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A quicker way to solve this problem?

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A quicker way to solve this problem?

by shoot4greatness » Wed Jun 08, 2011 7:52 pm
This one is from Kaplan CAT (I find it increasingly that most math questions from Kaplan take more than 2 minutes to solve!)

if x + y^2 = ( x + y^2)^2 , what is the value of y?

1. x=y^2
2. xy^2=0 (only y is squared, not xy)

OA c
Last edited by shoot4greatness on Wed Jun 08, 2011 7:59 pm, edited 1 time in total.
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by shoot4greatness » Wed Jun 08, 2011 7:57 pm
1) x and y are either 1 or 0 (only two integers that satisfy the equation)

2) x or y is 0

1 and 2) x and y are 0

Is my reasoning correct?
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by phanideepak » Wed Jun 08, 2011 8:00 pm
Hey didn't u expand the eqn and then substitute it? It was taking under 2 mins for me..
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by phanideepak » Wed Jun 08, 2011 8:02 pm
shoot4greatness wrote:1) x and y are either 1 or 0 (only two integers that satisfy the equation)

2) x or y is 0

1 and 2) x and y are 0

Is my reasoning correct?
Except that y can be 1,-1
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by manpsingh87 » Wed Jun 08, 2011 8:06 pm
shoot4greatness wrote:This one is from Kaplan CAT (I find it increasingly that most math questions from Kaplan takes more than 2 minutes to solve!)

if x + y^2 = ( x + y^2)^2 , what is the value of y?

1. x=y^2
2. xy^2=0 (only y is squared, not xy)

OA c
x + y^2 = ( x + y^2)^2----1)

1) x=y^2;
put x=y^2 in 1) we have;
y^2 + y^2 = ( y^2 + y^2)^2;
2y^2=4y^4;
2y^2(2y^2-1)=0;
so either 2y^2=0; or (sqrt(2)y)^2-1=0;
therefore y=0; or y=1/sqrt(2); or y=-1/sqrt(2);
as different values are possible for y hence 1 alone is not sufficient to answer the question.

2) xy^2=0;
either x=0; or y^2=0; i.e. y=0;
putting x=0 in 1) we have;
0 + y^2 = ( 0 + y^2)^2;
y^2=y^4;
y^2(y^2-1);
so y= -1,0,1;

hence 2 alone is also not sufficient to answer the question.

combining 1 and 2 we have only 1 possible common solution i.e. y=0; hence answer should be C
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by cans » Wed Jun 08, 2011 8:29 pm
x + y^2 = ( x + y^2)^2
(x+ y^2)(x+ y^2 -1)=0
thus either x=-y^2 --- eqn1 or x+y^2=1 --- eqn2
a)x=y^2
thus y^2=0 or y=0 (from eqn1)
2*y^2=1 -> y=+-root(o.5) (from eqn2)
insufficient
b)x*y^2=0
either y^2=0 ->y=0
or x=0 then from eqn1, y=0
then form eqn2. y=+-1
Insufficient
a&b) y=0 is the only common value.
Sufficient
IMO C
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by bubbliiiiiiii » Thu Jun 09, 2011 12:26 am
shoot4greatness wrote:This one is from Kaplan CAT (I find it increasingly that most math questions from Kaplan take more than 2 minutes to solve!)

if x + y^2 = ( x + y^2)^2 , what is the value of y?

1. x=y^2
2. xy^2=0 (only y is squared, not xy)

OA c
From the question, x + y^2 = ( x + y^2)^2, is only possible when x + y^2 is either 0 or 1.

Thus, we have to find if x + y^2 is either 0 or 1.

(i)x=y^2

x + y^2 = ( x + y^2)^2
2x = (2x)^4 => no information on X, insufficient.

(ii) xy^2=0 (only y is squared, not xy)

either x is 0 or y^2 is 0 => we have information only about one variable and know nothing about the other, Thus, insufficient.

Combined,

From (i) either x can be 0 0r 1
From (ii) either x is 0 or y^2 is 0

we can conclude that x is 0.

Thus Sufficient and answer C.
Regards,

Pranay
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by shoot4greatness » Thu Jun 09, 2011 11:50 am
phanideepak wrote:
shoot4greatness wrote:1) x and y are either 1 or 0 (only two integers that satisfy the equation)

2) x or y is 0

1 and 2) x and y are 0

Is my reasoning correct?
Except that y can be 1,-1
nice one phanideepak, forgot about the -1. Yes I didn't bother to expand it. Just seeing the two prompts, I knew than the only possibilities for x and y were 0,1, (and -1). Combining the two, y must be 0. The possibility of x and y being 1 went out with second prompt and 0=0^2.

BTW, this question was one of a few questions I found in Kaplan where applied math logic worked. I find Powerprep less hard because I can actually use applied math to quickly move on to next questions, whereas Kaplan problems actually require me to solve the question.
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