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If x^3 = 25, y^4 = 64, and z^5 = 216, and xy > 0, which of
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Updated on: 16 Apr 2013, 06:11
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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
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Originally posted by emmak on 16 Apr 2013, 06:07.
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
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16 Apr 2013, 06:53
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
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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
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20 Jun 2014, 01:43
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
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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
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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
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16 Mar 2018, 10:57
Hi All, This is something of a convoluted, layered "math" question, and you're not likely to see it on Test Day. It can be solved with comparative math though  instead of calculating the exact values of X, Y and Z, you can deduce which is bigger or smaller by pattern comparison. First, let's do a quick estimation… 3^3 = 27 3^4 = 81 3^5 = 243 X^3 = 25, so X is a little less than 3 Y^4 = 64, so Y is a little less than 3 (or a little bigger than 3) Z^5 = 216, so Z is a little less than 3 We're told that XY > 0, so we're forced to consider only the positive value of Y. X, Y and Z are all pretty close to one another, so we have to look for something that will differentiate them (and help us to figure out which is bigger when we look at any 2 of them) X^3 = (X)(X)(X) = 25 Y^4 = (Y)(Y)(Y)(Y) = 64 Since the values of X and Y are pretty close, multiplying by the "extra" Y is what turns 25 into 64….. 64/25 = about 2.5 This does NOT mean that Y = 2.5, but it DOES mean that Y MUST be farther away from 3 than X is. So X > Y Y^4 = (Y)(Y)(Y)(Y) = 64 Z^5 = (Z)(Z)(Z)(Z)(Z) = 216 The values of Y and Z are also pretty close, so multiplying by the "extra" Z is what turns 64 into 216…. 216/64 = more than 3 This does NOT mean that Z is greater than 3, but it DOES mean that Z MUST be closer to 3 than Y is. So Z > Y From here, the answer choices provide us with a great way "out" of this question. Since Y is smaller than both X and Z, the only answer that makes sense is…. GMAT assassins aren't born, they're made, Rich
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Re: If x^3 = 25, y^4 = 64, and z^5 = 216, and xy > 0, which of
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16 Mar 2018, 10:57






