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Ravi05885
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Data Sufficiency Question 25 Mar 2020, 04:34
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Confused with plugging in.

How we got x=1 & y=1/2?

Question

Are both X & Y positive?

a) 2x-2y=1
b) x/y>1

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Data Sufficiency Question 25 Mar 2020, 09:09
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Ravi05885
Confused with plugging in.

How we got x=1 & y=1/2?

Question

Are both X & Y positive?

a) 2x-2y=1
b) x/y>1
Show more

Are both X & Y positive?

a) 2x-2y=1
2x = 2y + 1
If y=-1 ; x= -.5; No
But if y=1; x=1.5; Yes
NOT SUFFICIENT

b) x/y>1
x/y - 1 >0
(x-y)/y>0
If y>0; x-y>0; x>y>0; Yes
But if y<0; x-y<0; x<y<0; No
NOT SUFFICIENT

a) 2x-2y=1
x = y+.5
b) x/y>1
(y+.5)/y>1
1 + .5/y >1
.5/y > 0
y>0
x = y+.5>y>0
x>y>0
SUFFICIENT

IMO C
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Data Sufficiency Question 25 Mar 2020, 09:42
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[quote="Ravi05885"]Confused with plugging in.

How we got x=1 & y=1/2?

Question

Are both X & Y positive?

a) 2x-2y=1
b) x/y>1[/quote]

(1) 2x-2y=1
—> x - y = 1/2
—> x = y + 1/2
—> x > y ALWAYS
Possible values of (x y) = {(1, 1/2), (-1/2, -1), . . . }
—> No definite outcome —> [b]Insufficient[/b]

(2) x/y>1
—> x & y are both positive or negative
Possible values of (x y) = {(1, 1/2), (-1, -1/2), . . . }
—> No definite outcome —> [b]Insufficient[/b]

Combining (1) & (2),
Possible values of (x, y) = (1, 1/2), (2, 3/2), (5/2, 2), . . .
—> Both x & y are positive

Option C

[size=80][b][i]Posted from my mobile device[/i][/b][/size]
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Data Sufficiency Question 25 Mar 2020, 21:47
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given only
1) 2x - 2y =1
or x-y = 1/2
x and y can both be negative or positive no sufficient

2) x/y >1
x = -2 and y =-1 satisfy
also x = 2 and y = 1 does
not sufficient
combine both
both should b of same sign
and x>y
thus
x and y are both positive
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Data Sufficiency Question 27 Mar 2020, 04:54
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Hello Ravi,

This is a YES-NO type of DS question. In this type of a DS question, our main objective is to find a unique YES or a unique NO, using the data given in the statements.

Let’s have a look at the main question. What does the question ask of us - Are both X and Y positive? Think about how we can answer it.

We could say – YES, both X and Y are positive OR we could say – NO, both X and Y are not positive.

Now, with respect to the second case, you need to understand that when you say both X and Y are not positive, it can mean one of the following:

    X is positive AND Y is negative OR
    X is negative AND Y is positive OR
    X and Y are both negative.
    One of them is ZERO and the other one is either positive or negative

Hope this makes sense!

Now that we know what are the answers we are looking for, let’s go about analyzing the statements to see if they are sufficient to give us a definite YES or a NO.

From statement I alone, 2x-2y = 1. This can be simplified to x-y = ½. This tells us two things:
1) x is greater than y since x-y is positive.
2) x is ½ more than y.

So, can we substitute x=1 and y=\(\frac{1}{2}\)? Will it satisfy the equation above? YES. Do you now understand why we have taken x=1 and y = ½? You don’t GET these values from the equation, you TEST out these values. For these values of x and y, are x and y both positive? YES.

Can we take x = 0 and y = -\(\frac{1}{2}\)? Will it satisfy the equation above? YES. But for these values of x and y, are x and y both positive? NO.

So, if we just know that x-y = ½, will we be able to conclusively say if x and y are both positive ALWAYS? The answer is a clear no.
Statement I alone is insufficient. Answer options A and D can be eliminated. The possible answer options are B, C or E.

From statement II alone, we know that \(\frac{x}{y}\)>1. What does this mean? This means the following:

1) x and y are of the same signs i.e. either both positive or both negative.
2) If x and y are both positive, x>y
3) If x and y are both negative, x<y.

Is this data enough for us to conclusively say if both X and Y are positive always? NO. Why? They could both be negative as well.
Statement II alone is insufficient. Answer option B can be eliminated. The possible answer options are C or E.

Combining both statements I and II, we have the following:

From statement II, we know that x and y have to be of the same signs; additionally we know that x>y if both are positive while x<y if both are negative.

From statement I, we know that x=y+\(\frac{1}{2}\). What does this mean? This essentially means that x>y.

Now that we know x>y and \(\frac{x}{y}\)>1, what’s the only possible conclusion which satisfies both? That both x and y HAVE to be positive.

At this stage, if you are not convinced, you can take a set of negative values for x and y and check if it satisfies both statements. For example, let’s assume that y = -3 and x = -2.5 since x = y + ½/. What’s the value of \(\frac{x}{y}\) now?

\(\frac{x}{y}\) = \(\frac{-2.5}{-3}\) = \(\frac{5}{6}\). Is \(\frac{5}{6}\) greater than 1? Clearly not.

This should tell you that both x and y have to be POSITIVE. The combination of statements is sufficient to answer the question with a definite YES. Answer option E can be eliminated.
The correct answer option is C.

So, the most important learning here for you is to spend time on understanding the question stem and trying to figure out what the expected answers are. Once you know this, it will become easy for you to look for specific pieces of information that will help you solve the question.
Also note that you cannot OBTAIN a value for x and y if you have only one equation like x-y = ½. You will have to test values and see which values satisfy the condition.

Hope that helps!
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Data Sufficiency Question 10 May 2020, 14:15
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Ravi05885
Confused with plugging in.

How we got x=1 & y=1/2?

Question

Are both X & Y positive?

a) 2x-2y=1
b) x/y>1
Show more

This is a GMAT Prep question, which is discussed here: are-x-and-y-both-positive-1-2x-2y-1-2-x-y-63377-40.html Hope it helps.

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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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Data Sufficiency Question 08 Aug 2020, 06:19
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Ravi05885
Confused with plugging in.

How we got x=1 & y=1/2?

Question

Are both X & Y positive?

a) 2x-2y=1
b) x/y>1
Show more

This question is discussed here: https://gmatclub.com/forum/are-x-and-y- ... 63377.html. In case of questions please post there. Hope it helps.

Also, please read carefully and follow our posting rules: https://gmatclub.com/forum/rules-for-po ... 33935.html Thank you.

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