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# M06 Q34

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Intern
Joined: 12 Oct 2008
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01 Apr 2012, 11:15
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If a, b, c and 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

from 1 we get b^2-a^2=23.

(b+a)(b-a) =23, as 23 a prime number, b+a =23 and b-a = 1, => b=12 and a =11. so c=9

Hence statement 1 is sufficient to answer the question.

Is the solution incorrect? if yes why? Help Please.

Kudos [?]: 4 [0], given: 2

Math Expert
Joined: 02 Sep 2009
Posts: 42248

Kudos [?]: 132539 [1], given: 12324

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01 Apr 2012, 13:04
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AmarSharma wrote:
If a, b, c and 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

from 1 we get b^2-a^2=23.

(b+a)(b-a) =23, as 23 a prime number, b+a =23 and b-a = 1, => b=12 and a =11. so c=9

Hence statement 1 is sufficient to answer the question.

Is the solution incorrect? if yes why? Help Please.

The problem with your solution is that you assume, with no ground for it, that variables represent integers only. From (b+a)(b-a)=23 you cannot say that b+a=23 and b-a=1, because for example b+a can be 46 and b-a can be 1/2.

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

(1) b^2+c^2=225. Not sufficient on its own.
(2) a^2+b^2=265. Not sufficient on its own.

(1)+(2) Subtract a^2+c^2=202 from b^2+c^2=225: b^2-a^2=23. Now, sum this with a^2+b^2=265: 2b^2=288 --> b^2=144 --> b=12 (since given that b is a positive number). Since b=12 then from b^2-a^2=23 we get that a=11 and from a^2+c^2=202 we get that c=9. Sufficient.

Hope it's clear.
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Kudos [?]: 132539 [1], given: 12324

Intern
Joined: 12 Oct 2008
Posts: 13

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01 Apr 2012, 21:43
Bunuel wrote:
AmarSharma wrote:
If a, b, c and 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

from 1 we get b^2-a^2=23.

(b+a)(b-a) =23, as 23 a prime number, b+a =23 and b-a = 1, => b=12 and a =11. so c=9

Hence statement 1 is sufficient to answer the question.

Is the solution incorrect? if yes why? Help Please.

The problem with your solution is that you assume, with no ground for it, that variables represent integers only. From (b+a)(b-a)=23 you cannot say that b+a=23 and b-a=1, because for example b+a can be 46 and b-a can be 1/2.

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

(1) b^2+c^2=225. Not sufficient on its own.
(2) a^2+b^2=265. Not sufficient on its own.

(1)+(2) Subtract a^2+c^2=202 from b^2+c^2=225: b^2-a^2=23. Now, sum this with a^2+b^2=265: 2b^2=288 --> b^2=144 --> b=12 (since given that b is a positive number). Since b=12 then from b^2-a^2=23 we get that a=11 and from a^2+c^2=202 we get that c=9. Sufficient.

Hope it's clear.

Thanks. I have to be more careful with my assumptions.

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13 Apr 2012, 06:19
Bunuel wrote:
AmarSharma wrote:
If a, b, c and 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

from 1 we get b^2-a^2=23.

(b+a)(b-a) =23, as 23 a prime number, b+a =23 and b-a = 1, => b=12 and a =11. so c=9

Hence statement 1 is sufficient to answer the question.

Is the solution incorrect? if yes why? Help Please.

The problem with your solution is that you assume, with no ground for it, that variables represent integers only. From (b+a)(b-a)=23 you cannot say that b+a=23 and b-a=1, because for example b+a can be 46 and b-a can be 1/2.

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

(1) b^2+c^2=225. Not sufficient on its own.
(2) a^2+b^2=265. Not sufficient on its own.

(1)+(2) Subtract a^2+c^2=202 from b^2+c^2=225: b^2-a^2=23. Now, sum this with a^2+b^2=265: 2b^2=288 --> b^2=144 --> b=12 (since given that b is a positive number). Since b=12 then from b^2-a^2=23 we get that a=11 and from a^2+c^2=202 we get that c=9. Sufficient.

Hope it's clear.

Bunuel - I have a question - The question states that a,b,c are positive. It doesn't state that they are positive integers. Essentially, I could have a=10.99 etc... Do you think that the answer would be E) then?

Thanks
Voodoo

Kudos [?]: 238 [0], given: 78

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Joined: 02 Sep 2009
Posts: 42248

Kudos [?]: 132539 [0], given: 12324

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13 Apr 2012, 06:29
voodoochild wrote:
Bunuel wrote:
AmarSharma wrote:
If a, b, c and 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

from 1 we get b^2-a^2=23.

(b+a)(b-a) =23, as 23 a prime number, b+a =23 and b-a = 1, => b=12 and a =11. so c=9

Hence statement 1 is sufficient to answer the question.

Is the solution incorrect? if yes why? Help Please.

The problem with your solution is that you assume, with no ground for it, that variables represent integers only. From (b+a)(b-a)=23 you cannot say that b+a=23 and b-a=1, because for example b+a can be 46 and b-a can be 1/2.

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

(1) b^2+c^2=225. Not sufficient on its own.
(2) a^2+b^2=265. Not sufficient on its own.

(1)+(2) Subtract a^2+c^2=202 from b^2+c^2=225: b^2-a^2=23. Now, sum this with a^2+b^2=265: 2b^2=288 --> b^2=144 --> b=12 (since given that b is a positive number). Since b=12 then from b^2-a^2=23 we get that a=11 and from a^2+c^2=202 we get that c=9. Sufficient.

Hope it's clear.

Bunuel - I have a question - The question states that a,b,c are positive. It doesn't state that they are positive integers. Essentially, I could have a=10.99 etc... Do you think that the answer would be E) then?

Thanks
Voodoo

We have a system of equations, which gives us fixed values of a, b and c:

From b^2=144 --> b=12 (since b>0 then b=-12 is not a valid solution);
From b^2-a^2=23 --> 144-a^2=23 --> a^2=121 --> a=11 (since a>0 then a=-11 is not a valid solution);
From a^2+c^2=202 --> 121+c^2=202 --> c^2=81 --> c=9 (since c>0 then c=-9 is not a valid solution).

Now, we are not told that a, b, and c are integers but how did this affect the solution? How can a solution of a^2=121 be a=10.99 or any other value but a=11 (or a=-11 which we discarded because of a>0)?
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13 Apr 2012, 08:15
Thanks Bunuel. You are correct. I didn't think about the system of equations.

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Re: M06 Q34   [#permalink] 13 Apr 2012, 08:15
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# M06 Q34

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