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Bunuel
3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a

(1) (x + y)^2 = 9a --> x^2+2xy+y^2=9a. Clearly insufficient.

(2) (x – y)^2 = a --> x^2-2xy+y^2=a. Clearly insufficient.

(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x,y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.

Answer: E.
Hi Bruel,

For (1): Because (x + y)^2 <= (1^2+1^2)(x^2+y^2) => x^2+y^2 >= 4.5a > 4a. So A is sufficient?

Could you help me find out whether there is any mistake in my solution?
Thanks Bruel,

Dear camlan1990

The highlighted part in your solution above is wrong.

The correct expansion for \((x+y)^2 = x^2 + y^2 + 2xy\).

You on the other hand have wrongly written: \((x+y)^2 = 2(x^2 + y^2)\).

Hope this helped. :)

Best Regards

Japinder

Dear Japinder,

As I highlighted, (x+y)^2 is smaller or equal 2(x^2 + y^2)
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camlan1990

Hi Bruel,

For (1): Because (x + y)^2 <= (1^2+1^2)(x^2+y^2) => x^2+y^2 >= 4.5a > 4a. So A is sufficient?

Could you help me find out whether there is any mistake in my solution?
Thanks Bruel,

Dear camlan1990

The highlighted part in your solution above is wrong.

The correct expansion for \((x+y)^2 = x^2 + y^2 + 2xy\).

You on the other hand have wrongly written: \((x+y)^2 = 2(x^2 + y^2)\).

Hope this helped. :)

Best Regards

Japinder

Dear Japinder,

As I highlighted, (x+y)^2 is smaller or equal 2(x^2 + y^2)

Oops camlan1990, that was my bad! I didn't notice the '<' sign in the "<=" :-D

Now I do see how you got to x^2+y^2 >= 4.5a

But please note that x = 0, y = 0 and a = 0 is one set of values that satisfies this inequality. For these values of x, y and a, the answer to the question 'Is x^2 + y^2 > 4a' is NO

For all other values of x, y and a, the answer will be YES.

Since we are not able to rule out x, y, a = 0, we cannot infer a unique answer to the posed question using St. 1 alone. So, St. 1 is not sufficient.

Hope this helped.

Japinder
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Bunuel
3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a

(1) (x + y)^2 = 9a --> x^2+2xy+y^2=9a. Clearly insufficient.

(2) (x – y)^2 = a --> x^2-2xy+y^2=a. Clearly insufficient.

(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x,y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.

Answer: E.

Hi Bunuel,

When you combine both statements 1 and 2, why do we need to substitute any value? This is the only part which I didn't understand. Could you please explain why '0' has to be substituted as the question just asks whether x^2 + y^2 = 4a. Nothing is mentioned about what 'a' is.

Awaiting your response.

Thanks.
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Bunuel
3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a

(1) (x + y)^2 = 9a --> x^2+2xy+y^2=9a. Clearly insufficient.

(2) (x – y)^2 = a --> x^2-2xy+y^2=a. Clearly insufficient.

(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x,y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.

Answer: E.

Hi Bunuel,

When you combine both statements 1 and 2, why do we need to substitute any value? This is the only part which I didn't understand. Could you please explain why '0' has to be substituted as the question just asks whether x^2 + y^2 = 4a. Nothing is mentioned about what 'a' is.

Awaiting your response.

Thanks.

The question asks whether x^2 + y^2 > 4a. If x = y = a = 0, then the answer is NO but if this is not so then the answer is YES. Please read the whole thread. This was discussed many, many, many times before.
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4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.

(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

One of the approaches:
\(2x-2y=1\) --> \(x=y+\frac{1}{2}\)
\(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.

Answer: C.

Hi Bunuel ,

I tried substituting values for x and y and seem to be getting a different ans.

If we take x=-1 and y=-1.5 the ans to the question is NO, whereas if we take x=1.5 & y=1 the ans is YES. Since we are unable to ans the question,

shouldn't the ans be E? if my reasoning is flawed can you please point out the flaw.. Thanks in advance
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Bunuel
4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.

(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

One of the approaches:
\(2x-2y=1\) --> \(x=y+\frac{1}{2}\)
\(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.

Answer: C.

Hi Bunuel ,

I tried substituting values for x and y and seem to be getting a different ans.

If we take x=-1 and y=-1.5 the ans to the question is NO, whereas if we take x=1.5 & y=1 the ans is YES. Since we are unable to ans the question,

shouldn't the ans be E? if my reasoning is flawed can you please point out the flaw.. Thanks in advance

x = -1 and y = -1.5 does not satisfy x/y > 1.
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Bunuel
4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.

(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

One of the approaches:
\(2x-2y=1\) --> \(x=y+\frac{1}{2}\)
\(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.

Answer: C.

Hi Bunuel ,

I tried substituting values for x and y and seem to be getting a different ans.

If we take x=-1 and y=-1.5 the ans to the question is NO, whereas if we take x=1.5 & y=1 the ans is YES. Since we are unable to ans the question,

shouldn't the ans be E? if my reasoning is flawed can you please point out the flaw.. Thanks in advance

Hi NavinRK,

The Flaw is

You can NOT take Values x=-1 and y=-1.5 after combining the two statements because in that case second statement x/y>1 will NOT be satisfied.

Hence the only set of values that all allowed to be taken are the one in which
Absolute value of x is greater than Absolute value of y

and

the Sign of both x and y must be same

so only Positive values are acceptable now


I hope it answers your query.
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Thanks for such post Bunuel,

Pls explain for Q #10 part B) how N will have only negative value when n<1/lnl

Thanks
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Thanks for such post Bunuel,

Pls explain for Q #10 part B) how N will have only negative value when n<1/lnl

Thanks

Hi Lipsi18,

The Explanations is as mentioned below:

10. If n is not equal to 0, is |n| < 4 ?
(1) n^2 > 16
(2) 1/|n| > n


Statement 2: 1/|n| > n

Multiplying |n| which is a positive value on both sides of the inequation, we get,

1> n*|n|
i.e. n*|n| < 1

Case 1: n is positive:
i.e. n^2 <1
i.e. 0<n<1

Case 2: n is negative
n*|n| <1 is true for all negative values of n

But in context of question this statement is NOT SUFFICIENT as it doesn't provide us any specific value of n
and for n=-5, |n| >4
and for n=-3, |n| <4
INCONSISTENT answer
Hence, NOT SUFFICIENT

I hope it answers your query!!!
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Bunuel
4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.

(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

One of the approaches:
\(2x-2y=1\) --> \(x=y+\frac{1}{2}\)
\(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.

Answer: C.

Also - Viewing graphically
A. 2x - 2y = 1
Put x = 1, you get y = 1.
Put x = 2, you get y = 1.5.
Make a rough plot. This is a line which agrees to negative values of both x and y as well. Insufficient.

B. x/y > 1
Both x an y can be negative. Insufficient

A + B

which means for all values x > y, which 2x - 2y = 1. This is possible only in the first quadrant whenever the value of x > 1. Sufficient.
Answer C

sorry for the flipped image :)
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Bunuel
4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.

(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

One of the approaches:
\(2x-2y=1\) --> \(x=y+\frac{1}{2}\)
\(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.

Answer: C.

I found it easy to rule out (1) and (2) as individually being insufficient. but the conclusion I drew from (2) was obviously that the absolute value of \(x\) has to be bigger than \(y\) (and of course that they are the same size), so regardless of the sign \(2x\) had to be of greater magnitude than \(2y\), and the only way for \(2x-2y=1\) was if they were both positive... I know this isn't ground breaking but it's the very simple way I arrived at the correct answer without getting too "mathsy"
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Bunuel
12. Is r=s?
(1) -s<=r<=s
(2) |r|>=s

This one is tough.

(1) -s<=r<=s, we can conclude two things from this statement:
A. s is either positive or zero, as -s<=s;
B. r is in the range (-s,s) inclusive, meaning that r can be -s as well as s.
But we don't know whether r=s or not. Not sufficient.

(2) |r|>=s, clearly insufficient.

(1)+(2) -s<=r<=s, s is not negative, |r|>=s --> r>=s or r<=-s. This doesn't imply that r=s, from this r can be -s as well.
Consider: s=5, r=5 --> -5<=5<=5 |5|>=5
s=5, r=-5 --> -5<=-5<=5 |-5|>=5
Both statements are true with these values. Hence insufficient.

Answer: E.

I thought of this one graphically...

____________________-s____________s____________________________
(1) xxxxxxxxxxxxxxxxxrrrrrrrrrrrrrrrrrrrrxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
(2) rrrrrrrrrrrrrrrrrrrrrrrrrxxxxxxxxxxxxxrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr
(1)&(2) xxxxxxxxxxxxxrxxxxxxxxxxxxxrxxxxxxxxxxxxxxxxx

(1) Everything between and including \(-s\) and \(s\) INSUFF
(2) Everything outside of but still including \(-s\) and \(s\) INSUFF
(1) & (2) \(r = -s\) or \(s\)... still INSUFF
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Bunuel
12. Is r=s?
(1) -s<=r<=s
(2) |r|>=s

This one is tough.

(1) -s<=r<=s, we can conclude two things from this statement:
A. s is either positive or zero, as -s<=s;
B. r is in the range (-s,s) inclusive, meaning that r can be -s as well as s.
But we don't know whether r=s or not. Not sufficient.

(2) |r|>=s, clearly insufficient.

(1)+(2) -s<=r<=s, s is not negative, |r|>=s --> r>=s or r<=-s. This doesn't imply that r=s, from this r can be -s as well.
Consider: s=5, r=5 --> -5<=5<=5 |5|>=5
s=5, r=-5 --> -5<=-5<=5 |-5|>=5
Both statements are true with these values. Hence insufficient.

Answer: E.

Can we eliminate B on the basis that it is just a reworded form of Statement A.

My understanding is |x| <1 means -1<x<1. Similarly, -s>=r>=s, which is not possible.
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Bunuel
12. Is r=s?
(1) -s<=r<=s
(2) |r|>=s

This one is tough.

(1) -s<=r<=s, we can conclude two things from this statement:
A. s is either positive or zero, as -s<=s;
B. r is in the range (-s,s) inclusive, meaning that r can be -s as well as s.
But we don't know whether r=s or not. Not sufficient.

(2) |r|>=s, clearly insufficient.

(1)+(2) -s<=r<=s, s is not negative, |r|>=s --> r>=s or r<=-s. This doesn't imply that r=s, from this r can be -s as well.
Consider: s=5, r=5 --> -5<=5<=5 |5|>=5
s=5, r=-5 --> -5<=-5<=5 |-5|>=5
Both statements are true with these values. Hence insufficient.

Answer: E.

Can we eliminate B on the basis that it is just a reworded form of Statement A.

My understanding is |x| <1 means -1<x<1. Similarly, -s>=r>=s, which is not possible.

Hi SauravPathak27

I'm no expert, but hope I might be able to help.

your understanding that |x| <1 means -1<x<1 is correct.

(1) is telling us that \(r\) falls between \(-s\) and \(s\) INCLUSIVE of \(-s\) and \(s\)...... INSUFFICIENT
(2) is telling us that \(r\) falls outside \(-s\) and \(s\) INCLUSIVE of \(-s\) and \(s\)........ INSUFFICIENT
(1) & (2) together tells us that \(r\) must be equal to either \(-s\) or \(s\) but cannot determine which one........ INSUFFICIENT

Hopefully my freehand below makes it a little clearer.

If it helps.... Throw me a Kudos :)
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3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a


(1) (x + y)^2 = 9a
x^2 + y^2 + 2xy = 9a
x^2 + y^2 = 9a - 2xy

(2) (x – y)^2 = a
x^2 + y^2 - 2xy = a
x^2 + y^2 = a +2xy

So we can conclude that:
9a - 2xy = a + 2xy
8a = 4xy
4a = 2xy

Hence, in the first conclusion is:
x^2 + y^2 = 9a - 4a
x^2 + y^2 = 5a

Answer is C (Together, they can answer the initial question).

Am I wrong, or right?
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RafaelPina
3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a


(1) (x + y)^2 = 9a
x^2 + y^2 + 2xy = 9a
x^2 + y^2 = 9a - 2xy

(2) (x – y)^2 = a
x^2 + y^2 - 2xy = a
x^2 + y^2 = a +2xy

So we can conclude that:
9a - 2xy = a + 2xy
8a = 4xy
4a = 2xy

Hence, in the first conclusion is:
x^2 + y^2 = 9a - 4a
x^2 + y^2 = 5a

Answer is C (Together, they can answer the initial question).

Am I wrong, or right?

You are correct till \(x^2+y^2=5a\) but what if x=y=0 giving you a=0. In this case, \(x^2+y^2\) will be = 4a and NOT > 4a. This is the reason why E is the correct answer.

If you were given "is \(x^2+y^2 \geq 4a\) instead of just >4a", then yes, you would have marked C as the correct answer but you are asked >4a which may or may not be true.

Hope this helps.
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Bunuel
2. If y is an integer and y = |x| + x, is y = 0?
(1) x < 0
(2) y < 1

Note: as \(y=|x|+x\) then \(y\) is never negative. For \(x>{0}\) then \(y=x+x=2x\) and for \(x\leq{0}\) then (when x is negative or zero) then \(y=-x+x=0\).

(1) \(x<0\) --> \(y=|x|+x=-x+x=0\). Sufficient.

(2) \(y<1\), as we concluded y is never negative, and we are given that \(y\) is an integer, hence \(y=0\). Sufficient.

Answer: D.

Hi Bunuel,

I do agree with option 1 but donot agree with option 2, as the x value can be positive value such as 0.25 etc.so the Y value shall be 0.50, in this case how do we deduce.
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Bunuel
2. If y is an integer and y = |x| + x, is y = 0?
(1) x < 0
(2) y < 1

Note: as \(y=|x|+x\) then \(y\) is never negative. For \(x>{0}\) then \(y=x+x=2x\) and for \(x\leq{0}\) then (when x is negative or zero) then \(y=-x+x=0\).

(1) \(x<0\) --> \(y=|x|+x=-x+x=0\). Sufficient.

(2) \(y<1\), as we concluded y is never negative, and we are given that \(y\) is an integer, hence \(y=0\). Sufficient.

Answer: D.

Hi Bunuel,

I do agree with option 1 but donot agree with option 2, as the x value can be positive value such as 0.25 etc.so the Y value shall be 0.50, in this case how do we deduce.

Statement 2: \(y<1\)

Since we know that \(y = |x| + x\)

case 1: x>0.... In this case y = 2x and will be positive
case 2: x<0.... In this case y = 0
i.e. Y can never be Negative

This statement tells us that Y is an Integer less than 1 therefore 0 is the only possible value of y

Hence, SUFFICIENT

I hope this helps!
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