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m06#34

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m06#34 [#permalink] New post 13 Jul 2010, 08:47
If a , b , and c are positive and a^2 + c^2 = 202 , what is the value of b-a-c ?

1. b^2+c^2=225
2. a^2+b^2=265

OA:C
OE: Stmt 1 & 2 are insufficient.
Both together are sufficient.
Subtract S1 from a^2 + c^2 = 202 to get a^2-b^2= -23 Now add this equation to Stmt 2. It wil give a^2=121 & thus a=11 , with the help of a, b & c can also be found.

How can a=11, as its not given that a is an integer. a can be \sqrt{121}

Please help.
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Re: m06#34 [#permalink] New post 13 Jul 2010, 16:15
sorry - i am not sure why its not clear.....

as you stated - 2a^2 = 242 => a^2 = 121
so a can be +11 or -11

since they mention positive integers, we take a=11 and hence get answer....
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Re: m06#34 [#permalink] New post 13 Jul 2010, 20:35
pranrasvij wrote:
sorry - i am not sure why its not clear.....

as you stated - 2a^2 = 242 => a^2 = 121
so a can be +11 or -11

since they mention positive integers, we take a=11 and hence get answer....


Highlighted part is not mentioned in the question. Its not mentioned that a is an integer. So a can be \sqrt{121},& this will give a^2=121
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Re: m06#34 [#permalink] New post 14 Jul 2010, 21:21
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Re: m06#34 [#permalink] New post 18 Jul 2010, 14:45
Expert's post
Hussain15 wrote:
If a , b , and c are positive and a^2 + c^2 = 202 , what is the value of b-a-c ?

1. b^2+c^2=225
2. a^2+b^2=265

OA:C
OE: Stmt 1 & 2 are insufficient.
Both together are sufficient.
Subtract S1 from a^2 + c^2 = 202 to get a^2-b^2= -23 Now add this equation to Stmt 2. It wil give a^2=121 & thus a=11 , with the help of a, b & c can also be found.

How can a=11, as its not given that a is an integer. a can be \sqrt{121}

Please help.


You are thinking too much about this problem as there is no reason for confusion: \sqrt{121}=11 as 11^2=121 (the same way as \sqrt{4}=2).

Given: a>0 and a^2=121 --> so either a=-\sqrt{121}=-11, but this solution is out as a>0 OR a=\sqrt{121}=11. So a=11.

Hope it's clear.
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Re: m06#34 [#permalink] New post 11 Sep 2012, 18:18
Isn't it possible to: (-A^2 - B^2 = -202) + (B^2 + C^2 = 225) = B^2 - A^2 = 23 which gives us (B - A)*(B+ A)=23 which implies A=11 B=12 no? So wouldn't 1 be enough to solve? Please explain, thank you.




Bunuel wrote:
Hussain15 wrote:
If a , b , and c are positive and a^2 + c^2 = 202 , what is the value of b-a-c ?

1. b^2+c^2=225
2. a^2+b^2=265

OA:C
OE: Stmt 1 & 2 are insufficient.
Both together are sufficient.
Subtract S1 from a^2 + c^2 = 202 to get a^2-b^2= -23 Now add this equation to Stmt 2. It wil give a^2=121 & thus a=11 , with the help of a, b & c can also be found.

How can a=11, as its not given that a is an integer. a can be \sqrt{121}

Please help.


You are thinking too much about this problem as there is no reason for confusion: \sqrt{121}=11 as 11^2=121 (the same way as \sqrt{4}=2).

Given: a>0 and a^2=121 --> so either a=-\sqrt{121}=-11, but this solution is out as a>0 OR a=\sqrt{121}=11. So a=11.

Hope it's clear.
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Re: m06#34 [#permalink] New post 12 Sep 2012, 01:00
Expert's post
ctiger100 wrote:
Isn't it possible to: (-A^2 - B^2 = -202) + (B^2 + C^2 = 225) = B^2 - A^2 = 23 which gives us (B - A)*(B+ A)=23 which implies A=11 B=12 no? So wouldn't 1 be enough to solve? Please explain, thank you.




Bunuel wrote:
Hussain15 wrote:
If a , b , and c are positive and a^2 + c^2 = 202 , what is the value of b-a-c ?

1. b^2+c^2=225
2. a^2+b^2=265

OA:C
OE: Stmt 1 & 2 are insufficient.
Both together are sufficient.
Subtract S1 from a^2 + c^2 = 202 to get a^2-b^2= -23 Now add this equation to Stmt 2. It wil give a^2=121 & thus a=11 , with the help of a, b & c can also be found.

How can a=11, as its not given that a is an integer. a can be \sqrt{121}

Please help.


You are thinking too much about this problem as there is no reason for confusion: \sqrt{121}=11 as 11^2=121 (the same way as \sqrt{4}=2).

Given: a>0 and a^2=121 --> so either a=-\sqrt{121}=-11, but this solution is out as a>0 OR a=\sqrt{121}=11. So a=11.

Hope it's clear.


Notice that we are not told that A and B are positive integers, so from B^2-A^2=23 it's possible for instance that A=1 and B=\sqrt{24}.

Hope it's clear.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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: m06#34 [#permalink] New post 12 Sep 2012, 10:55
a^2+ c^2 =202 -->1
b^2 + c^2 =225 -->2

2-1

b^2 - a^2 = 23
(b+a)(b-a)=23

since 23 is prime and all are positive

b+a =23
and b-a =1

solving b=12 a=11

substituing in 1 we will get c= 9


for second statement following the same approach there will be multiple possible values..

so the answer is A
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Re: m06#34   [#permalink] 12 Sep 2012, 10:55
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