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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.

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.

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?

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)? _________________