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What is the area of rectangular region R? [#permalink]
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17 Jan 2014, 01:31
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Re: What is the area of rectangular region R? [#permalink]
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17 Jan 2014, 01:32
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Re: What is the area of rectangular region R? [#permalink]
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17 Jan 2014, 10:50
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Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionWhat is the area of rectangular region R? (1) Each diagonal of R has length 5. (2) The perimeter of R is 14. Let the length and breadth of the rectangle by L and B Statement 1) Diagonal is 5. So \(L^2\) + \(B^2\) = 25. We can get multiple values of L and B and subsequently the area will also vary. Statement 2) The perimeter = 14 So 2(L+B) = 14 or L+B = 7 Not Sufficient as various values of L and B are possible. Combining the two statements: L + B = 7 and \(L^2\) + \(B^2\) = 25. The first pythagorus triplet (L = 4, B = 3)satisfies the two equation and hence Sufficient. Option C)
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Re: What is the area of rectangular region R? [#permalink]
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Re: What is the area of rectangular region R? [#permalink]
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20 Mar 2014, 03:51
Bunuel,
isn't this a special Pythagorean triangle? with a,b,c 3,4,5? Hence it would be A not C...



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Re: What is the area of rectangular region R? [#permalink]
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20 Mar 2014, 04:30
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Re: What is the area of rectangular region R? [#permalink]
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20 Mar 2015, 10:46
Bunuel wrote: SOLUTION (1) Each diagonal of R has length 5 > as the diagonals in a rectangle are the hypotenuses for the sides then: \(x^2+y^2=5^2\), but we can not get the value of \(xy\) from this info. Not sufficient.
Is there a quick rule of thumb to know that we can't get the value of \(xy\) from \(x^2+y^2=5^2\)? (I find myself trying to think of multiple combinations of x and y that would fulfill the equation, but that tends to take quite a bit of time.)



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Re: What is the area of rectangular region R? [#permalink]
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20 Mar 2015, 14:31
Hi swaggerer, The rule that you're describing is actually an Algebra rule. If you have just one equation (with no other restrictions) and two variables, then there will be more than one set of values that will "fit" the variables. Here's a simple example: X + Y = 10 What COULD (X)(Y) equal? Come up with a few examples. Note: there are no other restrictions here, so X and Y could be nonintegers, 0, negatives, square roots, etc. You'll come to find that there are LOTS of possible values for (X)(Y). That same concept exists in this prompt. GMAT assassins aren't born, they're made, Rich
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Re: What is the area of rectangular region R? [#permalink]
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18 Jun 2015, 05:28
Can the triangle made by the diagonal be considered as a 306090 degree triangle? If yes, then the length and width can be found out by using the 1 : root(3) : 2 ratios for the 3 sides of the triangle. And hence, answer choice A would be sufficient.



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Re: What is the area of rectangular region R? [#permalink]
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18 Jun 2015, 13:56
Hi saar, Unfortunately, we do NOT know what the angles will be when we draw a diagonal line through that rectangle. While it's possible that the resulting triangles could 30/60/90s, it's also possible (and highly likely) that they will be some OTHER type of triangle. As such, neither of the two Facts provides enough information to answer the question on its own. GMAT assassins aren't born, they're made, Rich
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What is the area of rectangular region R? [#permalink]
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25 Nov 2015, 04:29
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Bunuel wrote: Bunnuel  Why isn't it B then ? I marked it as D. However, after you explained, I understood why A isn't valid. But why isn't it B? 2L + 2B = 14 L + B = 7 The only combination that suits this is 4 and 3, and hence we can find the area? Also, given the logic you explained for Statement A, the reverse logic has been used in some of the problems in Manhattan GMAT and GMAT Prep. How can we solve this ? Thanks.



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Re: What is the area of rectangular region R? [#permalink]
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25 Nov 2015, 09:57
Hi torreadortorment, Using the information in Fact 2, you've correctly deduced that... L+B = 7 The question asks for the AREA of the rectangle though... What's the area when.... L = 1, B = 6 L = 2, B = 7 L = 2.5, B = 4.5 L = 3, B = 4 Etc. If the area changes, then the Fact is INSUFFICIENT. Once you COMBINE the two Facts though, then you know that you're dealing with a 3/4/5. GMAT assassins aren't born, they're made, Rich
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Re: What is the area of rectangular region R? [#permalink]
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26 Nov 2015, 07:09
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. What is the area of rectangular region R? (1) Each diagonal of R has length 5. (2) The perimeter of R is 14. There are 2 variables (width, length) and 2 equations are given by the conditions, so there is high chance (C) will be the answer, Looking at the conditions together, l^2+w^2=5^2, 2(l+w)=14 (l,w)=(3,4)=(4,3) area=3*4=12. This is sufficient and the answer becomes (C). For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: What is the area of rectangular region R? [#permalink]
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04 Aug 2016, 11:21
I have one question here. let's say if the hypotenuse of any right angle triangle is 5 then can I safely assume the lengths of other two sides are 3 and 4? If not, why ? could some explain in detail examples.



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Re: What is the area of rectangular region R? [#permalink]
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04 Aug 2016, 13:05
Hi santro789, The simple answer to your question is NO. The reason why is because there are two variables that you have to account for: (A^2) + (B^2) = 25 There are LOTS of different combinations for A and B that would complete this equation. eg. A = 3, B = 4 A = 1, B = (root24) A = 2, B = (root21) A = (root3), B = (root22) Etc. Now, IF you have specific types of additional information, then that could limit the possibilities. For example, IF we're told that A and B are both INTEGERS, then the triangle would have to be a 3/4/5 right triangle (although we wouldn't know which variables was the 3 and which was the 4). GMAT assassins aren't born, they're made, Rich
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