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

Question

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 · Accepted Answer

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 . 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.

WoundedTiger · Answer

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

GmatDestroyer2013 · Answer

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

Madhavi1990 · Answer

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

EMPOWERgmatRichC · Answer

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…. E

GMAT assassins aren't born, they're made, 
Rich

Legallyblond · Answer

@ Bunuel can we solve this thinking that all are almost near to the 3 raise to something like 3^3 is 27>25 so x would be like very very cloase to 3, 3^4 81>64 far from 3 AND 3^5 243>216 far but closer to 3 than 3^4 thing as if the base is same the exponent will have more power kinda now we know y is the smallest

emmak

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

Bunuel · Answer

z^5 = 216 --> z ≈ 2.930...
x^3 = 25 --> x ≈ 2.924...
y^4 = 64 --> y ≈ 2.828...

The numbers are too close to each other, so your reasoning is not precise enough to rank them.

Legallyblond · Answer

Y is smaller in both the case like if you try x and y you know y is farther from 3 then x and z is near to 3 then y because of higher exponent... so y is smallest

arushi118 · Answer

x = 5^2/3 , y = 2^3/2, z=6^3/5
Raise all these to the power of 30  
So we get x = 5^20, y = 2^45, z=6^18

comparing x and y: 
x=(5^4)^5 = 625^5
y=(2^9)^5 = 512^5
This clearly tells us that x>y

comparing y and z:
y=(2^5)^9 = 32^9
z=(6^2)^9 = 36^9
this clearly tells us that z>y

From the options, only  E  is the one where we have both x>y and z>y.

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

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