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# Is (y-10)^2 > (x+10)^2?

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15 Dec 2012, 11:31
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Is (y-10)^2 > (x+10)^2?

(1) -y > x+5
(2) x > y
[Reveal] Spoiler: OA

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Re: Is (y-10)^2 > (x+10)^2? [#permalink]

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17 Dec 2012, 00:34
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JJ2014 wrote:
Is (y-10)^2 > (x+10)^2?

(1) -y > x+5
(2) x > y

Hi JJ2014,

1. From St 1 , we get x+y<-5

X=10
Y= -4
Then St under consideration is not true i.e (y-10)^2> (x+10)^2

X=-11
y= 5

Then st under consideration is true

So A and D ruled out.

St 2 alone is not sufficient.

Ex X=3, y=2, St not true
X=-2, y=-3, St is true

So B ruled out.

Combining both the statements we get

x+y<-5 and x>y
We see that y<0 as otherwise if y>0 then the eqn X>y is not true.
Hence ans C
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Last edited by WoundedTiger on 02 Jan 2014, 01:19, edited 1 time in total.

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Re: Is (y-10)^2 > (x+10)^2? [#permalink]

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21 Dec 2012, 02:34
Karishma / Bunuel,

Could you throw some light on the below query.

I always have trouble in questions like this as to what numbers to pick. I end up going way beyond 3 minutes.

Do tell me the certain set of numbers/range of numbers that one should always begin testing with

P.S. I know it could vary from question to question but want a general set.
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Re: Is (y-10)^2 > (x+10)^2? [#permalink]

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22 Dec 2012, 06:23
we need to find if (y-10)^2 > (x+10)^2?
or, is y^2 - 20y + 100 > x^2 + 20x + 100?
or, is y^2 - 20y > x^2 + 20x?
or, is y^2 - x^2 > 20(x + y)?
or, is (y+x)(y-x) > 20(x+y)?
here, because of inequality, we cannot divide by (x+y) on both sides, as we don't know whether (x+y) is positive or negative. But we can break this in 2:

(1). if (x+y) is +ve, is (y-x) > 20?
(2). if (x+y) is -ve, is (y-x) < 20?

Let's see the 1st option:

-y > x+5
or, x+y < -5
or, x+y is -ve
So, eq (2) needs to be proved. is (y-x) < 20? Can't say. This does not prove eq (2).

Let's see the 2nd option:

x > y
or, x-y > 0
or y-x < 0

So, irrespective of whether (x+y) is +ve or -ve, (y-x) is less than 0. So, the inequality can be answered by this statement alone. The answer should be B. I am not sure why the spoiler says C as the answer.

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Re: Is (y-10)^2 > (x+10)^2? [#permalink]

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27 Dec 2012, 06:27
anandrajakrishnan wrote:
we need to find if (y-10)^2 > (x+10)^2?
or, is y^2 - 20y + 100 > x^2 + 20x + 100?
or, is y^2 - 20y > x^2 + 20x?
or, is y^2 - x^2 > 20(x + y)?
or, is (y+x)(y-x) > 20(x+y)?
here, because of inequality, we cannot divide by (x+y) on both sides, as we don't know whether (x+y) is positive or negative. But we can break this in 2:

(1). if (x+y) is +ve, is (y-x) > 20?
(2). if (x+y) is -ve, is (y-x) < 20?

Let's see the 1st option:

-y > x+5
or, x+y < -5
or, x+y is -ve
So, eq (2) needs to be proved. is (y-x) < 20? Can't say. This does not prove eq (2).

Let's see the 2nd option:

x > y
or, x-y > 0
or y-x < 0

So, irrespective of whether (x+y) is +ve or -ve, (y-x) is less than 0. So, the inequality can be answered by this statement alone. The answer should be B. I am not sure why the spoiler says C as the answer.

hey Anand
when x > y we get y - x < 0
How can y - x < 0 by itself be sufficient to solve (y+x)(y-x) > 20(x+y) ?
I feel C is the correct answer .. we need both A and B to determine. Please clarify

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Re: Is (y-10)^2 > (x+10)^2? [#permalink]

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27 Dec 2012, 19:40
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anandrajakrishnan wrote:
we need to find if (y-10)^2 > (x+10)^2?
or, is y^2 - 20y + 100 > x^2 + 20x + 100?
or, is y^2 - 20y > x^2 + 20x?
or, is y^2 - x^2 > 20(x + y)?
or, is (y+x)(y-x) > 20(x+y)?
here, because of inequality, we cannot divide by (x+y) on both sides, as we don't know whether (x+y) is positive or negative. But we can break this in 2:

(1). if (x+y) is +ve, is (y-x) > 20?
(2). if (x+y) is -ve, is (y-x) < 20?

Let's see the 1st option:

-y > x+5
or, x+y < -5
or, x+y is -ve
So, eq (2) needs to be proved. is (y-x) < 20? Can't say. This does not prove eq (2).

Let's see the 2nd option:

x > y
or, x-y > 0
or y-x < 0

So, irrespective of whether (x+y) is +ve or -ve, (y-x) is less than 0. So, the inequality can be answered by this statement alone. The answer should be B. I am not sure why the spoiler says C as the answer.

hey Anand
when x > y we get y - x < 0
How can y - x < 0 by itself be sufficient to solve (y+x)(y-x) > 20(x+y) ?
I feel C is the correct answer .. we need both A and B to determine. Please clarify

Hmm. I need some retrospection here:

As I mentioned that the inequality can be divided in 2:

(1). if (x+y) is +ve, is (y-x) > 20?
(2). if (x+y) is -ve, is (y-x) < 20?

With the 2nd option, we get y-x < 0.
This doesn't resolve the inequality as we still need to prove which side of zero does (x+y) lies. So, 1st part is necessary.
Option C is the right choice. Thanks for clarifying.

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Re: Is (y-10)^2 > (x+10)^2? [#permalink]

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13 Jan 2013, 23:39
C.
Individually each one is insuffient by plugging values.

Now question stem:

Is (y-10)^2 > (x+10)^2?

= Is (y+x)(y-x-20) > 0 on simplification

1. gives (y+x)<-5 => (y+x) is negative.

Using 1, therefore the stem becomes Is (y-x-20) < 0 ?

or Is (y-x) < 20 ?

On using 2. if x >y => (y-x) is negative and hence (y-x) < 20 and the answer to the

stem is Yes. (combining 1. & 2.) Hence C.

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Re: Is (y-10)^2 > (x+10)^2? [#permalink]

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01 Jan 2014, 08:41
eaakbari wrote:
Karishma / Bunuel,

Could you throw some light on the below query.

I always have trouble in questions like this as to what numbers to pick. I end up going way beyond 3 minutes.

Do tell me the certain set of numbers/range of numbers that one should always begin testing with

P.S. I know it could vary from question to question but want a general set.

Karisha/Bunuel and other math experts

Would you please me so kind to advice on best approach for this question. I've been trying to grind it for several minutes but still unable to find an elegant way

Cheers!
J

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Re: Is (y-10)^2 > (x+10)^2? [#permalink]

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01 Jan 2014, 11:35
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JJ2014 wrote:
Is (y-10)^2 > (x+10)^2?

(1) -y > x+5
(2) x > y

Hi,

I answered using the following process:

If (y-10)^2 > (x+10)^2 is true, then the difference between the two Y and X needs to be over 10 (Example: Y=-6 and X=5)

With 1) We have the beginning of the information

If -y > x+5 then it will be true for some numbers BUT, if Y=-5 and X=-100) Then you will have (-15)^2 > (-90)^2 which is impossible. You need X>Y

With 2) Used alone, the statement is not correct. If Y=10 and X=10255 than the equation (y-10)^2 > (x+10)^2 is false.
if you select other numbers than the equation will be right.

Then you comfirm the statement one with statement two.

Hope it helps!
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Re: Is (y-10)^2 > (x+10)^2? [#permalink]

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02 Jan 2014, 01:34
JJ2014 wrote:
Is (y-10)^2 > (x+10)^2?

(1) -y > x+5
(2) x > y

Here is another method I could throw in....We can simplify the given equation and we get

y^2 -20y>x^2-20x ------->y^2-x^2 -20 (y-x)>0
Taking (y-x) common we get -------> y-x (y+x -20)>0 -----> Eq 1
Now the Eq 1 will hold true if we have one of the 2 conditions satisfied ie.

y>x and y+x>20----------->Condition 1
or
y<x and (y+x)< 20 ---------->Condition 2

Now from St 1 we have that x+y<-5 -----This meets partially our condition 2 but we don't know whether x>y ( so A and D Ruled out ) so statement alone is not sufficient

From St 2, we have that x>y------> This statement also partially meets our condition2 but we don't know whether y+x<20 (So B ruled out ) so statement alone is not sufficient

Combining both the equations we get x>y and x+y<20 (because x+y<-5 and hence less than 20) and hence Equation 1 is true.

Ans C
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Re: Is (y-10)^2 > (x+10)^2? [#permalink]

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02 Jan 2014, 10:30
JJ2014 wrote:
Is (y-10)^2 > (x+10)^2?

(1) -y > x+5
(2) x > y

here is my solution:

(1) Not sufficient - examples were mentioned above
(2) Not sufficient - y can be large negative value and x can be small positive value or y can be small positive/negative value and x can be large positive value

(1) + (2)

Inequalities with the same sign can be added

$$-y+x > x+5+y$$ implies $$y< -5$$

We got above $$y< -5$$ implies $$-y>5$$ Substitute this value in statement (1) we get $$x<=0$$

So ranges of x,y are $$y<-5$$ and $$y<x<=0$$

Now consider the question: Is $$(y-10)^2 > (x+10)^2$$?
y is always negative so adding a more negative value (-10) is more negative and squaring is a positive big value. On the RHS, x is negative and you are adding a positive value making the number (without sign) smaller. So square will surely be less than LHS. Even if we take highest value of x which is 0.... RHS will be less than LHS...

Hence (C)

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Re: Is (y-10)^2 > (x+10)^2? [#permalink]

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02 Jan 2014, 11:49
JJ2014 wrote:
Is (y-10)^2 > (x+10)^2?

(1) -y > x+5
(2) x > y

Statement I is insufficient
- y > x + 5
x = -3, y<-2

x y (y-10)^2 (x-10)^2 result (YES/NO)
-3 -3 169 169 NO
-3 -4 196 169 YES

Statement II is insufficient
x y (y-10)^2 (x-10)^2 result (YES/NO)
-3 -4 196 169 YES
4 3 36 49 NO

Combining is sufficient (Approach should be algebraic while combining)
x > y
-y>x+5
________(Adding both the inequalities we get)
x - y > x + y + 5
y < -2.5
-y>2.5
Maximum value for -y is 2.6

2.6 > or equal to x + 5
-2.4 > or equal to x OR -2.5 > x

Since we know both the numbers are negative and x is greater than y always hence the square value will always be greater for (y-10)^2.

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Re: Is (y-10)^2 > (x+10)^2? [#permalink]

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03 Jan 2014, 04:46
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Expert's post
JJ2014 wrote:
Is (y-10)^2 > (x+10)^2?

(1) -y > x+5
(2) x > y

Responding to a pm:

The algebraic solution to this has been provided by Bunuel here: is-y-10-2-x-163368.html#p1294316

You can use the concept of mods and visualize it on the number line too.

Is (y-10)^2 > (x+10)^2?
Is |y-10| > |x+10|?

Is distance of y from 10 greater than distance of x from -10?

(1) -y > x+5
x+y < -5

________-10________x____y__________0__________________10
Yes, distance of y from 10 is greater than distance of x from -10.

__x_________________________-10______________________0____________y______10
No, distance of y from 10 is not greater than distance of x from -10.

Not Sufficient.

(2) x > y
y is to the left of x on the number line.
________-10________y____x_______0___________________10
Yes, distance of y from 10 is greater than distance of x from -10.

________-10____________________0_________y___x______10
No, distance of y from 10 is not greater than distance of x from -10.

Not Sufficient.

Using Both, y will be to the left of x and y+x < -5

________-10________y____x_______0___________________10
Yes, distance of y from 10 is greater than distance of x from -10.

__y________-10___________________0_______x_____________10
Yes, distance of y from 10 is greater than distance of x from -10.

So in any case, distance of y from 10 will be greater than distance of x from -10. Sufficient

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Re: Is (y-10)^2 > (x+10)^2? [#permalink]

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09 Oct 2015, 07:37
PerfectScores wrote:
JJ2014 wrote:
Is (y-10)^2 > (x+10)^2?

(1) -y > x+5
(2) x > y

Statement I is insufficient
- y > x + 5
x = -3, y<-2

x y (y-10)^2 (x-10)^2 result (YES/NO)
-3 -3 169 169 NO
-3 -4 196 169 YES

Statement II is insufficient
x y (y-10)^2 (x-10)^2 result (YES/NO)
-3 -4 196 169 YES
4 3 36 49 NO

Combining is sufficient (Approach should be algebraic while combining)
x > y
-y>x+5
________(Adding both the inequalities we get)
x - y > x + y + 5
y < -2.5
-y>2.5
Maximum value for -y is 2.6

2.6 > or equal to x + 5
-2.4 > or equal to x OR -2.5 > x

Since we know both the numbers are negative and x is greater than y always hence the square value will always be greater for (y-10)^2.

Hi,
Seems like some typo error.

Thanks
Sumit kumar
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Re: Is (y-10)^2 > (x+10)^2? [#permalink]

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09 Oct 2015, 07:41
sumitsinha4u wrote:

Hi,
Seems like some typo error.

Thanks
Sumit kumar

The post you are quoting is quite old and thus you will not get a reply. Look at VeritasPrepKarishma 's solution at is-y-10-2-x-144191.html#p1312490
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Re: Is (y-10)^2 > (x+10)^2?   [#permalink] 09 Oct 2015, 07:41
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