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m06#34

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m06#34 [#permalink]

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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]

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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]

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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]

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New post 14 Jul 2010, 21:21

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Re: m06#34 [#permalink]

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New post 18 Jul 2010, 14:45
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]

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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]

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New post 12 Sep 2012, 01:00
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.
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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:
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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.


What are GMAT Club Tests?
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Re: m06#34 [#permalink]

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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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