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

Great explanation once again. Thank you

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.

Answer C

Hope it helps!

Responding to a pm:

The algebraic solution to this has been provided by Bunuel here: https://gmatclub.com/forum/is-y-10-2-x- ... l#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

Answer (C)

Question: \(y^2-20y+10 > x^2+20x+100\)--> \((y-x)(y+x) > 20(y+x)\)?
(1) \(x+y < -5\) means we can divide the expression above while changing the sign --> \(y-x < 20\), we cannot go any further here. Not sufficient

(2) We have too many combinations considering this information. Just insert some values in the first expression \((y - 10)^2 > (x + 10)^2\)
\(x=1, y=-90\): \((-100)^2 > (1+10)^2?\) YES
\(x=1, y=2\): \((2-10)^2 > (1+10)^2\)? No, hence not sufficient

(1) + (2)
From (1) \(y-x < 20\) and from (2) \(y-x < 0\)the answer is YES as, if \(y-x\) is negative, then it diffenetelly will be less than 20.
Answer C

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution

Is (y−10)^2>(x+10)^2

(1) −y>x+5
(2) x>y

If we modify the original condition and the question, the question, (y-10)^2>(x+10)^2?, can be changed to (y-10)^2-(x+10)^2>0?. Then, if we further modify, (y-10-x-10)(y-10+x+10)>0? Thus, the question is essentially asking if (y-x-20)(y+x)>0?
There are 2 variables (x and y) in the original condition. In order to match the number of variables and the number of equations, we need 2 equations. Using both the condition 1) and 2), we get y-x<0. Then, y-x-20<0 and x+y<-5<0. So, we get (y-x-20)(y+x)>0 and the answer is always ‘yes’. The condition, thus, is sufficient, and the correct answer is C.

For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

VERY nice problem. Please note that we will use two different rephrasings (one for each (1) and (2) alone, the other for (1+2)):

\({\left( {y - 10} \right)^2}\,\,\mathop > \limits^? \,\,{\left( {x + 10} \right)^2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ \matrix{\\
\,\left| {y - 10} \right|\,\,\,\mathop > \limits^? \,\,\,\left| {x + 10} \right|\,\,\,\,\,\,\,\,\left( {\rm{I}} \right) \hfill \cr \\
\,\,\,{\rm{OR}} \hfill \cr \\
\,\left( {x + y} \right)\left( {y - x - 20} \right)\,\,\,\mathop > \limits^? \,\,\,0\,\,\,\,\,\,\,\,\left( {{\rm{II}}} \right)\,\,\,\,\,\left( * \right) \hfill \cr} \right.\)

\(\left( * \right)\,\,\,\left\{ \matrix{\\
{\left( {y - 10} \right)^2}\,\,\mathop > \limits^? \,\,{\left( {x + 10} \right)^2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{\left( {y - 10} \right)^2} - \,{\left( {x + 10} \right)^2}\,\,\mathop > \limits^? \,\,\,0 \hfill \cr \\
\,{\left( {y - 10} \right)^2} - \,{\left( {x + 10} \right)^2} = \left[ {\left( {y - 10} \right) + \left( {x + 10} \right)} \right]\left[ {\left( {y - 10} \right) - \left( {x + 10} \right)} \right] = \left( {x + y} \right)\left( {y - x - 20} \right)\,\,\,\mathop > \limits^? \,\,\,0 \hfill \cr} \right.\)


\(\left( 1 \right)\,\,\,x + y < - 5\,\,\,\,\left( {\rm{I}} \right)\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {0, - 6} \right)\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr \\
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( { - 16,10} \right)\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.\)

\(\left( 2 \right)\,\,x > y\,\,\,\,\left( {\rm{I}} \right)\,\,\,\,\,\,\left\{ \matrix{\\
\,\left( {{\mathop{\rm Re}\nolimits} } \right){\rm{Take}}\,\,\left( {x,y} \right) = \left( {0, - 6} \right)\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr \\
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {1,0} \right)\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.\)

\(\left( {1 + 2} \right)\,\,\,\,\,\,\left( {{\rm{II}}} \right)\,\,\,\left\{ \matrix{\\
\,x + y < 0\,\,\,{\rm{by}}\,\,\,\left( 1 \right) \hfill \cr \\
\,y - x - 20 = \left( {y - x} \right) - 20\mathop < \limits^{{\rm{by}}\,\,\left( 2 \right)} - 20\, < 0 \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {\rm{C}} \right)\,\,\,\,\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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