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M06 Q34

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M06 Q34 [#permalink] New post 01 Apr 2012, 10:15
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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.
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Re: M06 Q34 [#permalink] New post 01 Apr 2012, 12:04
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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.

Answer: C.

Hope it's clear.

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Re: M06 Q34 [#permalink] New post 01 Apr 2012, 20:43
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.
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Re: M06 Q34 [#permalink] New post 13 Apr 2012, 05:19
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
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Re: M06 Q34 [#permalink] New post 13 Apr 2012, 05:29
Expert's post
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)?

_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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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Get the best GMAT Prep Resources with GMAT Club Premium Membership

Manager
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Re: M06 Q34 [#permalink] New post 13 Apr 2012, 07:15
Thanks Bunuel. You are correct. I didn't think about the system of equations.
Re: M06 Q34   [#permalink] 13 Apr 2012, 07:15
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