If x, y, and z are distinct single-digit non-prime positive integers, and \sqrt {\frac{27}{125}} < \sqrt {\frac{9}{25}} < \sqrt {\frac{3}{5}} , what is the value of x + y + z? A. 13 B. 14 C. 15 D. 16 E. 17 M46-01

(3/5)^(3/9x) 2y>3x x, y z must be from 1,4,6,8,9 x=1 y=4 z=9 Sum = 14

Firstly I have solved the all the roots into 3/5 form and received 1/3x,1/2y and 1/z So after comparing if it 3/5^1/3x 1/2Y. So after solving this I have received only pair has 9,4,1 with distinct non prime positive integers and x+y+z is equal to 14

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12 Days of Christmas GMAT Competition 2025 - Day 1: If x, y, and z

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DhanyaAbhirami

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(3/5)^(3/9x)<(3/5)^(2/4y)<(3/5)^(1/z)
1/z<1/2y<1/3x
z>2y>3x

x, y z must be from 1,4,6,8,9
x=1
y=4
z=9
Sum = 14

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x,y,z are distinct single digit non-prime positive integers (1,4,6,8,9)
(27/125)^(1/9x) < (9/25)^(1/4y) < (3/5)^(1/z)
Find: x+y+z
Reduce all fractions to simplest forms
(3/5)^(3/9x) < (3/5)^(2/4y) < (3/5)^(1/z)
The base are all same and 0<3/5<1. So the function (3/5)^p decreases as p increases, hence 1/3x>1/2y>1/z
Lets choose values from (1,4,6,8,9)
Let, x=1...then next smallest is 4=y, try to put them in equation .... (1/3)>(1/8)>(1/9) ....True as we solve we need to choose highest value of x=9.... So no other possibilities.
x+y+z= 1+4+9= 14

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Bunuel

If x, y, and z are distinct single-digit non-prime positive integers, and $\sqrt[9x]{\frac{27}{125}} < \sqrt[4y]{\frac{9}{25}} < \sqrt[z]{\frac{3}{5}}$, what is the value of x + y + z?

A. 13
B. 14
C. 15
D. 16
E. 17

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Here, two concepts are used,
1. Exponents
2. When you deal with numbers between 0 and 1.

After creating a same base of 3/5, we can then use formula saying if bases are same power can be taken out. But because the base was a number between 0 and 1.
The greater would be the power number become smaller,so signs will get changed and the equation will become like this,

1/3x > 1/2y > 1/z

Now, after adjusting reciprocal we can write,

3x < 2y < z (because if 1/2>1/3 then 2<3)

Now, since only positive non prime numbers are llowed then we only have option {1,4,6,8,9}

By putting this we get z = 9, y = 4 and x = 1

Hence, option B is correct

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16 Dec 2025, 05:27

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Its single digit no prime number. So only options= 1,4,6,8,9

Based on combination
1,4,6= 11
1,4,8=13
1,4,9=14
4,6,8=18
4,6,9=19
6,8,9=23

Only option 13= Option A

Bunuel

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I've attached my initial transformations as a photo for the ease of illustration, but the point is that the original equation can be eventually written as this:
$0.6^{1/3x} < 0.6^{1/2y} < 0.6^{1/z}$

Since 0.6 is less than 1, the higher the power, the less the number will become. Therefore, our powers need to be related in reverse to the final numbers, that is:
$1/3x > 1/2y > 1/z$

And since these variables are in the denominator of the fraction, they again bear an inverse relationship to the numerator, so we need to make one final inequality twist:
$3x < 2y < z$

Now let's deal with the numbers we actually have. Non-prime single-digits are: 1, 4, 6, 8, 9.
Obviously, since Z has to be one digit, $3x = 1*3$, since if we take 4 three times, we'll land at 12.
Therefore, the smallest value Y can take is 4, and it means $z > 2y = 8$, so $z=9.$

Hence, our final sum is $S = x + y + z = 1 + 4 + 9 = 14$, and the answer is B.

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We need to find x + y + z

Given,
(27/125)^(1/9x) < (9/25)^(1/4x) < (3/5)^(1/x)

= (3/5)^(3/9x) < (3/5)^(2/4x) < (3/5)^(1/x)
= (3/5)^(1/3x) < (3/5)^(1/2x) < (3/5)^(1/x)

We know that,
0 < 3/5 < 1
=> inequality on exponents inverses
=> 1/3x > 1/2y > 1/z
=> 3x < 2y < z

It is given that only single distinct non prime integers are allowed = {1,4,6,8,9}
We try them out,
x = 1, y=4 , z= 8 (z=6 not possible since z >8 for y=4) => 3 < 8 < 9
WE see that other values with x = 1 is not possible

Now we try for x = 4,
=> 3x = 12
=> y = 8 (since 2y> 12)
But z > 16 not possible
=> higher values of x fail to meet the requirements

Thus the required values of x y and z = 1, 4 and 9

x+y+z = 1+4+9 = 14

B.14

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The first pattern recognition to be done here is see how each of them have powers of 3/5, so lets simplify that.
Each of them then leads to the underlying fraction to be the same, 3/5.
Post that, rewrite them by bringing the exponents into fractional powers themselves, such that it becomes 3/9x, 2/4y, 1/z. (1)
Now, its important to understand how signs work with fractions, and then exponents. In fractions, larger power = smaller number.
Hence, according to the inequality provided, the exponents must follow the following rule:
1/3x > 1/2y > 1/z from (1)
Next, lets simplify this to 3x < 2y < z. The logic behind this is all have to be positive numbers, hence we can take a reciprocal. Also since the reciprocal is a decreasing function, we flip the signs. ( Think of numbers 1,2,3 and if you took the reciprocal, how the inequality would change )
Let us now play with the values, identify that the values in place are {1,4,6,8,9} and z needs to be the largest, and then x needs to be smallest, and you get 1, 4, 9. ( You cannot have 6 for y as 2y=12 which is greater than 9 )
Hence, 14!

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Bunuel

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Only option available for x, y, and z are 1,4,6, 8 and 9.

Compare all the parts by making base same that is 3/5..

The power will become 1/3x, 1/2y, 1/z . since 3/5 is less than 1 Greater the power -- lesser the total value will be hence we can directly compare powers instead -- 1/3x > 1/2y > 1 / z or 3x<2y<z

Upon trying multiple value from given options we arrive to x=1, y=4, z = 9 .... x+y+z =14 answer B

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Convert each fraction into simpler form:
27/125 = (3/5)^3
9/25 = (3/5)^2

now powers will become:
1/3x,1/2y,1/z
Base <1:
1/3x > 1/2y > 1/z which implies, 3x<2y<z

Now, single digit non prime positive numbers are 1,4,6,8,9
When we put in above equation, we will get:
x=1,y=4,z=9

So, x+y+z = 1+4+9 = 14. Answer

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Firstly I have solved the all the roots into 3/5 form and received 1/3x,1/2y and 1/z So after comparing if it 3/5^1/3x <3/5^1/2y then 1/3X>1/2Y. So after solving this I have received only pair has 9,4,1 with distinct non prime positive integers and x+y+z is equal to 14

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A root of a number can be written as a fractional power. Eg. Cube root of, 3_/5 can be written as 5^1/3.

Similarly,
9x_/27/125 can be written as (_/3/5)^3/9x, i.e. (_/3/5)^1/3x

Similarly, 4y_/9/25 would be (_/3/5)^2/4y, i.e. (_/3/5)^1/2y

And, z_/3/5 would be (_/3/5)^1/z

Since all the base of all values are equal, 1/3x > 2y > z
OR, 3x < 2y < z

Since the numbers are non-prime single digits, the options are 1,4,6,8,9
Substituting the values, the only options that fits best is z = 9, y = 4 and x = 1
Hence, x+y+z = 14

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Given : x,y,z non prime integers
Simplifying the inequaities
27/125= (3/5) ^ 3/9x = (3/5)^1/3x
9/25=(3/5)^1/2y
& (3/5)^1/z

Given: (3/5)^1/3x < (3/5)^1/2y < (3/5)^1/z
Since 3/5 < 1 the powers will be reversed
1/3x> 1/2y> 1/z
= 3x<2y<z
if x = 1 next possible valyue for y = 4 and z = 9
x+y+z= 1+4+9= 14

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I think the answer should be A since any value of x more than 1 would not hold the condition as true.

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*: 3/5 <1, so as the exponent increases, the value of (3/5)^(1/3x) decreases. A large exponent gives a smaller result

**: when numerators equal, bigger denominator give smaller fraction

Consider: 3x < 2y < z, where x, y, z are all one-digit non-prime numbers.

Since 3x<z => 3x cannot exceed 9 => x has 3 options 1, 2, 3 => x=1 (eliminate 2 and 3, which are prime numbers)

Similarly, 3*1 < 2y < z => 2y cannot exceed 9 => y has 3 options 2, 3, 4 => y=4 (eliminate 2 and 3, which are prime number)

Therefore, 3*1 < 2*4 < z => z=9

Ultimately, x + y + z = 1 + 4 + 9 = 14

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Expand the equation and make all inside root values as of 3/5. Then we get 1/3x>1/2y>1/z inequality sign changes because 3/5 is less than 1 so exponent must be a greater value for the given Equation to be valid.
Again denominator values become finally 3x<2y<z ....
Check for distinct number which is single digit non prime positive we have[1,4,6,8,9]
The only possible value we can have is x=1, y=4, z=9
So x+y+z = 14

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could you explain how to spot this?

"Since x, y, and z are distinct single-digit non-prime positive numbers"

why cant we use the prime positive numbers as the x,y,z?

Thanks

Bunuel

GMAT Club Official Explanation:

Notice that 27/125 = (3/5)^3 and 9/25 = (3/5)^2, so:

$\sqrt[9x]{\frac{27}{125}} < \sqrt[4y]{\frac{9}{25}} < \sqrt[z]{\frac{3}{5}}$

$\sqrt[(3*3x)]{\frac{27}{125}} < \sqrt[(2*2y)]{\frac{9}{25}} < \sqrt[z]{\frac{3}{5}}$

$\sqrt[3x]{(\frac{27}{125})^{\frac{1}{3}}} < \sqrt[2y]{(\frac{9}{25} )^{\frac{1}{2}}} < \sqrt[z]{\frac{3}{5}}$

$\sqrt[3x]{\frac{3}{5}} < \sqrt[2y]{\frac{3}{5}} < \sqrt[z]{\frac{3}{5}}$

Here’s the key idea: when you take a positive integer root of a number between 0 and 1, the result becomes larger than the original number, and the higher the root, the closer the value moves toward 1.

For example, $\sqrt{\frac{1}{2}} < \sqrt[3]{\frac{1}{2}} < \sqrt[4]{\frac{1}{2}} < \sqrt[5]{\frac{1}{2}}...$

Thus, $\sqrt[3x]{\frac{3}{5}} < \sqrt[2y]{\frac{3}{5}} < \sqrt[z]{\frac{3}{5}}$ implies:

3x < 2y < z

Since x, y, and z are distinct single-digit non-prime positive numbers, they can be selected only from {1, 4, 6, 8, 9}.

Now, x cannot be 4 or greater, because in that case 3x becomes 12 or greater and cannot be less than z, which is at most 9. Thus, x = 1.

Similarly, y cannot be 6 or greater, so y = 4.

And finally, this forces z to be 9.

Thus: x = 1, y = 4, and z = 9.

(3 * 1) < (2 * 4) < 9

Therefore, x + y + z = 1 + 4 + 9 = 14.

Answer: B.

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sunshineeee

could you explain how to spot this?

"Since x, y, and z are distinct single-digit non-prime positive numbers"

why cant we use the prime positive numbers as the x,y,z?

Thanks

Please read the question:

If x, y, and z are distinct single-digit non-prime positive integers, and $\sqrt[9x]{\frac{27}{125}} < \sqrt[4y]{\frac{9}{25}} < \sqrt[z]{\frac{3}{5}}$, what is the value of x + y + z?

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17 Dec 2025, 07:45

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thanks Bunuel!! Please read the question carefully, me!!

Bunuel

Please read the question:

If x, y, and z are distinct single-digit non-prime positive integers, and $\sqrt[9x]{\frac{27}{125}} < \sqrt[4y]{\frac{9}{25}} < \sqrt[z]{\frac{3}{5}}$, what is the value of x + y + z?

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$\sqrt[9x]{\frac{27}{125}} < \sqrt[4y]{\frac{9}{25}} < \sqrt[z]{\frac{3}{5}}\\
<=> ((\frac{3}{5})^3)^{\frac{1}{9x}} < ((\frac{3}{5})^2)^{\frac{1}{4y}} < (\frac{3}{5})^{\frac{1}{z}}\\
<=> (\frac{3}{5)}^{\frac{1}{3x}} < (\frac{3}{5})^{\frac{1}{2y}} < (\frac{3}{5})^{\frac{1}{z}} $

Since $\frac{3}{5} < 1 => {\frac{1}{3x}} > {\frac{1}{2y}} > {\frac{1}{z}}$
Given x,y,z are single-digit and non-prime integers, x,y,z can be {1,4,6,8,9}
=> x = 1, y = 4, z = 9
=> x+y+z = 14

Answer: B.

Bunuel

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