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If x^3 = 25, y^4 = 64, and z^5 = 216, and xy > 0, which of

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If x^3 = 25, y^4 = 64, and z^5 = 216, and xy > 0, which of [#permalink]

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If x^3 = 25, y^4 = 64, and z^5 = 216, and xy > 0, which of the following is true?

(A) x > y > z
(B) y > x > z
(C) y > z > x
(D) z > y > x
(E) z > x > y
[Reveal] Spoiler: OA

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Last edited by Bunuel on 16 Apr 2013, 06:11, edited 1 time in total.
Edited the question.

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Re: If x^3 = 25, y^4 = 64, and z^5 = 216, and xy > 0, which of [#permalink]

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New post 16 Apr 2013, 06:53
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emmak wrote:
If x^3 = 25, y^4 = 64, and z^5 = 216, and xy > 0, which of the following is true?

(A) x > y > z
(B) y > x > z
(C) y > z > x
(D) z > y > x
(E) z > x > y


Given that x = \(5^{2/3}, y = 2^{3/2}.\) As xy>0, they both have the same sign and also as odd power of x is positive, thus x>0--> Both x,y>0.'

Now lets assume that x>y, thus as both the sides are positive, we can safely raise to the power of 6. Thus, we assume that \(5^{4}>2^{9}\)

or 625>512; this is infact correct. Thus eliminate all the options wherein x<y. Thus we are left with A and E.

Now \(z = 2^{3/5}*3^{3/5}\). We again start with the assumption that z>y[We take y and z as there is a common factor of 2]. Again, as the odd power of z is positive, thus z is also positive. Raising the inequality z>y to the 10th power, we have \(2^{6}*3^{6}>2^{15}\)

or \(3^6>2^9\)

or 729>512, which is again correct. Thus the order is : z>x>y

E.
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Re: If x^3 = 25, y^4 = 64, and z^5 = 216, and xy > 0, which of [#permalink]

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New post 27 Dec 2013, 03:09
emmak wrote:
If x^3 = 25, y^4 = 64, and z^5 = 216, and xy > 0, which of the following is true?

(A) x > y > z
(B) y > x > z
(C) y > z > x
(D) z > y > x
(E) z > x > y


Mau5 method is better. I have a little more crued way but it gets you by...

here it

We know x= Cuberoot 25 (or 5^2)
y= 4th root of 64 (or 2^6) and
z= 5th root of 216 (or 2^3 *3^3)

Note that x>0 so y also is greater then 0 (because xy>0)
Consider x and y at first,
y can be written as 2*4th root of 4 and 4th root of 1 is 1 so 4th root of 4 will be ~ close to 1 so value of y will be close to 2*(Something close to 1) = 2

For x we have Cube root of 25...if we divide and multiply by 5 so x = 5/ (cuberoot of 5)....Now Cube root of 1 is 1 so cube root of 5 will be value less than 2 but more than 1.....less than 2 because cube root of 8 is 2 and hence possible values x will be between 5 and 2.5 but more close to 2.5

So x>y


Coming to z ie 5th root of 216....Can be written as 5th root of (2^3*3^3). Now multiply and divide by 36(2^2*3^2) in the numerator and denominator, we get 6/ (5th root of 36).....Now 2^5 = 32 or 2 = 5th root of 32 and 3^5 =243 or 3=5th root of 216......so 5th root of 216 will be between 2 and 3 and hence value of z will be close to 2.(Something) and therefore z little less than 3


Therefore z>x>y
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Re: If x^3 = 25, y^4 = 64, and z^5 = 216, and xy > 0, which of [#permalink]

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New post 20 Jun 2014, 01:43
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I approached this ques with approximation .. though I like mau's above mentioned method :
x cube = 25 this means x<3 bcoz cube 3 = 27
y raise to power 4 = 64 will result in 2.8 or -2.8 since its given x and y has similar signs hence consider only 2.8
z raise to power 5 = 216 which is cube 6 ... 3*3*3*4*2 which will be closer to 3 but greater than x. reason being the 4*2 =8 just one digit less to be perfect root 5, but 25 is 2 digits less than perfect cube root.

hence the desired order will be
z>x>y

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If x^3 = 25, y^4 = 64, and z^5 = 216, and xy > 0, which of [#permalink]

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New post 12 Jul 2015, 10:32
emmak wrote:
If x^3 = 25, y^4 = 64, and z^5 = 216, and xy > 0, which of the following is true?

(A) x > y > z
(B) y > x > z
(C) y > z > x
(D) z > y > x
(E) z > x > y


\(3^{3}=27\) --> x is approaching 3
\(8^{2}=4^{3}=2^{4}=64\) --> \(y =2\)
\(3^{5}=243\) ---> z is approaching 3

--> y is the smallest number. Among the options, E offers y as the smallest number. --> Answer: E

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Re: If x^3 = 25, y^4 = 64, and z^5 = 216, and xy > 0, which of [#permalink]

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New post 06 Dec 2017, 16:39
Was really flummoxed for the first 30 seconds.
Then:
X, y we know will be fractions,though positive. And, x>y (root x = basically more than 3, y is 2)
z is also positive, but with highest value --> z^5 = 216 --> but root z value is approaching 6, greater than 5)

Hence E --> z> x> y

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Re: If x^3 = 25, y^4 = 64, and z^5 = 216, and xy > 0, which of   [#permalink] 06 Dec 2017, 16:39
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