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Why is B not the answer for this problem?

Given information in 2), I solved as follows:

Step 1. Square root both sides -> x - 50 = y - 50
Step 2. Add 50 to both sides -> x=y
Therefore 2 is sufficient and B is the answer
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Why is B not the answer for this problem?

Given information in 2), I solved as follows:

Step 1. Square root both sides -> x - 50 = y - 50
Step 2. Add 50 to both sides -> x=y
Therefore 2 is sufficient and B is the answer

Hi,

You are only considering the result to be positive when doing the square root on both sides e.g. sqrt(4) can be either 2 or -2.

Although I think the method you have chosen is not the quickest one, if you still want to go with that you will have to consider all the cases.

Case 1: sqrt of (x-50)^2 = +(x-50) and sqrt of (y-50)^2 = +(y-50) => x = y
Case 2: sqrt of (x-50)^2 = -(x-50) and sqrt of (y-50)^2 = +(y-50) => x + y = 100 ( x may or may not be equal to y)
Case 3: sqrt of (x-50)^2 = +(x-50) and sqrt of (y-50)^2 = -(y-50) => x + y = 100 ( x may or may not be equal to y)
Case 4: sqrt of (x-50)^2 = -(x-50) and sqrt of (y-50)^2 = -(y-50) => x = y

Therefore, statement 2 is insufficient.

Hope this helps.

I would suggest to simplify the equation.

(x-50)^2 = (y-50)^2
x^2 -100x + 2500 = y^2 -100y + 2500
x^2 - 100x = y^2 -100y
x^2 - y^2 -100x -100y = 0
x+y-100 = 0 or x-y = 0
x+y = 100 or x = y

Therefore, insufficient.
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ChiLeafsFan
Why is B not the answer for this problem?

Given information in 2), I solved as follows:

Step 1. Square root both sides -> x - 50 = y - 50
Step 2. Add 50 to both sides -> x=y
Therefore 2 is sufficient and B is the answer

Hi,

You are only considering the result to be positive when doing the square root on both sides e.g. sqrt(4) can be either 2 or -2.

Although I think the method you have chosen is not the quickest one, if you still want to go with that you will have to consider all the cases.

Case 1: sqrt of (x-50)^2 = +(x-50) and sqrt of (y-50)^2 = +(y-50) => x = y
Case 2: sqrt of (x-50)^2 = -(x-50) and sqrt of (y-50)^2 = +(y-50) => x + y = 100 ( x may or may not be equal to y)
Case 3: sqrt of (x-50)^2 = +(x-50) and sqrt of (y-50)^2 = -(y-50) => x + y = 100 ( x may or may not be equal to y)
Case 4: sqrt of (x-50)^2 = -(x-50) and sqrt of (y-50)^2 = -(y-50) => x = y

Therefore, statement 2 is insufficient.

Hope this helps.

I would suggest to simplify the equation.

(x-50)^2 = (y-50)^2
x^2 -100x + 2500 = y^2 -100y + 2500
x^2 - 100x = y^2 -100y
x^2 - y^2 -100x -100y = 0
x+y-100 = 0 or x-y = 0
x+y = 100 or x = y

Therefore, insufficient.


Thank you! I see what I wasn't thinking though now.
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Hi All,

This DS prompt is based around the 'thoroughness' of your thinking. You can likely see an 'obvious' answer when dealing with each of the two Facts, but can you see any OTHER possibiilties... We can TEST VALUES to get to the solution.

We're asked if X equals Y. This is a YES/NO question.

Fact 1: X + Y = 100

IF...
X = 50
Y = 50
Then the answer to the question is YES.

IF...
X = 0
Y = 100
Then the answer to the question is NO.
Fact 1 is INSUFFICIENT.

Fact 1:\((X - 50)^{2}\) = \((Y - 50)^{2}\)

IF...
X = 50
Y = 50
Then the answer to the question is YES.

IF...
X = 0
Y = 100
Then the answer to the question is NO.
Fact 2 is INSUFFICIENT.

Combined, we have two TESTS that already fit BOTH Facts and provide 2 different answers (one YES and one NO), so there's no more work required.
Combined, INSUFFICIENT.

Final Answer:
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St1. x+y=100, INSUFF

St2. just say |x-50|=|y-50|, so distance from x to 50 is equal to distance from y to 50, so INSUFF.

St1+St2. Can be multiple options where x=y and x not equal y. INSUFF

E
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Why is B not the answer for this problem?

Given information in 2), I solved as follows:

Step 1. Square root both sides -> x - 50 = y - 50
Step 2. Add 50 to both sides -> x=y
Therefore 2 is sufficient and B is the answer

Because A can be 75 and B 25, still the squares will match.
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shyind
Is x = y ?

(1) x + y = 100

(2) \((x − 50)^2 = (y − 50)^2\)


Kindly help me understand where am I going wrong here?

For (2)

(x-50)^2=(y-50)^2

or, (x-50)^2-(y-50)^2=0
or, (x-50+y-50) (x-50-y+50)=0
or, (x+y-100)(x-y)=0

so, either x=y or x=100-y.

From (1) & (2) X=100-y. Suff (C).

Where am I going wrong?

Please Help!!
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Hi SOUMYAJIT,

In DS questions, you have to make sure that you're not missing any of the 'possibilities' and that you're answering the question that is ASKED.

You did enough work to determine that Fact 1 and Fact 2 were both insufficient on their own, but when combined, there is more than one answer to the question "Is X = Y?"

Here are two solutions to the equations given:
X=50
Y=50
and the answer to the question is YES.

X=0
Y=100
and the answer to the question is NO.

The are two different answers to the given question, so combined/insufficient (E) is the answer.

GMAT assassins aren't born, they're made,
Rich
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EMPOWERgmatRichC
Hi SOUMYAJIT,

In DS questions, you have to make sure that you're not missing any of the 'possibilities' and that you're answering the question that is ASKED.

You did enough work to determine that Fact 1 and Fact 2 were both insufficient on their own, but when combined, there is more than one answer to the question "Is X = Y?"

Here are two solutions to the equations given:
X=50
Y=50
and the answer to the question is YES.

X=0
Y=100
and the answer to the question is NO.

The are two different answers to the given question, so combined/insufficient (E) is the answer.

GMAT assassins aren't born, they're made,
Rich


Got it thanks!!

I seem to have this problem, where in :

(1) I do not forget the first options condition when I goto solve the second DS point, thereby actually checking for C and not B.


or


(2) In the process of thinking, I forget the question that was asked to me in the first place.

Sometimes, even in geometry when I make transformations in a figure to solve a question, I forget the real figure that was there in the question in the first place.

Do you have a way that I can handle this?
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