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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.

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)? _________________