NEW HERE? LEARN MORE
closeclose
Last visit was: 25 Apr 2024, 05:22 It is currently 25 Apr 2024, 05:22
Decision Tracker
My Rewards
New posts
New comers' posts

Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

M06 Q34

User avatar
AmarSharma
Intern
Intern
Joined: 12 Oct 2008
Posts: 11
Own Kudos [?]: 4 [0]
Given Kudos: 2
Send PM
M06 Q34 [#permalink] New post   01 Apr 2012, 11:15
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.

Archived Topic
Hi there,
Archived GMAT Club Tests question - no more replies possible.
Where to now? Try our up-to-date Free Adaptive GMAT Club Tests for the latest questions.
Still interested? Check out the "Best Topics" block below for better discussion and related questions.
Thank you for understanding, and happy exploring!
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92912
Own Kudos [?]: 618924 [1]
Given Kudos: 81595
Send PM
CAT Tests Forum Quiz Mode
Re: M06 Q34 [#permalink] New post   01 Apr 2012, 13:04
1
Kudos
Expert Reply
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.

Answer: C.

Hope it's clear.
User avatar
AmarSharma
Intern
Intern
Joined: 12 Oct 2008
Posts: 11
Own Kudos [?]: 4 [0]
Given Kudos: 2
Send PM
Re: M06 Q34 [#permalink] New post   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.

Answer: C.

Hope it's clear.



Thanks. I have to be more careful with my assumptions.
User avatar
voodoochild
Manager
Manager
Joined: 16 Feb 2011
Posts: 148
Own Kudos [?]: 919 [0]
Given Kudos: 78
Schools:ABCD
Send PM
Re: M06 Q34 [#permalink] New post   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.

Answer: C.

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?

Please help me :(

Thanks
Voodoo
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92912
Own Kudos [?]: 618924 [0]
Given Kudos: 81595
Send PM
CAT Tests Forum Quiz Mode
Re: M06 Q34 [#permalink] New post   13 Apr 2012, 06:29
Expert Reply
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.

Answer: C.

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?

Please help me :(

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)?
User avatar
voodoochild
Manager
Manager
Joined: 16 Feb 2011
Posts: 148
Own Kudos [?]: 919 [0]
Given Kudos: 78
Schools:ABCD
Send PM
Re: M06 Q34 [#permalink] New post   13 Apr 2012, 08:15
Thanks Bunuel. You are correct. I didn't think about the system of equations.

Archived Topic
Hi there,
Archived GMAT Club Tests question - no more replies possible.
Where to now? Try our up-to-date Free Adaptive GMAT Club Tests for the latest questions.
Still interested? Check out the "Best Topics" block above for better discussion and related questions.
Thank you for understanding, and happy exploring!
GMAT Club Bot
gmatclubot
Re: M06 Q34 [#permalink]
13 Apr 2012, 08:15
Moderator:
Bunuel
Math Expert
92912 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions