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If AE = EC, what is the length of AB?
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Updated on: 24 Sep 2018, 19:45
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If AE = EC, what is the length of AB? (1) AC =10 (2) x = 30 Attachment:
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Originally posted by prernamalhotra on 10 May 2014, 07:46.
Last edited by Bunuel on 24 Sep 2018, 19:45, edited 4 times in total.
Renamed the topic, edited the question, added the OA and moved to DS forum.



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If AE = EC, what is the length of AB?
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10 May 2014, 08:29
If AE = EC, what is the length of AB?Since AE = EC, then \(\angle{EAC}=\angle{ECA}=x\). (1) AC =10. We can squeeze or stretch side AB vertically not violating any of the conditions given, thus AB can be of any length. Not sufficient. (2) x = 30. We can find all the angles in the figure, though we don't know any lengths. Not sufficient. (1)+(2) We have that \(\angle{EAC}=\angle{ECA}=30\) and AC =10. So, the triangle AEC is fixed: it can have only one shape. This means that we can find its height from vertex E, which would be equal to AB. Here we don't really care about the actual length of AB, the important part is that it can have only one length. Sufficient. Answer: C. Figure for reference: P.S. Please read carefully and follow: http://gmatclub.com/forum/rulesforpos ... 33935.html Pay attention to rules 2, 3, and 7. Thank you. Attachment:
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Re: If AE = EC, what is the length of AB?
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11 May 2014, 20:55
Hi Bunuel,
Thank you for the explanation. Have another query, if this was a PS question, how would we solve this for the actual length.
Thank you, Prerna



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Re: If AE = EC, what is the length of AB?
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11 May 2014, 23:39
prernamalhotra wrote: Hi Bunuel,
Thank you for the explanation. Have another query, if this was a PS question, how would we solve this for the actual length.
Thank you, Prerna Good questions: Attachment:
Untitled.png [ 69.18 KiB  Viewed 2972 times ]
So, we need to find the length of EF (which is equal to AB). Notice that triangle AFE is a 306090 triangle, where AF = 5. MUST KNOW FOR THE GMAT: • A right triangle where the angles are 30°, 60°, and 90°.This is one of the 'standard' triangles you should be able recognize on sight. A fact you should commit to memory is: The sides are always in the ratio \(1 : \sqrt{3}: 2\). Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°). • A right triangle where the angles are 45°, 45°, and 90°. This is one of the 'standard' triangles you should be able recognize on sight. A fact you should also commit to memory is: The sides are always in the ratio \(1 : 1 : \sqrt{2}\). With the \(\sqrt{2}\) being the hypotenuse (longest side). This can be derived from Pythagoras' Theorem. Because the base angles are the same (both 45°) the two legs are equal and so the triangle is also isosceles. For more check Triangles chapter of our Math Book: mathtriangles87197.htmlAccording to the above \(EF:AF = 1 : \sqrt{3}\) > \(EF:5 = 1 : \sqrt{3}\) > \(EF= 5 : \sqrt{3}\). Hope it's clear.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: If AE = EC, what is the length of AB?
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12 May 2014, 01:18
Got it! Thank you!
Regards, Prerna



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Re: If AE = EC, what is the length of AB?
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24 Sep 2018, 19:21
Bunuel wrote: prernamalhotra wrote: Hi Bunuel,
Thank you for the explanation. Have another query, if this was a PS question, how would we solve this for the actual length.
Thank you, Prerna Good questions: So, we need to find the length of EF (which is equal to AB). Notice that triangle AFE is a 306090 triangle, where AF = 5. MUST KNOW FOR THE GMAT: • A right triangle where the angles are 30°, 60°, and 90°.This is one of the 'standard' triangles you should be able recognize on sight. A fact you should commit to memory is: The sides are always in the ratio \(1 : \sqrt{3}: 2\). Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°). • A right triangle where the angles are 45°, 45°, and 90°. This is one of the 'standard' triangles you should be able recognize on sight. A fact you should also commit to memory is: The sides are always in the ratio \(1 : 1 : \sqrt{2}\). With the \(\sqrt{2}\) being the hypotenuse (longest side). This can be derived from Pythagoras' Theorem. Because the base angles are the same (both 45°) the two legs are equal and so the triangle is also isosceles. According to the above \(EF:AF = 1 : \sqrt{3}\) > \(EF:5 = 1 : \sqrt{3}\) > \(EF= 5 : \sqrt{3}\). Hope it's clear. Hi Bunuel, I hope you're well and apologies for replying to your comment a mere 4 years later (I'm just beginning my GMAT journey). I had a quick question with regards to the PS question asked.. The triangle of the matter, as mentioned, is a 30:60:90 triangle which follows the ratios 1:√3:2. So my question is, having that we already know that AF=5, why wouldn't EF=5 as well? I say this because side EF is aligned with the ratio 1, and we already know AF=5.. so wouldn't EF be equal to 1*5? I understand that there's a flaw in my reasoning, but I cannot wrap my head around it. Thanks in advance!



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Re: If AE = EC, what is the length of AB?
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24 Sep 2018, 19:54
Euphor1a wrote: Bunuel wrote: prernamalhotra wrote: Hi Bunuel,
Thank you for the explanation. Have another query, if this was a PS question, how would we solve this for the actual length.
Thank you, Prerna Good questions: So, we need to find the length of EF (which is equal to AB). Notice that triangle AFE is a 306090 triangle, where AF = 5. MUST KNOW FOR THE GMAT: • A right triangle where the angles are 30°, 60°, and 90°.This is one of the 'standard' triangles you should be able recognize on sight. A fact you should commit to memory is: The sides are always in the ratio \(1 : \sqrt{3}: 2\). Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°). • A right triangle where the angles are 45°, 45°, and 90°. This is one of the 'standard' triangles you should be able recognize on sight. A fact you should also commit to memory is: The sides are always in the ratio \(1 : 1 : \sqrt{2}\). With the \(\sqrt{2}\) being the hypotenuse (longest side). This can be derived from Pythagoras' Theorem. Because the base angles are the same (both 45°) the two legs are equal and so the triangle is also isosceles. According to the above \(EF:AF = 1 : \sqrt{3}\) > \(EF:5 = 1 : \sqrt{3}\) > \(EF= 5 : \sqrt{3}\). Hope it's clear. Hi Bunuel, I hope you're well and apologies for replying to your comment a mere 4 years later (I'm just beginning my GMAT journey). I had a quick question with regards to the PS question asked.. The triangle of the matter, as mentioned, is a 30:60:90 triangle which follows the ratios 1:√3:2. So my question is, having that we already know that AF=5, why wouldn't EF=5 as well? I say this because side EF is aligned with the ratio 1, and we already know AF=5.. so wouldn't EF be equal to 1*5? I understand that there's a flaw in my reasoning, but I cannot wrap my head around it. Thanks in advance! How can both AF and EF be 5 if the ratio of their length is 1:√3? Is 5:5 = 1:√3 ? In AEF: \(EF:AF:AE=1:√3:2\); \(EF:AF = 1 : \sqrt{3}\); \(EF:5 = 1 : \sqrt{3}\); \(EF= 5 : \sqrt{3}\).
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New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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