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Square countertop

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Square countertop

by acorra » Sat Oct 25, 2008 7:26 am
A square countertop has a square tile inlay in the center, leaving an untiled strip of uniform width around the tile. If the ratio of the tiled area to the untiled area is 25 to 39, which of the following could be the width, in inches, of the strip?

I. 1
II. 3
III. 4

A) I only

B) II only

C) I and II only

D) I and III only

E) I, II and III

Could you please also tell me what level is this question? (700+; 650 etc)

thanks
Andrew
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by 2009wish » Sat Oct 25, 2008 10:58 pm
IMO its 3 . whats the OA?

ratio of the square tile inside to the total area of the larger tile will be -

25x / ( 25+39)x = 25x / 64x

x prefarably should be a perfect square number

1. let x be 4 ------- then the sides will be 10 and 16 ---- width will be 3

2.let x be 9 -----------sides will be 15 and 24 ---- width will be 3 again

METHOD 2 ----

sqrt 64x = 2( widht of the strip ) + sqrt 25x
solving this we get --

9x =4 (width ) ^2
wher e width is 3
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by stop@800 » Sun Oct 26, 2008 1:30 pm
I am lost :)

I think all are possible.
There is no point which says width of squares (countertop and tile) has to be integer.
If these are non integers I can get any legth.


ratio as 2009wish found
25x : 64x

x = 4
width = 3

x = 4/9
width = 1
here ratio will be

100/9 : 256/9
so sides will be
10/3 and 16/3

so width = 1


similarly x=8/3 will give width as 4.

Whats the OA?
I am with E

btw, this is one of the most difficult Q I have seen.
I almost gave up :)
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by acorra » Sun Oct 26, 2008 2:45 pm
OA is E. Is one of the hardest I ever found.... OG Verbal, 175

rgds
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by stop@800 » Mon Oct 27, 2008 4:41 am
acorra wrote:OA is E. Is one of the hardest I ever found.... OG Verbal, 175
OG quant :)

rgds
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Re: Square countertop

by Ian Stewart » Mon Oct 27, 2008 10:55 am
acorra wrote:A square countertop has a square tile inlay in the center, leaving an untiled strip of uniform width around the tile. If the ratio of the tiled area to the untiled area is 25 to 39, which of the following could be the width, in inches, of the strip?

I. 1
II. 3
III. 4

A) I only

B) II only

C) I and II only

D) I and III only

E) I, II and III

Could you please also tell me what level is this question? (700+; 650 etc)

thanks
Andrew
I suppose a reasonable question, when reading the above, would be "why would any of the values be impossible?" Really, the width of the strip is just telling you how big the diagram is; if the question can be answered at all, then the width could have any positive value -- just scale the diagram appropriately. No calculations are actually required.

To illustrate:

We have a smaller square inside a larger square. If the ratio of the area of the smaller square to the area of the strip is 25 to 39, that means the area of the smaller square is 25/64 of the area of the large square. So, the smaller square could be 5x5, and the larger square could be 8x8, leaving a strip of width (8-5)/2 = 3/2 = 1.5 inches. What happens if we double all of the dimensions? We get a 10x10 square inside a 16x16 square; the ratio is then 10*10/16*16 = (2*5)(2*5)/(2*8)(2*8) = 25/64 (the 2's cancel), and the width of the strip has doubled. Similarly, you can make the strip any width you like by scaling the dimensions appropriately.
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by gmat009 » Tue Oct 28, 2008 6:03 am
Can someone plz. help me in this question. I am not getting width=3 when x=4.
The way I am doing this is:-

Let x be the side of smaller square . If w is width around tile, then side of larger square is x+2w

Therefore
(x^2)/((x+2w)^2) = 25/64
(x)/(x+2w)=5/8
w=3x/10

If x=1,3,4 then I am getting different width of 3/10, 9/10, 6/5.

Can someone plz. help me where I am going wrong...
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Re: Square countertop

by logitech » Tue Oct 28, 2008 8:20 am
acorra wrote:A square countertop has a square tile inlay in the center, leaving an untiled strip of uniform width around the tile. If the ratio of the tiled area to the untiled area is 25 to 39, which of the following could be the width, in inches, of the strip?

I. 1
II. 3
III. 4

A) I only

B) II only

C) I and II only

D) I and III only

E) I, II and III

Could you please also tell me what level is this question? (700+; 650 etc)

thanks
Andrew
Well the question is actually an EASY tough question, like a skinny fat people. ( People who have excess body fat but not really HEAVY ) :lol:

I attached a picture showing what the problem is talking about

And we can see that X/Y = bla bla and Width is bla bla but whatever the width ration we can multiply it by a NUMBER and make it: 1, 4, 128, 212718 , and so on..

And so EVERYTHING could be the answer. And remember when we are talking about RATIOS ..we have a freedom to define any number we want as long as it is proportional.

So I would say E as in EVEREST 8)
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by logitech » Tue Oct 28, 2008 8:22 am
gmat009 wrote:Can someone plz. help me in this question. I am not getting width=3 when x=4.
The way I am doing this is:-

Let x be the side of smaller square . If w is width around tile, then side of larger square is x+2w

Therefore
(x^2)/((x+2w)^2) = 25/64
(x)/(x+2w)=5/8
w=3x/10

If x=1,3,4 then I am getting different width of 3/10, 9/10, 6/5.

Can someone plz. help me where I am going wrong...
X does NOT need to be an integer! :!:
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by gmat009 » Tue Oct 28, 2008 10:31 am
logitech wrote:
X does NOT need to be an integer! :!:
I understand that X doesnot need to be an integer but my question is how is everyone getting width=3 when x=4.
I am getting width=6/5 when x=4
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by logitech » Tue Oct 28, 2008 11:10 am
gmat009 wrote:
logitech wrote:
X does NOT need to be an integer! :!:
I understand that X doesnot need to be an integer but my question is how is everyone getting width=3 when x=4.
I am getting width=6/5 when x=4

Because your X and their X are NOT the same :)

Your X refers to the length of the one side of inner SQUARE

Their X is the variable in 25x/64x

Got it ?
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by pbanavara » Tue Oct 28, 2008 1:23 pm
Tough one - I assumed integers and came up with 3 .. Duh Goes to say how wrong it is to assume things on GMAT

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by santosh_surathkal » Mon Feb 22, 2010 9:07 am
gmat009 wrote:Can someone plz. help me in this question. I am not getting width=3 when x=4.
The way I am doing this is:-

Let x be the side of smaller square . If w is width around tile, then side of larger square is x+2w

Therefore
(x^2)/((x+2w)^2) = 25/64
(x)/(x+2w)=5/8
w=3x/10

If x=1,3,4 then I am getting different width of 3/10, 9/10, 6/5.

Can someone plz. help me where I am going wrong...



This can be only multiplier of 3 ... because whatever you do with 5 and 8 the width is always the multiplier of 1.5.
@Ian Stewart could you check your calculation once again????
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