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Quote:

What is the value of 2x/3y ?

(1) x^2/y^2 = 36/25
(2) x^5/y^5 > 1

Sol:

St 1 says x^2/y^2 = 36/25 or we have x/y=6/5 or -6/-5 So 2x/3y value will be same So St1 is sufficient

St 2 gives you ratio of x>y and there are many numbers possible for ex x=2 and y=1 or x=1000, y=2...Value of 2x/3y will change.

Ans A


Oops...I missed the case where x=6,y=-5 or x=-6 and y=5.. :beat
St 1 is not sufficient

St 2 is important here

we have that \((x^5-y^5)/y^5 >0\)

so we have 2 cases

x^5-y^5 >0 and y^5>0 Both x and y have same sign

or

x^5-y^5<0 and y^5<0... y is negative and x<y so x is negative...

so we have that x and y have same sign..

Combining we see that x and y have same sign so we can find unique value of 2x/3y

Ans C
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Bunuel:

Still not able to get , how can |x|/|y| = -6/-5,
as sqrt36=sqrt6^2=|6|.
Right??
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Bunuel:

Still not able to get , how can |x|/|y| = -6/-5,
as sqrt36=sqrt6^2=|6|.
Right??

If you are unclear about the text in red above, look at it this way,

Given, \(\frac{x^2}{y^2}\) = \(\frac{36}{25}\) --->> \(\frac{x}{y} = \pm \frac{6}{5}\)

Thus, x =6k, y =5k or x=-6k, y = 5k , giving you 2 different answers for 2x/3y. Hence this statement is not sufficient.

Hope this helps.
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Engr2012 :

Here we are taking square roots of 36 and 25 right??
Here it is written in GMAT square roots always refer to positive roots.
square-root-always-positive-114114.html
VeritasPrepKarishma

Am i missing anything??


Here
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Engr2012 :

Here we are taking square roots of 36 and 25 right??
Here it is written in GMAT square roots always refer to positive roots.
square-root-always-positive-114114.html
VeritasPrepKarishma

Am i missing anything??


Here

Yes, what you are missing is the how you define your equation.

When you write x^2/y^2 = 36/25 , this equation is of degree 2 in (x/y) and thus you should get 2 solutions (it is a maths rule that you will get n solutions for nth degree equation, be it real or non real solutions!)

But if you write the above equation as , (x/y) = \(\sqrt\frac{36}{25}\)] ---> this equation is 1 degree in (x/y) and hence can only have 1 solution.

That is what VeritasPrepKarishma has written at the link posted by you.

Hope this clears your doubt.
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Engr2012 :

Here we are taking square roots of 36 and 25 right??
Here it is written in GMAT square roots always refer to positive roots.
square-root-always-positive-114114.html
VeritasPrepKarishma

Am i missing anything??


Here

When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice though, that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).
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We need to find 2x:3y = ?

(1) x^2 : Y^2 = 36:25
Now, x can be ±6 and y can be ±5;

If x and y are of opposite sign, 2x/3y = -12/15;
But if x and y are of the same sign, then 2x/3y = 12/15; Hence, Insufficient.

(2) x^5/y^5 > 1;
This can be true if x and y both are positive or negative. So, we can say both are of same sign.
But we don't have any values for x and y. Hence, Insufficient.

(1) and (2) together,
x = ±6 and y = ±5 and both are of same sign (from point 2).
Hence, 2x/3y = 12/15; Sufficient.

Answer C.
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Bunuel

Tough and Tricky questions: Algebra.



What is the value of 2x/3y ?

(1) x^2/y^2 = 36/25
(2) x^5/y^5 > 1

\(\frac{2}{3}*\frac{x}{y}=?\)

(1) \((\frac{x}{y})^2=(\frac{6}{5})^2\)
\(\frac{x}{y}=±\frac{6}{5}\) => INSUFFICIENT

(2) \((\frac{x}{y})^5>1\)
\(\frac{x}{y}>1\) => INSUFFICIENT

1 & 2 => \(\frac{x}{y}=\frac{6}{5}\)
SUFFICIENT.

C.
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Bunuel

Tough and Tricky questions: Algebra.



What is the value of 2x/3y ?

(1) x^2/y^2 = 36/25
(2) x^5/y^5 > 1

St 1

X could negative and y could be positive- because either possibility would still give you 36/25

St 2

x and y must be positive but no info about the actual values

St 1 and st 2

eliminates possibility of x and y being positive and negative- must be positive

suff

C
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Bunuel

Tough and Tricky questions: Algebra.



What is the value of 2x/3y ?

(1) x^2/y^2 = 36/25
(2) x^5/y^5 > 1

What is the ratio of 2x to 3y?

(1) The ratio of x^2 to y^2 is equal to 36/25.

(2) The ratio of x^5 to y^5 is greater than 1.
.

(1) x^2/y^2 = 36/25 => deducing this we can get x/y values as +6/+5 , -6/-5, +6/-5, -6/+5 . Square any of these 4 and you end up with 36/25
(2) x^5/y^5 > 1 => deducing this we get x/y > 1 or -x/-y > 1 essentially saying the same thing

if you look at (1) ALONE, 2x/3y can yield multiple values .x/y can be +(6/5) or -(6/5) . So no definite answer A, D are out

if you look at (2) ALONE, x/y > 1 , 2x/3y can yield multiple values as x and y can take any number as long as x/y >1 . So no definite answer . Therefore B is out

we are now left with C and E

Let's analyze C (COMBINING 1 & 2)

x/y can only take +6/+5 >1 or -6/-5 >1

so 2x/3y = (2/3) X (x/y) will only give one value. So definite answer . C is the correct answer
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