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plz help me solve this sum

Expert replies
by sakan » Thu Jun 19, 2008 7:37 pm
The perimeter of square S and rectangular R are equal,if the sides of r are in the ratio 2:3,what is the ratio of the region R to the area of region S?

1)25:16

2)24:25

3)5:6

4)4:5

5)4:9
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Source: — Problem Solving |

by chidcguy » Thu Jun 19, 2008 7:56 pm
4S=2l+2w

l/w=2/3

lw/s^2 =??

lw= l X 3l /2

4S=5l S=5l/4 S^2 = 25l^2/16

lw/s^2 = (3/2)/(25/16) =24/25
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by kishore » Thu Jun 19, 2008 9:01 pm
The perimeter of square S and rectangular R are equal,if the sides of r are in the ratio 2:3,what is the ratio of the region R to the area of region S?


Perimter of square = 4*S

Perimater of rectangle = 2(l+b)

Both the perimeters are equal

=> 4*S = 2(l+b)

Since, the 2 sides of R are in the ratio 2:3
l = 2*k and b = 3*k


=> 4* S = 2(2*k + 3*k)
=> 4* S = 2(5 *k)
=> 4* S = 10* K
=> S = 5/2 * K


ratio of the region R to the area of region S?
Area of rectangle R = l*b
Area of square S = S ^2

R:S = l*b: s^2
= (2* k) *(3*k) : (5/2 * k) ^2
= 6 * k^2 : 25/4 * k^2
= 6:25/4
multiply both by 4

= 6 * 4 : 25/4 * 4
= 24: 25
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by AleksandrM » Fri Jun 20, 2008 9:40 am
At what level of difficulty is this problem?

It took me almost 3 minutes to brainstorm it and carefully write out the arithmetic to avoid making a stupid mistake.
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by kishore » Fri Jun 20, 2008 10:01 am
This is a simple problem and might take max of 1 min.

Medium level of difficulty i guess....
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by atlantic » Fri Jun 20, 2008 10:20 am
Well guys, I think GMAT is about choosing the right answer, not spending time in finding algebric solutions.

How I did it......

We know that for a given perimeter, the square is the geometric form that maximize the area. So the ratio between a rectangle and square with the same perimeter must be less than 1. Eliminate 1), which is greater.

Then by picking numbers, for instance 2 and 3 for the sides of the rectangle one will find that the ratio must be very close to 1 (0,9....).

The only choice so close to 1 is 24/25.

I took me 1,5min.

IMO is a 600 level question.
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by AleksandrM » Fri Jun 20, 2008 2:22 pm
I just saw why my answer took so long, I was too focused on using algebra.

When I used a faster method, I solved this problem in a little over a minute.

since the ratio of the sides of the rectangle are in the form of 2:3, then you can just assume the following:

4s = 2(3) + 2(6)

4s = 10

s = 10/4

Then the ratio of areas will be:

(2)(3)/(10/4)^2 =

6/100/16, which can be reduced to 6/25/4 = 24/25.

This is certainly an easy problem. I need to prevent myself from getting carried away with algebra.
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by gmataug08 » Fri Jun 20, 2008 8:17 pm
perimeter of square = perimeter of rectangle
and rectangle sides are in ratio 2:3

by applying numbers to it,
if we take rectangle sides 2 & 3

rectangle perimeter = 10 that would make each square sides 2.5

so used a more convienient numbers 4 & 6 for rectangle sides
that makes each side of aquare as 5 , that gives the area ratio between them as 24:25
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