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M27-05

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M27-05  [#permalink]

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New post 16 Sep 2014, 01:26
2
14
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

34% (01:16) correct 66% (01:29) wrong based on 187 sessions

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Re M27-05  [#permalink]

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New post 16 Sep 2014, 01:26
1
1
7
Official Solution:


This is a 700+ question.

(1) \(a-b=15\). Must know for the GMAT: the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. So, \(a+b \gt c \gt 15\), which means that \(a+b+c \gt 30\). Sufficient.

(2) The area of the triangle is 50. For a given perimeter equilateral triangle has the largest area. Now, if the perimeter were equal to 30 then it would have the largest area if it were equilateral. Let's find what this area would be: \(Area_{equilateral}=s^2*\frac{\sqrt{3}}{4}=(\frac{30}{3})^2*\frac{\sqrt{3}}{4}=25*\sqrt{3} \lt 50\). Since even an equilateral triangle with perimeter of 30 cannot produce the area of 50, then the perimeter must be more than 30. Sufficient.


Answer: D
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Re: M27-05  [#permalink]

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New post 01 Aug 2016, 05:19
I think this is a high-quality question and I agree with explanation.
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Re: M27-05  [#permalink]

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New post 03 Mar 2017, 09:34
1
Bunuel wrote:
Official Solution:


This is a 700+ question.

(1) \(a-b=15\). Must know for the GMAT: the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. So, \(a+b \gt c \gt 15\), which means that \(a+b+c \gt 30\). Sufficient.

(2) The area of the triangle is 50. For a given perimeter equilateral triangle has the largest area. Now, if the perimeter were equal to 30 then it would have the largest area if it were equilateral. Let's find what this area would be: \(Area_{equilateral}=s^2*\frac{\sqrt{3}}{4}=(\frac{30}{3})^2*\frac{\sqrt{3}}{4}=25*\sqrt{3} \lt 50\). Since even an equilateral triangle with perimeter of 30 cannot produce the area of 50, then the perimeter must be more than 30. Sufficient.




Answer: D


Could you please explain statement 2 in detail. I am unable to understand the solution for statement 2.
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Re: M27-05  [#permalink]

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New post 04 Mar 2017, 03:12
daboo343 wrote:
Bunuel wrote:
Official Solution:


This is a 700+ question.

(1) \(a-b=15\). Must know for the GMAT: the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. So, \(a+b \gt c \gt 15\), which means that \(a+b+c \gt 30\). Sufficient.

(2) The area of the triangle is 50. For a given perimeter equilateral triangle has the largest area. Now, if the perimeter were equal to 30 then it would have the largest area if it were equilateral. Let's find what this area would be: \(Area_{equilateral}=s^2*\frac{\sqrt{3}}{4}=(\frac{30}{3})^2*\frac{\sqrt{3}}{4}=25*\sqrt{3} \lt 50\). Since even an equilateral triangle with perimeter of 30 cannot produce the area of 50, then the perimeter must be more than 30. Sufficient.




Answer: D


Could you please explain statement 2 in detail. I am unable to understand the solution for statement 2.


From (2) we have that if even an equilateral triangle with perimeter of 30 cannot have the area of 50, then the perimeter must be more that 30.
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Re: M27-05  [#permalink]

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New post 14 Jun 2017, 06:16
Hi Sir,

I believe it is the same if I assume this way right?

For a give area, equilateral triangle has the smallest perimeter.

S^2 (sq.rt3/4)=50 when simplified will show clearly that a side is >12.

Hence sufficient.
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Re: M27-05  [#permalink]

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New post 16 Oct 2017, 11:04
Nice answer, in my opinion doing it in your way would be more useful for other problems!
Thanks! :thumbup:
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M27-05  [#permalink]

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New post 13 Sep 2018, 08:15
EDIT: Got it. Number 2 is sufficient that perimeter will always be >30.
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Re: M27-05  [#permalink]

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New post 13 Sep 2018, 10:52
1
Statement 1 can be easily shown as sufficient as the difference between the two sides must be less than the third side. So the third side would be at least 15 and as given in statement the difference between the two sides is 15 so their sum would obviously be greater than 15. So perimeter will be greater than 30.

Statement 2: Area of a triangle is given to be 50. As it is a fact that for a given perimeter the equilateral triangle will have maximum area. So let us assume the side of a triangle is 10 then (root3/4)*(10)2 then we will get the area as 43.25 which less than 50. hence the perimeter has to be more than 30. Sufficient.

So answer is D
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Re: M27-05  [#permalink]

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New post 13 Sep 2018, 23:34
Bunuel wrote:
Official Solution:


This is a 700+ question.

(1) \(a-b=15\). Must know for the GMAT: the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. So, \(a+b \gt c \gt 15\), which means that \(a+b+c \gt 30\). Sufficient.

(2) The area of the triangle is 50. For a given perimeter equilateral triangle has the largest area. Now, if the perimeter were equal to 30 then it would have the largest area if it were equilateral. Let's find what this area would be: \(Area_{equilateral}=s^2*\frac{\sqrt{3}}{4}=(\frac{30}{3})^2*\frac{\sqrt{3}}{4}=25*\sqrt{3} \lt 50\). Since even an equilateral triangle with perimeter of 30 cannot produce the area of 50, then the perimeter must be more than 30. Sufficient.


Answer: D


HI,

I understood Statement 1 is sufficient.

However, I have issues understanding statement 2.

How can we confirm that "For a given perimeter equilateral triangle has the largest area.".

Can it be proved by a theorem?

Thanks
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Re: M27-05  [#permalink]

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New post 13 Sep 2018, 23:50
honneeey wrote:
Bunuel wrote:
Official Solution:


This is a 700+ question.

(1) \(a-b=15\). Must know for the GMAT: the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. So, \(a+b \gt c \gt 15\), which means that \(a+b+c \gt 30\). Sufficient.

(2) The area of the triangle is 50. For a given perimeter equilateral triangle has the largest area. Now, if the perimeter were equal to 30 then it would have the largest area if it were equilateral. Let's find what this area would be: \(Area_{equilateral}=s^2*\frac{\sqrt{3}}{4}=(\frac{30}{3})^2*\frac{\sqrt{3}}{4}=25*\sqrt{3} \lt 50\). Since even an equilateral triangle with perimeter of 30 cannot produce the area of 50, then the perimeter must be more than 30. Sufficient.


Answer: D


HI,

I understood Statement 1 is sufficient.

However, I have issues understanding statement 2.

How can we confirm that "For a given perimeter equilateral triangle has the largest area.".

Can it be proved by a theorem?

Thanks


Here is a proof of this theorem: https://www.mathalino.com/reviewer/diff ... -perimeter

Hope it helps.
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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

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Re: M27-05  [#permalink]

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New post 13 Sep 2018, 23:54
Bunuel wrote:
honneeey wrote:
Bunuel wrote:
Official Solution:


This is a 700+ question.

(1) \(a-b=15\). Must know for the GMAT: the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. So, \(a+b \gt c \gt 15\), which means that \(a+b+c \gt 30\). Sufficient.

(2) The area of the triangle is 50. For a given perimeter equilateral triangle has the largest area. Now, if the perimeter were equal to 30 then it would have the largest area if it were equilateral. Let's find what this area would be: \(Area_{equilateral}=s^2*\frac{\sqrt{3}}{4}=(\frac{30}{3})^2*\frac{\sqrt{3}}{4}=25*\sqrt{3} \lt 50\). Since even an equilateral triangle with perimeter of 30 cannot produce the area of 50, then the perimeter must be more than 30. Sufficient.


Answer: D


HI,

I understood Statement 1 is sufficient.

However, I have issues understanding statement 2.

How can we confirm that "For a given perimeter equilateral triangle has the largest area.".

Can it be proved by a theorem?

Thanks


Here is a proof of this theorem: https://www.mathalino.com/reviewer/diff ... -perimeter

Hope it helps.


Useful video:


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Resources:
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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: M27-05  [#permalink]

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New post 20 Sep 2018, 11:16
Bunuel wrote:
Official Solution:


This is a 700+ question.

(1) \(a-b=15\). Must know for the GMAT: the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. So, \(a+b \gt c \gt 15\), which means that \(a+b+c \gt 30\). Sufficient.

(2) The area of the triangle is 50. For a given perimeter equilateral triangle has the largest area. Now, if the perimeter were equal to 30 then it would have the largest area if it were equilateral. Let's find what this area would be: \(Area_{equilateral}=s^2*\frac{\sqrt{3}}{4}=(\frac{30}{3})^2*\frac{\sqrt{3}}{4}=25*\sqrt{3} \lt 50\). Since even an equilateral triangle with perimeter of 30 cannot produce the area of 50, then the perimeter must be more than 30. Sufficient.


Answer: D

This is a damn tricky one for sure
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Re: M27-05 &nbs [#permalink] 20 Sep 2018, 11:16
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