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1. The ratio of the lengths of sides of the rectangle is 3:4:- This gives area of rectangle equal to 48 hence sufficient
2. The difference between the lengths of sides of the rectangle is smaller than 3: If a,b are the two sides then a:b=3:4 or a=3k,b=4k where k is any constant.
Difference =4k-3k=k<3 hence k=0,1,2 k cannot be equal to 0,1 as if k=1 then a=3,b=4 which is not possible as diameter of circle=10. if k=2 then a=6,b=8 and area=48 hence sufficient.
so IMO D
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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. Two different answers not sufficient. Answer: A. 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....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
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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 2503 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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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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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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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#p1078258Hope 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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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 Hope this answers your question. Cheers!! Happy GMATing!!!
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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.
My answer is A. 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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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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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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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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