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# If a, b, and c are positive and a^2+c^2=202, what is the val

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If a, b, and c are positive and a^2+c^2=202, what is the val [#permalink]

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04 Jul 2013, 17:59
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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 = 225
(2) a^2 + b^2 =265

M06-34
[Reveal] Spoiler: OA

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Re: If a, b, and c are positive [#permalink]

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04 Jul 2013, 22:21
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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. 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.

M06-34
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Re: If a, b, and c are positive [#permalink]

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08 Jul 2013, 02:02
I think the answer should be A since statement1 is enough by itself.

Lets consider statement 1

a^2+c^2=202 given in the question
b^2+c^2=225 given in the statement1

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

b^2-a^2=(b-a)*(b+a)=23 b+a must be 23 and b-a must be 1 since a,b,c given positive and 23 is prime number. b=12 and a=11

if we know a and b then we could calculate c by using a^2+c^2=202.

Many thanks,
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Re: If a, b, and c are positive [#permalink]

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08 Jul 2013, 02:28
hfiratozturk wrote:
I think the answer should be A since statement1 is enough by itself.

Lets consider statement 1

a^2+c^2=202 given in the question
b^2+c^2=225 given in the statement1

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

b^2-a^2=(b-a)*(b+a)=23 b+a must be 23 and b-a must be 1 since a,b,c given positive and 23 is prime number. b=12 and a=11

if we know a and b then we could calculate c by using a^2+c^2=202.

Many thanks,

It is not mentioned that a,b and c are integer values, so we cannot say that b+a =23 and b-a = 1

Regards
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Re: If a, b, and c are positive [#permalink]

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29 May 2014, 04:59
Bunuel 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. 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.

M06-34

Hi bunnel,

Following is my logic

following is my logic

here in question it is given that

a^2+c^2=202 ----1

from st1 b^2+c^2=225------2

subtract 2 from 1

a^2-b^2 = -23

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

here we are given that a.b.c are positive so (a+b) can not be -23 or -1 as some of two positive no can not be negative

a-b=-1
a+b=23

resolving this we get a=11, b=12

now I will put b=12 in equation so I can get value of C.
b^2+c^2=225

same I can get with st2.

Bunnel could you please clarify this. What is the issue with my logic as official ans. is different

Thanks.
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Re: If a, b, and c are positive [#permalink]

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29 May 2014, 06:51
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PathFinder007 wrote:
Bunuel 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. 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.

M06-34

Hi bunnel,

Following is my logic

following is my logic

here in question it is given that

a^2+c^2=202 ----1

from st1 b^2+c^2=225------2

subtract 2 from 1

a^2-b^2 = -23

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

here we are given that a.b.c are positive so (a+b) can not be -23 or -1 as some of two positive no can not be negative

a-b=-1
a+b=23

resolving this we get a=11, b=12

now I will put b=12 in equation so I can get value of C.
b^2+c^2=225

same I can get with st2.

Bunnel could you please clarify this. What is the issue with my logic as official ans. is different

Thanks.

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.
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Re: If a, b, and c are positive and a^2+c^2=202, what is the val [#permalink]

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29 May 2014, 08:36
Why not D?

Using bruteforce we can deduce each variable.

Initial equation a^2+c^2=202
202 is a sum of squares, so we need to find 'em:
201 and 1 - no
198 and 4 - no
193 and 9 - no
186 and 16 - no
177 and 25 - no
166 and 36 - no
153 and 49 - no
138 and 64 - no
121 and 81 - yes
102 and 100 - no

So a and c could be 11 or 9

Using the same method whith both statements:
(1) b^2 + c^2 = 225
The only pair is 81 and 144, so b and c could only be 9 or 12 => c=9, b=12, a=11

(2) a^2 + b^2 = 265
There are two pairs:
a) 256 and 9 => a and b could be 16 or 3, but we know that "a" could only be 11 or 9, so eliminating this pair.
b) 144 and 121 => a and b could be 12 or 11 => b=12, a=11 , c=9

Of course it is time consuming process and I run out of 2 minutes, but still it solvable using each statement alone.
So i'm little bit confused.
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Re: If a, b, and c are positive and a^2+c^2=202, what is the val [#permalink]

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29 May 2014, 08:51
Valrus wrote:
Why not D?

Using bruteforce we can deduce each variable.

Initial equation a^2+c^2=202
202 is a sum of squares, so we need to find 'em:
201 and 1 - no
198 and 4 - no
193 and 9 - no
186 and 16 - no
177 and 25 - no
166 and 36 - no
153 and 49 - no
138 and 64 - no
121 and 81 - yes
102 and 100 - no

So a and c could be 11 or 9

Using the same method whith both statements:
(1) b^2 + c^2 = 225
The only pair is 81 and 144, so b and c could only be 9 or 12 => c=9, b=12, a=11

(2) a^2 + b^2 = 265
There are two pairs:
a) 256 and 9 => a and b could be 16 or 3, but we know that "a" could only be 11 or 9, so eliminating this pair.
b) 144 and 121 => a and b could be 12 or 11 => b=12, a=11 , c=9

Of course it is time consuming process and I run out of 2 minutes, but still it solvable using each statement alone.
So i'm little bit confused.

Please check here: if-a-b-and-c-are-positive-and-a-2-c-2-202-what-is-the-val-155421.html#p1369107 We are NOT told that a, b, and c are integers! So, you cannot make the chart you made.
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If a, b, and c are positive and a^2 + c^2 = 202, what is the [#permalink]

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05 Jul 2014, 01:57
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

M06-34

Last edited by Bunuel on 05 Jul 2014, 05:05, edited 1 time in total.
Renamed the topic and edited the question.
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Re: If a, b, and c are positive and a^2 + c^2 = 202, what is the [#permalink]

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05 Jul 2014, 02:02
Though the OA is C, i got A as an answer.
Reasoning.

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 , the only way we can have 23 as a product of 23 and 1.

Solving ,
b-a=1
b+a=23

we get , b=12 ,a=11 and c=9.

Is there something i am missing?

Thanks,
Gaurav

------------------------------
Kudos??
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Re: If a, b, and c are positive and a^2 + c^2 = 202, what is the [#permalink]

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05 Jul 2014, 05:06
gaurav245 wrote:
Though the OA is C, i got A as an answer.
Reasoning.

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 , the only way we can have 23 as a product of 23 and 1.

Solving ,
b-a=1
b+a=23

we get , b=12 ,a=11 and c=9.

Is there something i am missing?

Thanks,
Gaurav

------------------------------
Kudos??

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 [?]: 9 [0], given: 27

Re: If a, b, and c are positive and a^2 + c^2 = 202, what is the [#permalink]

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05 Jul 2014, 05:43
Thanks Bunuel...
That makes sense..
I tend to make a lot of such mistakes in DS questions.. Anything that you can suggest so that these things become alot more visible?
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Kudos [?]: 85878 [0], given: 10256

Re: If a, b, and c are positive and a^2+c^2=202, what is the val [#permalink]

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05 Jul 2014, 07:20
gaurav245 wrote:
Thanks Bunuel...
That makes sense..
I tend to make a lot of such mistakes in DS questions.. Anything that you can suggest so that these things become alot more visible?

Well, for DS questions do not assume anything on your own. Use only the information given in the questions.

Biggest GMAT Mistakes You Should Avoid: biggest-gmat-mistakes-you-should-avoid-166353.html

DS strategies:
trouble-with-quant-some-tips-traps-that-might-help-you-34329.html
tips-on-approaching-ds-questions-104945.html
http://gmatclub.com/blog/2010/01/verita ... questions/
http://gmatclub.com/blog/2010/09/elimin ... fficiency/
http://gmatclub.com/blog/2011/08/gmat-d ... tion-type/
http://gmatclub.com/blog/2011/08/must-k ... trategies/
http://gmatclub.com/blog/2011/08/kaplan ... questions/
data-sufficiency-strategy-guide-130264.html
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Re: If a, b, and c are positive and a^2+c^2=202, what is the val   [#permalink] 05 Jul 2014, 07:20
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