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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. 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. 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.

M06-34

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

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.

M06-34

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.
_________________

Re: If a, b, and c are positive and a^2+c^2=202, what is the val [#permalink]

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29 May 2014, 07:36

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.

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.

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.

Re: If a, b, and c are positive and a^2 + c^2 = 202, what is the [#permalink]

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05 Jul 2014, 04:43

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?

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.

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