Register    Login    Search    Rss Feeds

 Page 1 of 1 [ 2 posts ] 



 
Author Message
 Post subject: The perimeters of square region S and rectangular region R
 Post Posted: Mon Feb 16, 2009 1:48 pm 
Offline


Posts: 1
Practice Test 1, MBA.com

I wanted to know if there was a faster way to approach the below problem, especially if i would not be able to think of appropriate numbers to pick right away:

The perimeters of square region S and rectangular region R are equal. If the sides of R are in the ratio 2:3, what is the ratio of the area of region R to the area of region S?
25:16
24: 25
5:6
4:5
4:9

Ans. B

I picked numbers:
SQUARE: let a side length of a sqaure = 5 thus perimeter 20
RECTANGLE: width = 4 and length 6 = 20 as the perimeter
Perimeters are equal; areas of rectangle is 24/25-area of square


Top 
 Post subject: Re: The perimeters of square region S and rectangular region R
 Post Posted: Wed Feb 18, 2009 8:29 am 
Offline
ManhattanGMAT Staff


Posts: 7146
pairadox_20 wrote:
Practice Test 1, MBA.com

I wanted to know if there was a faster way to approach the below problem, especially if i would not be able to think of appropriate numbers to pick right away:

The perimeters of square region S and rectangular region R are equal. If the sides of R are in the ratio 2:3, what is the ratio of the area of region R to the area of region S?
25:16
24: 25
5:6
4:5
4:9

Ans. B

I picked numbers:
SQUARE: let a side length of a sqaure = 5 thus perimeter 20
RECTANGLE: width = 4 and length 6 = 20 as the perimeter
Perimeters are equal; areas of rectangle is 24/25-area of square


--

well, you're picking numbers the hard way. you should pick the sides of the rectangle FIRST, according to the ratio, because it's harder to use the ratio if you're already picked other numbers.
since it's absolutely trivial to find the side of a square from its perimeter, you should leave the square for last.
always save the EASIEST or LEAST RESTRICTED number picking for LAST.

so, for instance, you could pick a rectangle whose measurements are literally 2 x 3. then the perimeter is 10. the square would thus be 2.5 units on a side.
since decimals suck, let's double everything: make the rectangle 4 x 6. then the perimeter is 20, so the square is 20/4 = 5 units on a side.
these are the same numbers you got, but it's easier to get them this way.

--

the other way to do this problem is to let the sides of the rectangle be 2x and 3x (standard approach to ratios: the "multiplier approach"). then the perimeter is 10x, so the side of the square is one-fourth of that, or 5x/2.
the areas are 6x^2 and (25/4)x^2. the ratio of 6 to 25/4 is 24/4 to 25/4, or 24/25.

picking numbers is easier.


Top 
Display posts from previous:  Sort by  
 
 Page 1 of 1 [ 2 posts ] 





Who is online

Users browsing this forum: No registered users and 1 guest

 
 

 
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to: