The perimeters of square region S and rectangular region R

[pairadox_20]

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

[RonPurewal]

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.

[vivinkarkera]

Hey,
i was wondering cant we solve this question just by looking at the answer choices

since its given that the sides of R are in ratio of 2:3
and we have to find the ratio of R:S

if we look at answer choices B is the only answer that can satisfy the ratio of R as 2:3

or am i terribly wrong somewhere

can someone please help me with this

thanks..

[RonPurewal]

Ah, I see"”you're pointing out that, if the sides of the rectangles are integers, then the area of R must be (multiple of 2)(multiple of 3) = multiple of 6.

This is actually a brilliant observation, but, I'm sure, relatively few people would be able to make it under the pressure of this exam. If you could, then that's awesome, of course"”but creative thinking is much easier in one's living room than in the testing room!

This reasoning is contingent on the notion that the sides can actually be integers. That's not too hard to see in this problem, I suppose, since the equal things are just perimeters (= sums of sides).
If we were dealing with equal areas, on the other hand, then there would be no guarantee that an all-integer solution would be available.

[vivinkarkera]

This is actually a brilliant observation, but, I'm sure, relatively few people would be able to make it under the pressure of this exam. .

thanks a lot, that totally made my day :)
guess after all i am stacey koprince's student

[RonPurewal]

Stacey has lots of brilliant insights, too. Makes sense all around.

Again, a very nice insight into the problem there.