rectangle.

For this comparison, it would probably be better/clearer to state that we want the following to be true:

(length of shorter side of ORIGINAL rectangle) : (length of longer side of ORIGINAL rectangle) = (length of shorter side of NEW rectangle) : (length of longer side of NEW rectangle)

Does that help?

Cheers,
Brent

greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/a-project-requires-a-rectangular-sheet-...

The question isn't worded very well.

However, for this question, the equation x : y = x : 2y is unsolvable (unless x = y = 0)

We're comparing the dimensions of the original rectangle to the dimensions of one of the newly-formed rectangle.

For this comparison, it would probably be better/clearer to state that we want the following to be true:

(length of shorter side of ORIGINAL rectangle) : (length of longer side of ORIGINAL rectangle) = (length of shorter side of NEW rectangle) : (length of longer side of NEW rectangle)

Does that help?

Cheers,
Brent

Hi,
I have understood this part (length of shorter side of ORIGINAL rectangle) : (length of longer side of ORIGINAL rectangle) = (length of shorter side of NEW rectangle) : (length of longer side of NEW rectangle)

But why x/y = y/(x/2) =2y/x won't it should be x/y = x/2y?

May I assume that to hold the above condition to be true it should be x/y = y/(x/2) =2y/x , so it will be as close to the original ratio

Plz clarify
greenlight-admin's picture

When we cut the original rectangle in half (with dimensions x by y), one of new rectangles will have the dimensions x/2 by y (since we're cutting the side with length x into two pieces.

So, based on the equation (length of shorter side of ORIGINAL rectangle) : (length of longer side of ORIGINAL rectangle) = (length of shorter side of NEW rectangle) : (length of longer side of NEW rectangle)...

...the equation becomes: y/x = (x/2)/y

We can take the ratio (x/2)/y and multiply top and bottom by 2 to create an EQUIVALENT ratio x/2y

So, our equation becomes y/x = x/2y

Of if we flip both ratios, we get: x/y = 2y/x (either is valid)

Cheers,
Brent

Understood. Thanks Brent

Hi, on the GRE if they mention the length and width of a rectangle, can we assume the length size is longer that the width side? I assume no.
greenlight-admin's picture

Great question!

On the GRE, the length need not be longer than the width.

Cheers,
Brent

Have a question about this video?

Post your question in the Comment section below, and a GRE expert will answer it as fast as humanly possible.

Change Playback Speed

You have the option of watching the videos at various speeds (25% faster, 50% faster, etc). To change the playback speed, click the settings icon on the right side of the video status bar.

Let me Know

Have a suggestion to make the course even better? Email us today!

Free “Question of the Day” emails!