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Completely lost

by reply2spg » Wed Sep 29, 2010 7:15 pm
Source - Grokit test

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[/img]

The ratio of the area of a square mirror to its frame is 16 to 33. If the frame has a uniform width (c) around the mirror, which of the following could be the value, in inches, of c?

I. 2
II. 3 1/2
III. 5

A I only
B III only
C I and II only
D I and III only
E I, II and III
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by limestone » Wed Sep 29, 2010 7:56 pm
Let's call the smaller square: S, the larger: L
x is the side of S, and y is the side of L

Area of S/ Area of L = 16/33
Then x^2/y^2 = 16/33
x/y = sqrt(16/33) = 4/sqrt(33)

As the frame has the uniform width around the mirror, then both the distance from the mirror to left side of the frame or to the right side of the frame are c.
then x+2c = y
Then c = (y-x)/2

However, we cannot find an exactly pair of x,y because x/y = 4/sqrt33 is insuff to do so.
Thus:
x can be 4*a, y can be a*sqrt33, c can be a*(sqrt33-4)/2
Where a is any positive number.

We see that c is dependent on a.
As a is not fixed, then c is not fixed too. c can be any positive number.
Then Pick E as I,II,III are all positive.
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by reply2spg » Wed Sep 29, 2010 7:59 pm
I am lost here, please explain :(
limestone wrote: We see that c is dependent on a.
As a is not fixed, then c is not fixed too. c can be any positive number.
Then Pick E as I,II,III are all positive.
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by Rahul@gurome » Wed Sep 29, 2010 8:04 pm
Let the length of the mirror be a
So area of mirror is a^2.
Length of frame is a+2c.
area of mirror : area of frame is 16 : 33.
Or area of mirror : area of (frame + mirror) is 16 : 16+33 which is 16 : 49.
Or a^2 : (a+2c)^2 is 16:49.
Or a : a+2c is 4 : 7.
Or 4*(a+2c) = 7a.
Or 4a+8c = 7a.
Or 8c = 3a.
Or c = 3a/8.
So depending on "a", "c" can take any value.
Hence all I, II, and III can be true.

The correct answer is hence (E).
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by reply2spg » Wed Sep 29, 2010 8:09 pm
Thanks. I guess I need to get my basics clear.

Stupid questions, but I am not getting answers for these.

Aren't we suppose to get 'c' as an integer?
How should we come to know that 'c' can be anything?
If I have any other value as 4th value, then will that also be the value of 'c'?

Please explain
Rahul@gurome wrote:Let the length of the mirror be a
So area of mirror is a^2.
Length of frame is a+2c.
area of mirror : area of frame is 16 : 33.
Or area of mirror : area of (frame + mirror) is 16 : 16+33 which is 16 : 49.
Or a^2 : (a+2c)^2 is 16:49.
Or a : a+2c is 4 : 7.
Or 4*(a+2c) = 7a.
Or 4a+8c = 7a.
Or 8c = 3a.
Or c = 3a/8.
So depending on "a", "c" can take any value.
Hence all I, II, and III can be true.

The correct answer is hence (E).
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by limestone » Wed Sep 29, 2010 8:17 pm
Sure,

I'll start from here
x can be 4*a, y can be a*sqrt33, c can be a*(sqrt33-4)/2
If I plug in an "a", there will be fixed value for c,x,y:
- plug in "a" to find c via this c= a*(sqrt33-4)/2
- plug in c and a to find x,y via this: x=4*a, y=a*sqrt33

Then if we can find an "a" that makes c = 2 (as in I) or 3 1/2 (in II) or 5 (in III), then those c is acceptable
Can we?

Yes, we just reverse the equation c= a*(sqrt33-4)/2 to
a = c*2/(sqrt33-4)
hence for each c in I,II,III we can find an "a" that is positive. So all the three cases can happen.

a must be positive because x=4*a >0, y = a*sqrt33>0 (x,y are the sides of the two squares)

To make it clearer:
A,B and C >0
A/B = 3/5
C= B- 5
What are the possible value of C?
as A can be any positive number, B can be any positive no. too ( as long as A/B=3/5)
And as B is unfixed, C is unfixed too. C can be anything. ( actually, the lower limit for A is 0, then that of B is zero too. C must be larger than -5)
Last edited by limestone on Wed Sep 29, 2010 8:23 pm, edited 1 time in total.
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by Rahul@gurome » Wed Sep 29, 2010 8:22 pm
Aren't we suppose to get 'c' as an integer?
How should we come to know that 'c' can be anything?
If I have any other value as 4th value, then will that also be the value of 'c'?
No it is not necessary that "c" has to be an integer.
"c" can be anything because if a is 1, c is 3/8.
If a is1.5 c is (3/8)*1.5 = 9/16.
If a is 1 1/3 or 4/3, c can be (3/8)*(4/3) = 1/2.
Yes any fourth positive value can also be c.
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by reply2spg » Wed Sep 29, 2010 8:23 pm
Thanks for your quick reply Rahul
Rahul@gurome wrote:
Aren't we suppose to get 'c' as an integer?
How should we come to know that 'c' can be anything?
If I have any other value as 4th value, then will that also be the value of 'c'?
No it is not necessary that "c" has to be an integer.
"c" can be anything because if a is 1, c is 3/8.
If a is1.5 c is (3/8)*1.5 = 9/16.
If a is 1 1/3 or 4/3, c can be (3/8)*(4/3) = 1/2.
Yes any fourth positive value can also be c.
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by reply2spg » Wed Sep 29, 2010 8:24 pm
Thanks Limestone, I think I need to see this again in the morning with fresh mind.

Thanks a lot.
limestone wrote:Sure,

I'll start from here
x can be 4*a, y can be a*sqrt33, c can be a*(sqrt33-4)/2
If I plug in an "a", there will be fixed value for c,x,y:
- plug in "a" to find c via this c= a*(sqrt33-4)/2
- plug in c and a to find x,y via this: x=4*a, y=a*sqrt33

Then if we can find an "a" that makes c = 2 (as in I) or 3 1/2 (in II) or 5 (in III), then those c is acceptable
Can we?

Yes, we just reverse the equation c= a*(sqrt33-4)/2 to
a = c*2/(sqrt33-4)
hence for each c in I,II,III we can find an "a" that is positive. So all the three cases can happen.

a must be positive because x=4*a >0, y = a*sqrt33>0 (x,y are the sides of the two squares)

To make it clearer:
A,B and C >0
A/B = 3/5
C= B- 5
What are the possible value of C?
as A can be any positive number, B can be any positive no. too ( as long as A/B=3/5)
And as B is unfixed, C is unfixed too. C can be anything. ( actually, the lower limit for A is 0, then that of B is zero too. C must be larger than -5)
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by limestone » Wed Sep 29, 2010 8:32 pm
Wow, Rahul@gurome, your approach is shorter and esier.

I didn't realize that 16+33 = 49 = 7^2, and I forgot to add the minor square area when finding the manor one's area.

Next time, I'll be more careful with this kind of problem.

Thanks.

@reply2spg : I made a mistake in my approaching, hence you should follow Rahul's one.
I just post my simple example to make what Rahul said clearer.

A/B = 3/5
C= B- 5
What are the possible value of C?
as A can be any number (integer, fraction, real ...) B can be any number too ( as long as A/B=3/5)
And as B is unfixed, C is unfixed too. C can be anything. ( as long as C is 5 units smaller than B)
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by GMATGuruNY » Thu Sep 30, 2010 3:15 am
reply2spg wrote:Source - Grokit test

Image
[/img]

The ratio of the area of a square mirror to its frame is 16 to 33. If the frame has a uniform width (c) around the mirror, which of the following could be the value, in inches, of c?

I. 2
II. 3 1/2
III. 5

A I only
B III only
C I and II only
D I and III only
E I, II and III
Since the problem is not restricted to integers, the width could be any positive value.

The correct answer is E.

No math is needed to solve this problem. Just use common sense.
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