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M27-12

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M27-12 [#permalink]

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(1) \(x^2+y^2=1\). Recall that \((x-y)^2 \ge 0\) (square of any number is more than or equal to zero). Expand: \(x^2-2xy+y^2 \ge 0\) and since \(x^2+y^2=1\) then: \(1-2xy \ge 0\). So, \(xy \le \frac{1}{2}\). Sufficient.

(2) \(x^2-y^2=0\). Re-arrange and take the square root from both sides: \(|x|=|y|\). Clearly insufficient.


Answer: A
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Re: M27-12 [#permalink]

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New post 23 Oct 2014, 09:13
The same applicable to (x+y)^2 ==> x^2 + y^2 + 2 x*y is greter than or equal to ZERO.
1 + 2 x*y is greter than or equal to ZERO.
xy greter than or equal -1/2...This can prove A is insufficient, right ???

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Re: M27-12 [#permalink]

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New post 23 Oct 2014, 09:23
eureka750 wrote:
The same applicable to (x+y)^2 ==> x^2 + y^2 + 2 x*y is greter than or equal to ZERO.
1 + 2 x*y is greter than or equal to ZERO.
xy greter than or equal -1/2...This can prove A is insufficient, right ???


You need to consider \((x-y)^2\geq{0}\) to get sufficiency.
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Re: M27-12 [#permalink]

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New post 01 May 2015, 10:46
Why did you start from : (x-y)^2>=0 ??

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Re: M27-12 [#permalink]

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New post 01 May 2015, 10:48
Why did you start from : (x-y)^2>=0 ??

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Re: M27-12 [#permalink]

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New post 08 Jun 2015, 09:29
Bunuel wrote:
eureka750 wrote:
The same applicable to (x+y)^2 ==> x^2 + y^2 + 2 x*y is greter than or equal to ZERO.
1 + 2 x*y is greter than or equal to ZERO.
xy greter than or equal -1/2...This can prove A is insufficient, right ???


You need to consider \((x-y)^2\geq{0}\) to get sufficiency.



Please explain why do we need to consider only (x-y)^2 to get sufficiency, as also in (x+y)^2 we use x^2 + y^ 2=1 premise, hence its not in all cases that(A) results in single conclusion or sufficiency.

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Re: M27-12 [#permalink]

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New post 20 Jun 2015, 06:40
yeah even i have a same doubt why not (x+y)^2 as harshalnamdeo88...

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Re: M27-12 [#permalink]

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New post 20 Jun 2015, 12:29
BCCLSN wrote:
Why did you start from : (x-y)^2>=0 ??

harshalnamdeo88 wrote:
Bunuel wrote:
eureka750 wrote:
The same applicable to (x+y)^2 ==> x^2 + y^2 + 2 x*y is greter than or equal to ZERO.
1 + 2 x*y is greter than or equal to ZERO.
xy greter than or equal -1/2...This can prove A is insufficient, right ???


You need to consider \((x-y)^2\geq{0}\) to get sufficiency.



Please explain why do we need to consider only (x-y)^2 to get sufficiency, as also in (x+y)^2 we use x^2 + y^ 2=1 premise, hence its not in all cases that(A) results in single conclusion or sufficiency.

shrutikasat wrote:
yeah even i have a same doubt why not (x+y)^2 as harshalnamdeo88...


Guys, the point is that if you consider \((x+y)^2 \ge 0\), you'll get \(xy\geq{-\frac{1}{2}}\), which is not helpful at all. Actually since both \(xy\geq{-\frac{1}{2}}\) (from \((x+y)^2 \ge 0\)) and \(xy \le \frac{1}{2}\) (from \((x-y)^2 \ge 0\)) are true, then we get that \(-\frac{1}{2} \leq{xy} \leq{\frac{1}{2}}\). But only the approach given in the solution above gives you the answer we are looking for, while another one gives you an inequality which is not helpful.

Hope it's clear.
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Re: M27-12 [#permalink]

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New post 07 Jul 2015, 09:02
Bunuel wrote:
Official Solution:


(1) \(x^2+y^2=1\). Recall that \((x-y)^2 \ge 0\) (square of any number is more than or equal to zero). Expand: \(x^2-2xy+y^2 \ge 0\) and since \(x^2+y^2=1\) then: \(1-2xy \ge 0\). So, \(xy \le \frac{1}{2}\). Sufficient.

(2) \(x^2-y^2=0\). Re-arrange and take the square root from both sides: \(|x|=|y|\). Clearly insufficient.


Answer: A


Hi Bunuel,

Could you help me if I am wrong?

What I did was first to convert question stem to 2xy - 1 < 0, then from (1) I concluded that in order to satisfy x^2+y^2=1, The only x and y possible were; x=1, y=0, x=-1, y=0, x=0, y=1, x=0, y=-1; so question stem 2xy - 1 < 0 is always negative... Is my thinking process wrong or is correct?

Thanks a lot.

Regards.

Luis Navarro
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Re: M27-12 [#permalink]

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luisnavarro wrote:
Bunuel wrote:
Official Solution:


(1) \(x^2+y^2=1\). Recall that \((x-y)^2 \ge 0\) (square of any number is more than or equal to zero). Expand: \(x^2-2xy+y^2 \ge 0\) and since \(x^2+y^2=1\) then: \(1-2xy \ge 0\). So, \(xy \le \frac{1}{2}\). Sufficient.

(2) \(x^2-y^2=0\). Re-arrange and take the square root from both sides: \(|x|=|y|\). Clearly insufficient.


Answer: A


Hi Bunuel,

Could you help me if I am wrong?

What I did was first to convert question stem to 2xy - 1 < 0, then from (1) I concluded that in order to satisfy x^2+y^2=1, The only x and y possible were; x=1, y=0, x=-1, y=0, x=0, y=1, x=0, y=-1; so question stem 2xy - 1 < 0 is always negative... Is my thinking process wrong or is correct?

Thanks a lot.

Regards.

Luis Navarro
Looking for 700


Your thought process is not precise. We are NOT told that x and y are integers, thus x^2 + y^2 = 1 will have infinitely many solutions, not just the ones you mentioned.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M27-12 [#permalink]

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New post 07 Jul 2015, 09:40
Bunuel wrote:
luisnavarro wrote:
Bunuel wrote:
Official Solution:


(1) \(x^2+y^2=1\). Recall that \((x-y)^2 \ge 0\) (square of any number is more than or equal to zero). Expand: \(x^2-2xy+y^2 \ge 0\) and since \(x^2+y^2=1\) then: \(1-2xy \ge 0\). So, \(xy \le \frac{1}{2}\). Sufficient.

(2) \(x^2-y^2=0\). Re-arrange and take the square root from both sides: \(|x|=|y|\). Clearly insufficient.


Answer: A


Hi Bunuel,

Could you help me if I am wrong?

What I did was first to convert question stem to 2xy - 1 < 0, then from (1) I concluded that in order to satisfy x^2+y^2=1, The only x and y possible were; x=1, y=0, x=-1, y=0, x=0, y=1, x=0, y=-1; so question stem 2xy - 1 < 0 is always negative... Is my thinking process wrong or is correct?

Thanks a lot.

Regards.

Luis Navarro
Looking for 700


Your thought process is not precise. We are NOT told that x and y are integers, thus x^2 + y^2 = 1 will have infinitely many solutions, not just the ones you mentioned.


Thanks a lot Bunuel.

Regards

Luis Navarro
Looking for 700

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Re: M27-12 [#permalink]

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New post 12 Mar 2016, 22:16
Could you kindly explain the expansion on the first statement? Txx

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New post 13 Mar 2016, 08:23

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Re: M27-12 [#permalink]

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New post 06 Jul 2016, 16:34
I used geometrical approach to solve this problem. Did I get lucky?

(X)^2 +(Y)^2=1

That would indicate a right angle triangle with x and y as height and base. Maximize the area of triangle would maximise the x*y product.
Thus X=Y=1/(root2) (not sure if this step is right.... I don't have a reason for this)

Thus the product will always be less than or equals 1/2

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Re: M27-12 [#permalink]

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New post 05 Feb 2017, 13:05
Hi Bunuel,

I don't understand why do we need this: "Recall that (x−y)2≥0(x−y)2≥0 (square of any number is more than or equal to zero)" to solve the equation. What's the relationship with x2+y2?

Thanks

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Re: M27-12 [#permalink]

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New post 16 Feb 2017, 04:07
For statement 1 I used a different approach: I considered X^2+Y^2=1 a circle with centre in (0,0) and radius 1. The product of xy is represented by a parallelogram whose area is maximized in the first quadrant when the parallelogram is a square. Hence xy reaches the max when x=y = 1/square root of 2. So the product can be max 1/2.

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Re: M27-12 [#permalink]

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New post 23 Aug 2017, 16:20
This is a hard question, in particular because knowing to use the square of a difference is a tricky detail.

You can also use the fact that the maximum product of \(ab\) given that \(a+b=c\) is the set of numbers where \(a\) and \(b\) are nearest each other and still sum to \(c\).

For example, if \(a+b=20\), the maximum product of \(ab=10*10=100\)

In this case the maximum product of \(ab\) where \(a+b=1\) is when \(a=b=\frac{1}{2}\)

In this case \(x^2=a\) and \(y^2=b\), so the the maximum product of \(xy\) is \(\sqrt{\frac{1}{2}}*\sqrt{\frac{1}{2}}=\frac{1}{2}\)

Answer A

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Re: M27-12   [#permalink] 23 Aug 2017, 16:20
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