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# m22#30

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25 Apr 2012, 08:07
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harshavmrg wrote:
Bunuel....Can u please explain the solution..( both options A and B)... i am quite confused about this problem of how is A sufficient and how is B insufficient

A rectangle is inscribed in a circle of radius 5. Is the area of the rectangle bigger than 48 ?

Look at the diagram below:
Attachment:

Rectangle.png [ 15.39 KiB | Viewed 6909 times ]

A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. Since a rectangle is made of two right triangles then a rectangle inscribed in a circle must have its diagonal as the diameter of the circle.

Given: $$radius=5$$ --> $$diameter=diagonal=10$$. Question: $$area=ab=?$$, where $$a$$ and $$b$$ are the sides of the rectangle.

(1) The ratio of the lengths of sides of the rectangle is 3:4 --> $$\frac{a}{b}=\frac{3x}{4x}$$, for some positive multiple $$x$$. So, $$diagonal^2=a^2+b^2=(3x)^2+(4x)^2$$ --> $$100=9x^2+16x^2$$ --> $$x=2$$ --> $$a=3x=6$$ and $$b=4x=8$$ --> $$area=ab=6*8=48$$. Sufficient.

Alternately, even not calculating, one can spot that since hypotenuse (diameter) is 10 and the legs are in the ratio of 3 to 4 then we have 6-8-10 right triangle (Pythagorean Triple).

(2) The difference between the lengths of sides of the rectangle is smaller than 3 --> $$b-a<3$$ --> square both sides: $$b^2-2ab+a^2<9$$. Now, since $$diagonal=10^2=a^2+b^2$$ then $$100-2ab<9$$ --> $$ab>45.5$$. So we have that $$area=ab>45.5$$, hence the are may or may not be more than 48. Not sufficient.

Hope it's clear.
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04 Jul 2012, 00:52
My attempt at a more elegant solution

So, we can rephrase the question as
75 (Approx 25pi) > xy> 48?

1) SUFFICIENT
Let's try multiples of the ratio of 3:4
(3,4) -- FALSE
(6,8) -- FALSE
(9,12) --FALSE
We can see that at any multiple the value the statement doesn't hold true.

2) INSUFFICIENT
Given: x-y<3, we can find two values
1) where x-y<3 and xy is greater than 75 e.g. 9,10
2) where x-y<3 and xy is in between 75 (Approx 25pi) > xy> 48 , e.g. 8,8
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19 Mar 2013, 12:09
Quote:
S2: b - a < 3
Squaring both the sides of the inequality, gives:
a^2 + b^2 - 2ab < 9 => 100 - 2ab < 9
Solving the inequality gives: ab > 45.5
Therefore, ab > 48 may or may not be true.

I need help- why does the inequality sign flip? Why don't we add 2ab and subtract 9 to both sides of 100 - 2ab + 2an -9 < 9 + 2ab -9 = 91 < 2AB, then divide by positive 2 = 45.5 < AB?
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19 Mar 2013, 13:36
shanek wrote:
Quote:
S2: b - a < 3
Squaring both the sides of the inequality, gives:
a^2 + b^2 - 2ab < 9 => 100 - 2ab < 9
Solving the inequality gives: ab > 45.5
Therefore, ab > 48 may or may not be true.

I need help- why does the inequality sign flip? Why don't we add 2ab and subtract 9 to both sides of 100 - 2ab + 2an -9 < 9 + 2ab -9 = 91 < 2AB, then divide by positive 2 = 45.5 < AB?

Aren't you getting the same result?

Anyway, check here for a solution: m22-73309-20.html#p1078258

Hope it helps.
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20 Mar 2013, 03:57
Bunuel wrote:
shanek wrote:
Quote:
S2: b - a < 3
Squaring both the sides of the inequality, gives:
a^2 + b^2 - 2ab < 9 => 100 - 2ab < 9
Solving the inequality gives: ab > 45.5
Therefore, ab > 48 may or may not be true.

I need help- why does the inequality sign flip? Why don't we add 2ab and subtract 9 to both sides of 100 - 2ab + 2an -9 < 9 + 2ab -9 = 91 < 2AB, then divide by positive 2 = 45.5 < AB?

Aren't you getting the same result?

Anyway, check here for a solution: m22-73309-20.html#p1078258

Hope it helps.

Wow. I guess your tests really are hard.. after 2 of them I went full retard. I was staring at this for an hour. After a good night's sleep, it's REALLY obvious, and I'm embarrassed.

Thank you Bunuel.. awesome website.
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20 Mar 2013, 04:37
In regards to question about flipping the sign on the inequality, the below is how I look at it:

Taking from inequality: 100 - 2ab < 9
Subtract 100 from both sides of inequality:
100 - 2ab < 9 - 100
=> -2ab < -91

Divide 2 from both sides of inequality:
(-2ab) / 2 < (-91) / 2
=> -ab < -45.5

Now, this is where the flipping of the inequality sign comes into play. The rule is that when you change the arithmetic sign from negative to positive in an inequality, you should flip the sign of the inequality.

=> ab > 45.5

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21 Mar 2013, 21:55
vshrivastava wrote:
Suppose sides of the ractangle are a and b.

S2: b - a < 3
Squaring both the sides of the inequality, gives:
a^2 + b^2 - 2ab < 9 => 100 - 2ab < 9
Solving the inequality gives: ab > 45.5
Therefore, ab > 48 may or may not be true.

Result: S2 is NOT sufficient to answer the question.

Vshrivastava,

Even though I got this question correct, but I do like the way you prove the second statement insufficient. Thanks so much.
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31 Mar 2013, 17:50
I often find myself doing unecessary calculations in DS. In my opinion working on this is the best way to get more comfortable with the time. On this question, for statement 1, wouldn't it suffice to notice that there is only one possible rectangle with sides of ratio 2:3 that can fit inside a circle of radius 5? What I mean is everyone in this thread calculated that the area of this rectangle is 48 but isn't this unecessary? We basicaly don't really care if it's 48, more than 48 or less than 48.. Am I making a mistake somewhere? thank you
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31 Mar 2013, 18:45
I've always believed that the DS problems are a lot about developing intuition on the sufficiency test for the given statements. One who can develop such intuition can sure gain on time; but one might give away on accuracy. For example, I'd be careful while evaluating sufficiency in problems involving inequalities and would perform necessary calculations before making up my mind on sufficiency. Hope this helps!!
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06 May 2013, 04:52
Hi Experts,

The Official Explanation for Statement B
says that -
The difference between the lengths of sides of the rectangle is smaller than 3. Given that b−a<3. Square both sides: b2−2ab+a2<9. Now, since $$diagonal=102=a2+b2$$ then 100−2ab<9, so$$ab>45.5$$. So we have that area=ab>45.5, hence the area may or may not be more than 48. Not sufficient

My doubt is -Can we Square the given Inequality(b−a<3), since rule says that until and unless you are sure about both terms being positive you cannot square the Inequality. How can we be sure that the LHS or RHS is positive. Please explain.

Regards,
H
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01 Jul 2013, 08:09
Bunuel wrote:
sahilkhurana06 wrote:
Answer is D (Correct me if I am wrong)

Statement 1 : Everybody knows that this is sufficient.
Statement 2 : It says that the difference between the lengths of sides of the rectangle is smaller than 3,so that means l-b < 3..Also it has a fact that the diagonal of the rectangle is 10 (since radius of circle is 5).Now,considering these values (l^2 + b^2 = 100 and l-b<3),we have only one possibility when the diagonal can be of 10 units i.e. when we have sides 8 and 6.Hence,this statement is also sufficient to answer.

Let me know what you guys think....

Statement (2) is not sufficient, see algebraic solutions on previous page or consider the following examples:

If $$a=8$$ and $$b=6$$ ($$a-b=2<3$$ and $$a^2+b^2=100$$) then $$area=ab=48$$ and the answer to the question "is $$area>48$$" is NO;

If $$a=b=\sqrt{50}$$ ($$a-b=0<3$$ and $$a^2+b^2=100$$) then $$area=ab=50$$ and the answer to the question "is $$area>48$$" is YES. Note that in this case inscribed figure is square, which is a special type of rectangle.

I think that the problem with your solution is that you assumed with no ground for it that the lengths of the sides of the rectangle are integers.

Hope it's clear.

Bunuel, I understand the solution given in your other explanation. Thanks for the help. My question is on this first explination. How do we get $$a=b=\sqrt{50}$$? Is it simply because $$a^2+b^2=100$$...I think I figured it out as I was typing this, but is that correct? I was originally trying to draw a connection to the radius and the sides of the square (25+25...) Anyway.
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01 Jul 2013, 08:30
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usmabama wrote:
Bunuel wrote:
sahilkhurana06 wrote:
Answer is D (Correct me if I am wrong)

Statement 1 : Everybody knows that this is sufficient.
Statement 2 : It says that the difference between the lengths of sides of the rectangle is smaller than 3,so that means l-b < 3..Also it has a fact that the diagonal of the rectangle is 10 (since radius of circle is 5).Now,considering these values (l^2 + b^2 = 100 and l-b<3),we have only one possibility when the diagonal can be of 10 units i.e. when we have sides 8 and 6.Hence,this statement is also sufficient to answer.

Let me know what you guys think....

Statement (2) is not sufficient, see algebraic solutions on previous page or consider the following examples:

If $$a=8$$ and $$b=6$$ ($$a-b=2<3$$ and $$a^2+b^2=100$$) then $$area=ab=48$$ and the answer to the question "is $$area>48$$" is NO;

If $$a=b=\sqrt{50}$$ ($$a-b=0<3$$ and $$a^2+b^2=100$$) then $$area=ab=50$$ and the answer to the question "is $$area>48$$" is YES. Note that in this case inscribed figure is square, which is a special type of rectangle.

I think that the problem with your solution is that you assumed with no ground for it that the lengths of the sides of the rectangle are integers.

Hope it's clear.

Bunuel, I understand the solution given in your other explanation. Thanks for the help. My question is on this first explination. How do we get $$a=b=\sqrt{50}$$? Is it simply because $$a^2+b^2=100$$...I think I figured it out as I was typing this, but is that correct? I was originally trying to draw a connection to the radius and the sides of the square (25+25...) Anyway.

Yes. $$a=8$$ and $$b=6$$ AND $$a=b=\sqrt{50}$$ are just examples to show that (2) is not sufficient. Notice that both satisfy $$a-b=2$$ and $$a^2+b^2=100$$.
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23 Jul 2013, 22:31
Another approach for Option B: Rectangle with largest area is square. So if square were to be fit in the circle, the area would be 50 ($$(5\sqrt{2})^2$$). With 8 and 6 as the sides (as 10 is the radius), the area would be 48. The difference in lengths here is less than 2. If the difference were to increase (suppose 2.5), the area would be below 48. So according to this option, it can be above or below 48.

@Bunuel - Is my approach correct?
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26 Oct 2013, 05:29
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Why would we need any of the two statements. The question itself is sufficient:

The rectangle is inscribed in the circle of radius 5; so diameter would be 10: This has to be the diagonal of the inscribed rectangle.

The sides has to be 6 & 8 and area would be equal to 48.

Correct me if i am missing something round here
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27 Oct 2013, 05:12
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amkabdul wrote:
Why would we need any of the two statements. The question itself is sufficient:

The rectangle is inscribed in the circle of radius 5; so diameter would be 10: This has to be the diagonal of the inscribed rectangle.

The sides has to be 6 & 8 and area would be equal to 48.

Correct me if i am missing something round here

You assume with no ground for it that the lengths of the sides are integers. Knowing that hypotenuse equals to 10 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 6:8:10. Or in other words: if $$a^2+b^2=10^2$$ DOES NOT mean that $$a=6$$ and $$b=8$$, certainly this is one of the possibilities but definitely not the only one. In fact $$a^2+b^2=10^2$$ has infinitely many solutions for $$a$$ and $$b$$ and only one of them is $$a=6$$ and $$b=8$$.

For example: $$a=1$$ and $$b=\sqrt{99}$$ or $$a=2$$ and $$b=\sqrt{96}$$ ...

Hope it's clear.
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27 Feb 2014, 01:04
Bunuel.. There was one question in gmat club tests, I dont remember which question it was! In which it was mentioned that rectangle is inscribed in circle.. bt OE was , its not neccessary rectangle wud cross through radius of circle or it can touch any point of circle..

Thats y i chose E for this question. because i thought rectangle inscribed in circle mean it can touch any point of circle.
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