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M11-21

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M11-21  [#permalink]

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New post 15 Sep 2014, 23:44
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

72% (00:51) correct 28% (01:07) wrong based on 109 sessions

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Re M11-21  [#permalink]

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New post 15 Sep 2014, 23:44
Official Solution:


Statement (1) by itself is insufficient. We can decrease the perimeter by shrinking the triangle (i.e. by choosing \(AB\) close to 7 and \(BC\) close to 0).

Statement (2) by itself is insufficient. There are no limits on the size of the triangle.

Statements (1) and (2) combined are sufficient. S1 and S2 combined provide a system of equations from which it follows that \(AB\) is 3, \(BC\) is 4, the hypotenuse is 5, and the perimeter is 12.


Answer: C
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M11-21  [#permalink]

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New post 18 Feb 2015, 18:26
Bunuel wrote:
Official Solution:


Statement (1) by itself is insufficient. We can decrease the perimeter by shrinking the triangle (i.e. by choosing \(AB\) close to 7 and \(BC\) close to 0).

Statement (2) by itself is insufficient. There are no limits on the size of the triangle.

Statements (1) and (2) combined are sufficient. S1 and S2 combined provide a system of equations from which it follows that \(AB\) is 3, \(BC\) is 4, the hypotenuse is 5, and the perimeter is 12.


Answer: C


Hello Bunuel,

Here is how approached this problem:

Statement 1: I considered that as ABC is a right angled triangle, I should look for pythagoras triplets - so AB and BC must be 3,4 to give 5 as hypotenuse.

Could you please tell me how is this approach wrong?

Thanks!
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Re: M11-21  [#permalink]

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New post 06 Jul 2016, 05:08
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Hello aimtoteach

Not all right angled triangles follow the Pythagorean Triplet theory
In the sense that AB could equal = 2 and BC = 5
still holding statement (1) AB + BC = 7 true

and the hypotenuse AC would equal (2^2 + 5^2)^.5 = 29^.5
= approximately between 5 and 6

& Perimeter would then = 2+5+5 = 15

And hence statement 1 is Not Sufficient

Thanks
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Re: M11-21  [#permalink]

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New post 06 Jul 2016, 19:50
Excellent question for today, thank you Bunuel and thank you ppnimkar for your explanation as well...I was wrong myself and chose A) with S1 being sufficient but I see the fallacy in my logic now!
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Re: M11-21  [#permalink]

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New post 31 Oct 2016, 01:13
if it is a right triangle does this not mean that 2 sides of the triangle should be equal , by that logic two sides are 3.5,3.5 and the third side can be found by pythagores theoram
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Re: M11-21  [#permalink]

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New post 31 Oct 2016, 01:21
yashrakhiani wrote:
if it is a right triangle does this not mean that 2 sides of the triangle should be equal , by that logic two sides are 3.5,3.5 and the third side can be found by pythagores theoram


No, it does not mean that at all. Why should ABC necessarily be an isosceles right triangle?
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Re: M11-21  [#permalink]

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Re: M11-21  [#permalink]

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New post 28 Mar 2017, 16:39
My 2 cents:
As in question it is NOT mentioned whether size of sides AB and BC is integer we CANNOT assume, based on AB + BC = 7, that AB = 3 or 4 & BC = 4 or 3 to form Pythagorean triplet or 3-4-5. As CORRECTLY shown above by others, AB = 6.5 and BC = 0.5.
Honestly, i made the same mistake in this problem. :(
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Re: M11-21  [#permalink]

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New post 29 Mar 2018, 01:06
Bunuel could you please explain why statement 1 is not sufficient if I use the Pythagorean triple theory?
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Re: M11-21  [#permalink]

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New post 29 Mar 2018, 01:11
sadikabid27 wrote:
Bunuel could you please explain why statement 1 is not sufficient if I use the Pythagorean triple theory?


So you are saying that the sides must be 3 and 4 and the hypotenuse comes out to the 5. But why? How do you know that the sides of this particular triangle must be Pythagorean triple? Why cannot they be 1 and 6, for example? Or why they cannot be 1.4 and 5.6? We are neither told that the sides are Pythagorean triples nor that the sides are integers.
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Re: M11-21  [#permalink]

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New post 29 Mar 2018, 01:28
One angle is 90 degree. so I took in triangle abc assuming it's a right triangle, ab^2+bc^2=ac^2 , ab+bc=7 so ab^2+bc^2=7^2 , then you can find ac=7, so ab+bc+ca=7+7=14 (that's perimeter), where I am going wrong? Bunuel
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Re: M11-21  [#permalink]

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New post 29 Mar 2018, 01:33
sadikabid27 wrote:
One angle is 90 degree. so I took in triangle abc assuming it's a right triangle, ab^2+bc^2=ac^2 , ab+bc=7 so ab^2+bc^2=7^2 , then you can find ac=7, so ab+bc+ca=7+7=14 (that's perimeter), where I am going wrong? Bunuel


I tried to explain it above. Here it is again. Yes, the given triangle is right angled at B and yes, AB^2 + BC^2 = AC^2. But AB + BC = 7 does not mean that AB^2 + BC^2 = 7^2. If you square AB + BC = 7 you'll get AB^2 + 2AB*BC + BC^2 = 7^2.
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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

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Re: M11-21 &nbs [#permalink] 29 Mar 2018, 01:33
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