Data Sufficiency

If x^3 - x^2 = 2 x and y = ¼ x^2, then what is the value of y?
I. x is a positive integer.
II. x and y are integers.



[spoiler]made up by Sanjeev K Saxena for Avenues Abroad[/spoiler]

sanju09 wrote:If x^3 - x^2 = 2 x and y = ¼ x^2, then what is the value of y?
I. x is a positive integer.
II. x and y are integers.



[spoiler]made up by Sanjeev K Saxena for Avenues Abroad[/spoiler]

it's d

question stem is simplified as x^2-x-2=0 or (x-2)(x+1)=0 with roots x=(2;-1); y=(x/2)^2 , find y?
st(1) Sufficient - we get one unique value for y, as x=2
st(2) Sufficient - we get one unique value for y, as x=-1 returns non-integer

[spoiler]?[/spoiler]

Your explanation is baffling for me.
values of x=2 and -1
Substituting same to get value of y (as asked in the question )
y=1/4 x^2
STATEMENT 1:x is positive .implies value of x would be 2
so y=1/4 2^2=>1
i.e y=1
SUFFICIENT
Statement 2:x and y are integers .
if x=2 y=>1
if x=-1 y=>1/4 not an integer

here its mentioned integers(not mentioned as negative or positive)
INSUFFICIENT

My ans :A

Didn't understood your explanation for statement 2 where you have mentioned x=-1 returns non integer.
Dhiren
[/quote]

dhiren8182 wrote: Didn't understood your explanation for statement 2 where you have mentioned x=-1 returns non integer.
Dhiren

[/quote]
when x=-1, y is non integer.
Are you sure your reasoning is correct? The question stem supplies two possibilities for x and we need additional information to determine the one (unique) value of y. The question is asking: What IS the value? . We need one value for y.

statement(2) x and y are integers. must be Sufficient

dhiren8182 wrote: Statement 2:x and y are integers .
if x=2 y=>1
if x=-1 y=>1/4 not an integer

here its mentioned integers(not mentioned as negative or positive)
INSUFFICIENT

By knowing that y is integer (regardless positive or negative) you decide about sufficiency here, as another value of y is non integer (that's what I meant)

[spoiler]this is very easy question.
since you hit the permutation of 'experts@BTG', i guess you've just over-analyzed[/spoiler]

Got it...
Silly mistake...
Dhiren

According to me the answer is E. After applying each of the statements also, I am not getting one specific value of y.

Hence E...


OA please...

pemdas wrote:

sanju09 wrote:If x^3 - x^2 = 2 x and y = ¼ x^2, then what is the value of y?
I. x is a positive integer.
II. x and y are integers.



[spoiler]made up by Sanjeev K Saxena for Avenues Abroad[/spoiler]

it's d

question stem is simplified as x^2-x-2=0 or (x-2)(x+1)=0 with roots x=(2;-1); y=(x/2)^2 , find y?
st(1) Sufficient - we get one unique value for y, as x=2
st(2) Sufficient - we get one unique value for y, as x=-1 returns non-integer

[spoiler]?[/spoiler]

If x^3 - x^2 = 2 x then x = 0, -1, and 2, hence y = 0, ¼, and 1.

I. If x is a positive integer then it must be 2, hence y = 1. Sufficient

II. We don't need to know again that x is an integer. What's important to note here is that y is an integer, hence y = 0, 1; two values! Insufficient



[spoiler]OA A[/spoiler]

If x^3 - x^2 = 2 x and y = ¼ x^2, then what is the value of y?
I. x is a positive integer.
II. x and y are integers.


after solving this sum very minutely, I got the answer as D because I really did not get x=0 and y=0 possibility...

Could sanju09 please explain how did it strike to you that it is also the possibility as i think we all missed out on it...

after getting the quadratic equation no one thought of going back to the original equation and putting the value as 0.

what is it that struck you for doing so... please explain and share it with us...

The answer is (A)

Statement 2 is not SUFFICIENT

We get two sets of values
(0,0) and (2,1)

[email protected] wrote:If x^3 - x^2 = 2 x and y = ¼ x^2, then what is the value of y?
I. x is a positive integer.
II. x and y are integers.


after solving this sum very minutely, I got the answer as D because I really did not get x=0 and y=0 possibility...

Could sanju09 please explain how did it strike to you that it is also the possibility as i think we all missed out on it...

after getting the quadratic equation no one thought of going back to the original equation and putting the value as 0.

what is it that struck you for doing so... please explain and share it with us...

If x^3 - x^2 = 2 x, then

x^3 - x^2 - 2 x = 0, (we can divide both sides by x, only once we're sure that x cannot be zero, otherwise that would be an undefined state to go any further),or

x (x^2 - x - 2) = 0, or x (x - 2) (x + 1) = 0, or

x = 0, -1, 2

and the rest is same as already stated.

sanju09 wrote:If x^3 - x^2 = 2 x and y = ¼ x^2, then what is the value of y?
I. x is a positive integer.
II. x and y are integers.


[spoiler]made up by Sanjeev K Saxena for Avenues Abroad[/spoiler]

hi all,

i understood the fact that A is sufficient since "2"is the only solution. my doubt is the question stem uses the word " IF" then does that mean we need/ to consider or try solving the given fact with other values also?

i did that way using 3 as option then obviously stmnt 1 is incorrect

thanks

pappueshwar wrote:

sanju09 wrote:If x^3 - x^2 = 2 x and y = ¼ x^2, then what is the value of y?
I. x is a positive integer.
II. x and y are integers.


[spoiler]made up by Sanjeev K Saxena for Avenues Abroad[/spoiler]

hi all,

i understood the fact that A is sufficient since "2"is the only solution. my doubt is the question stem uses the word " IF" then does that mean we need/ to consider or try solving the given fact with other values also?

i did that way using 3 as option then obviously stmnt 1 is incorrect

thanks

How is IF posing doubt here, pappueshwar?