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# m06#34

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Retired Moderator
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m06#34 [#permalink]  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 $$\sqrt{121}$$

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Re: m06#34 [#permalink]  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....
Retired Moderator
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Re: m06#34 [#permalink]  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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Retired Moderator
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Re: m06#34 [#permalink]  14 Jul 2010, 21:21
can any one answer my query?? I am waiting.
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Re: m06#34 [#permalink]  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 $$\sqrt{121}$$

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]  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 $$\sqrt{121}$$

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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Kudos [?]: 42120 [0], given: 5957

Re: m06#34 [#permalink]  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 $$\sqrt{121}$$

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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Re: m06#34 [#permalink]  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..

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Re: m06#34   [#permalink] 12 Sep 2012, 10:55
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# m06#34

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