Quadratic

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Quadratic

by vscid » Sat Feb 20, 2010 3:47 pm
If x2/9 - 4/y2 = 12, what is the value of x?

(1) x/3 + 2/y = 6

(2) x/3 - 2/y = 2
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by onedayi'll » Sat Feb 20, 2010 3:58 pm
IMO C
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by thephoenix » Sat Feb 20, 2010 6:14 pm
vscid wrote:If x2/9 - 4/y2 = 12, what is the value of x?

(1) x/3 + 2/y = 6

(2) x/3 - 2/y = 2

s1)squaring bth side=x^2/9+4/y^2+x^2/18y=36
but we dnt know x andy
insuff
s2) same as s1)
insuff

combining
multiplying bth statements
x^2/9-4/y^2=12 same as the eqn in q
hence two variables and one eqn----> no solution
hence insuff

IMO E

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by shashank.ism » Sat Feb 20, 2010 8:38 pm
vscid wrote:If x2/9 - 4/y2 = 12, what is the value of x?

(1) x/3 + 2/y = 6

(2) x/3 - 2/y = 2
x^2/9 - 4/y^ 2 = 12 --> (x/3 )^2- (2/y)^2 = 12 -->(x/3 + 2/y)(x/3 - 2/y) = 12

s.(1) x/3 + 2/y = 6 ---> (x/3 - 2/y) = 12/6 =2 two eq we can solve it ===== sufficient.
s.(2) x/3 - 2/y = 2 ---> (x/3 + 2/y) = 12/2 =6 two eq we can solve it ===== sufficient.

Ans D
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by shashank.ism » Sat Feb 20, 2010 8:40 pm
thephoenix wrote:
vscid wrote:If x2/9 - 4/y2 = 12, what is the value of x?

(1) x/3 + 2/y = 6

(2) x/3 - 2/y = 2

s1)squaring bth side=x^2/9+4/y^2+x^2/18y=36
but we dnt know x andy
insuff
s2) same as s1)
insuff

combining
multiplying bth statements
x^2/9-4/y^2=12 same as the eqn in q
hence two variables and one eqn----> no solution
hence insuff

IMO E
Phoenix I think u did a mistake..In this type of problem don't go on with squaring like
x/3 + 2/y = 6
x^2/9+4/y^2+x^2/18y=36

it will lead you in confusion and u will not be able to solve the problem..

Rather try to divide the given expression so thjat u can get simpler terms
x^2/9 - 4/y^ 2 = 12 --> (x/3 )^2- (2/y)^2 = 12 -->(x/3 + 2/y)(x/3 - 2/y) = 12

In this way u can easily reach to the Answer ...
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by thephoenix » Sat Feb 20, 2010 8:44 pm
shashank.ism wrote:
thephoenix wrote:
vscid wrote:If x2/9 - 4/y2 = 12, what is the value of x?

(1) x/3 + 2/y = 6

(2) x/3 - 2/y = 2

s1)squaring bth side=x^2/9+4/y^2+x^2/18y=36
but we dnt know x andy
insuff
s2) same as s1)
insuff

combining
multiplying bth statements
x^2/9-4/y^2=12 same as the eqn in q
hence two variables and one eqn----> no solution
hence insuff

IMO E
Phoenix I think u did a mistake..In this type of problem don't go on with squaring like
x/3 + 2/y = 6
x^2/9+4/y^2+x^2/18y=36

it will lead you in confusion and u will not be able to solve the problem..

Rather try to divide the given expression so thjat u can get simpler terms
x^2/9 - 4/y^ 2 = 12 --> (x/3 )^2- (2/y)^2 = 12 -->(x/3 + 2/y)(x/3 - 2/y) = 12

In this way u can easily reach to the Answer ...
thanks dost

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by shashank.ism » Sat Feb 20, 2010 8:45 pm
thephoenix wrote:
thanks dost
anytime phoenix....u could have thanked me through thanks button...anyway its fine.. :D
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by vscid » Mon Feb 22, 2010 2:09 pm
OA is D.
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by lunarpower » Tue Feb 23, 2010 6:21 am
there are some wrong answers in this thread.

you should be able to recognize the difference of squares pattern INSTANTLY. if you don't recognize this pattern in a few seconds, you should put it on a flash card, and drill it like crazy.

the expression x^2/9 - 4/y^2 is a difference-of-squares. you should factor it before you even turn on your brain!
so:
(x/3 - 2/y)(x/3 + 2/y) = 12.

if you have statement (1), you can DIVIDE this equation by statement (1), giving ... statement (2)!
if you have statement (2), you can DIVIDE this equation by statement (2), giving ... statement (1)!
so, basically, if you have either of the statements, you have them both.

if you have both statements, you can just line them up and add them, giving 2x/3 = 8. (the y's cancel.)
that's good enough, so the answer is (d).
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by sairamGmat » Thu Feb 25, 2010 8:19 am
D is the answer. The question is in the form of a^2-b^2 = <some constant> . so, it can be rewritten as (a+b) (a-b)

In both options, one of these , viz (a+b) and (a-b) are given. So, we can calculate the other word and hence a or b by using either of the options..

so, Answer is D :)