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VP
Status: The last round
Joined: 18 Jun 2009
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Concentration: Strategy, General Management
GMAT 1: 680 Q48 V34
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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=2252. a^2+b^2=265OA: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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Manager
Joined: 15 Apr 2010
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Kudos [?]:
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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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VP
Status: The last round
Joined: 18 Jun 2009
Posts: 1327
Concentration: Strategy, General Management
GMAT 1: 680 Q48 V34
Followers: 43
Kudos [?]:
383
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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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VP
Status: The last round
Joined: 18 Jun 2009
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Concentration: Strategy, General Management
GMAT 1: 680 Q48 V34
Followers: 43
Kudos [?]:
383
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GMAT Club team member
Joined: 02 Sep 2009
Posts: 11516
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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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Intern
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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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GMAT Club team member
Joined: 02 Sep 2009
Posts: 11516
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Kudos [?]:
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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.
_________________
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
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. NEW!!!
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. NEW!!!
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Intern
Joined: 02 Nov 2009
Posts: 45
Location: India
Concentration: General Management, Technology
GMAT Date: 04-21-2013
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WE: Information Technology (Internet and New Media)
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Kudos [?]:
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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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