Is (y−10)^2>(x+10)^2 ?
(1) −y>x+5
(2) x>y
OA: C
Source: Veritas
Is (y−10)^2>(x+10)^2? ----Veritas
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VERY nice problem. Please note we will use two different rephrasings (one for each (1) and (2) alone, the other for (1+2)):Mo2men wrote:Is (y−10)^2>(x+10)^2 ?
(1) −y>x+5
(2) x>y
Source: Veritas
$${\left( {y - 10} \right)^2}\,\,\mathop > \limits^? \,\,{\left( {x + 10} \right)^2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ \matrix{
\,\left| {y - 10} \right|\,\,\,\mathop > \limits^? \,\,\,\left| {x + 10} \right|\,\,\,\,\,\,\,\,\left( {\rm{I}} \right) \hfill \cr
\,\,\,{\rm{OR}} \hfill \cr
\,\left( {x + y} \right)\left( {y - x - 20} \right)\,\,\,\mathop > \limits^? \,\,\,0\,\,\,\,\,\,\,\,\left( {{\rm{II}}} \right)\,\,\,\,\,\left( * \right) \hfill \cr} \right.$$
$$\left( * \right)\,\,\,\left\{ \matrix{
{\left( {y - 10} \right)^2}\,\,\mathop > \limits^? \,\,{\left( {x + 10} \right)^2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{\left( {y - 10} \right)^2} - \,{\left( {x + 10} \right)^2}\,\,\mathop > \limits^? \,\,\,0 \hfill \cr
\,{\left( {y - 10} \right)^2} - \,{\left( {x + 10} \right)^2} = \left[ {\left( {y - 10} \right) + \left( {x + 10} \right)} \right]\left[ {\left( {y - 10} \right) - \left( {x + 10} \right)} \right] = \left( {x + y} \right)\left( {y - x - 20} \right)\,\,\,\mathop > \limits^? \,\,\,0 \hfill \cr} \right.$$
$$\left( 1 \right)\,\,\,x + y < - 5\,\,\,\,\left( {\rm{I}} \right)\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {0, - 6} \right)\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( { - 16,10} \right)\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.$$
$$\left( 2 \right)\,\,x > y\,\,\,\,\left( {\rm{I}} \right)\,\,\,\,\,\,\left\{ \matrix{
\,\left( {{\mathop{\rm Re}\nolimits} } \right){\rm{Take}}\,\,\left( {x,y} \right) = \left( {0, - 6} \right)\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {1,0} \right)\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.$$
$$\left( {1 + 2} \right)\,\,\,\,\,\,\left( {{\rm{II}}} \right)\,\,\,\left\{ \matrix{
\,x + y < 0\,\,\,{\rm{by}}\,\,\,\left( 1 \right) \hfill \cr
\,y - x - 20 = \left( {y - x} \right) - 20\mathop < \limits^{{\rm{by}}\,\,\left( 2 \right)} - 20\, < 0 \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {\rm{C}} \right)\,\,\,\,$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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x² < y² can be rephrased as follows:Mo2men wrote:Is (y−10)^2>(x+10)^2 ?
(1) −y>x+5
(2) x>y
|x| < |y|
Thus, the question stem can be rephrased as follows:
Is |x+10| < |y-10|?
|a-b| = the distance between a and b
|a+b| = |a-(-b)| = the distance between a and -b
Thus, the question stem can be further rephrased as follows:
Is the distance between x and -10 less than the distance between y and 10?
Statement 1, rephrased: x+y < -5
Case 1: y=-20 and x=-10
y= -20..........x=-10..........0..........10
In this case, the distance between x and -10 is less than the distance between y and 10, so the answer to the rephrased question stem is YES.
Case 2: x=-20 and y=10
x=-20..........-10..........0.........y=10
In this case, the distance between x and -10 is NOT less than the distance between y and 10, so the answer to the rephrased question stem is NO.
Since the answer is YES in Case 1 but NO in Case 2, INSUFFICIENT.
Statement 2: y < x
Case 1 also satisfies Statement 2,
In Case 1, the answer to the rephrased question stem is YES.
Case 3: y=10 and x=20
-10..........0..........y=10..........x=20
In this case, the distance between x and -10 is NOT less than the distance between y and 10, so the answer to the rephrased question stem is NO.
Since the answer is YES in Case 1 but NO in Case 3, INSUFFICIENT.
Statements combined:
Adding together x+y < -5 and y < x, we get:
x+y+y < x-5
2y < -5
y < -2.5
Case 4: y=-3 and x=-2.5, with the result that x+y < -5 and y < x
-10........y=-3.......x=-2.5......0..........10
In this case, the distance between x and -10 is less than the distance between y and 10, so the answer to the rephrased question stem is YES.
Case 5: y=-10 and x=4, with the result that x+y < -5 and y < x
y=-10............0...........x=4............10
In this case, the distance between x and -10 is less than the distance between y and 10, so the answer to the rephrased question stem is YES.
The cases above illustrate that -- when the two statements are combined -- the distance between x and -10 must be less than the distance between y and 10.
Thus, the answer to the rephrased question stem is YES.
SUFFICIENT.
The correct answer is C.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
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