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

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

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.

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.

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.

Answer: C.

Hope it's clear.

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.

The following links might help.

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
https://gmatclub.com/blog/2010/01/verita ... questions/
https://gmatclub.com/blog/2010/09/elimin ... fficiency/
https://gmatclub.com/blog/2011/08/gmat-d ... tion-type/
https://gmatclub.com/blog/2011/08/must-k ... trategies/
https://gmatclub.com/blog/2011/08/kaplan ... questions/
data-sufficiency-strategy-guide-130264.html

question stem says:
a=9, c=11
or, a=11, c=9

Statement 1:
c=12, b=9
or, c=9, b=12
---> not sufficient.

Statement 2: a=12, b=11
or, a=11, b=12
--->not sufficient

Statement 1+2:
a=11, b=12, c=9
Which gives a definite value of b-a-c. So, C is ok

I think it's a common mistake to assume that the variables would be integers. Hence, the sum of squares have also been kept as sum of squares of two integers to entice that error.

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