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

GMAT Club Official Explanation: Notice that 27/125 = (3/5)^3 and 9/25 = (3/5)^2, so: \sqrt {\frac{27}{125}} < \sqrt {\frac{9}{25}} < \sqrt {\frac{3}{5}} \sqrt {\frac{27}{125}} < \sqrt {\frac{9}{25}} < \sqrt {\frac{3}{5}} \sqrt {(\frac{27}{125})^{\frac{1}{3}}} < \sqrt {(\frac{9}{25} )^{\frac{1}{2}}} < \sqrt {\frac{3}{5}} \sqrt {\frac{3}{5}} < \sqrt {\frac{3}{5}} < \sqrt {\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 {\frac{1}{2}} < \sqrt {\frac{1}{2}} < \sqrt {\frac{1}{2}}... Thus, \sqrt {\frac{3}{5}} < \sqrt {\frac{3}{5}} < \sqrt {\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.

Deconstructing the Question We are given distinct single-digit non-prime positive integers x, y, z . The possible set of values for x, y, z is derived from integers 1-9 excluding primes (2, 3, 5, 7). Candidates: {1, 4, 6, 8, 9} . We are given the inequality: \sqrt {\frac{27}{125}} < \sqrt {\frac{9}{25}} < \sqrt {\frac{3}{5}} Target: Find x + y + z . Step 1: Simplify the Terms Express all terms with the base \frac{3}{5} : 1. \sqrt {\frac{27}{125}} = \left(\left(\frac{3}{5}\right)^3\right)^{\frac{1}{9x}} = \left(\frac{3}{5}\right)^{\frac{3}{9x}} = \left(\frac{3}{5}\right)^{\frac{1}{3x}} 2. \sqrt {\frac{9}{25}} = \left(\left(\frac{3}{5}\right)^2\right)^{\frac{1}{4y}} = \left(\frac{3}{5}\right)^{\frac{2}{4y}} = \left(\frac{3}{5}\right)^{\frac{1}{2y}} 3. \sqrt {\frac{3}{5}} = \left(\frac{3}{5}\right)^{\frac{1}{z}} Step 2: Analyze the Exponents The inequality is: \left(\frac{3}{5}\right)^{\frac{1}{3x}} \frac{1}{2y} > \frac{1}{z} Since x, y, z are positive, we can invert the fractions and reverse the signs again: 3x 8 . The only candidate is z = 9 . Check: 3 12 . No single digit available. So the only solution is x = 1, y = 4, z = 9 . Step 4: Calculate Sum x + y + z = 1 + 4 + 9 = 14 . Answer: B

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

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Bunuel

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Updated on: 12 Jan 2026, 05:44

Originally posted by Bunuel on 15 Dec 2025, 09:00.
Last edited by Bunuel on 12 Jan 2026, 05:44, edited 1 time in total.

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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

M46-01

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Bunuel

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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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simplifying this
take all the values to bases of 3/5
since it is < 1
higher the root, higher will be the value -
3x < 2y < z
z to be the greatest we need z =9
y = 4
x =1

hence 9 + 4 + 1 = 14

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Updated on: 16 Dec 2025, 10:44

Originally posted by ManifestDreamMBA on 15 Dec 2025, 09:22.
Last edited by ManifestDreamMBA on 16 Dec 2025, 10:44, edited 1 time in total.

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First term = (3/5)^3*1/9x = (3/5)^1/3x
Second term = (3/5)^2*1/4y = (3/5)^1/2y
Third term = (3/5)^1/z

Since the base is same and less than 1, we know higher the power lower the value
1/3x > 1/2y > z
Taking reciprocal
3x<2y<z

Since, x, y, z are single digit on prime positive integers, they can take values from : 1,4,6,8,9

If x=1, y=4, z=9; 3*1<2*4<9 is satisfied
Sum = x+y+z=1+4+9=14

Answer B

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$\frac{27}{125}^\frac{1}{9x} < \frac{9}{25}^\frac{1}{4y} < \frac{3}{5}^\frac{1}{z}$

$\frac{3}{5}^\frac{1}{3x} < \frac{3}{5}^\frac{1}{2y} < \frac{3}{5}^\frac{1}{z}$

Since 0 < 3/5 < 1

$\frac{1}{3x} > \frac{1}{2y} > \frac{1}{z}$
3x < 2y < z

Single-digit non-prime positive integers = {1,4,6,8,9}

x = 2; y = 4; z = 9
x + y + z = 2 + 4 + 9 = 15

IMO C

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First simplify the square roots the substitute, clear the denominators ad approximate the possible values, the final answer is 13

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This problem might seem intimidating at first but all we need to do is rewrite it in a more friendly way. First, get rid of roots as we know that square root is a number to the power of 1/2.

So, we get: $(\frac{27}{125})^{ {1/(9x)} }$ < $(\frac{9}{25})^{ {1/(4y)} }$ < $(\frac{3}{5})^{ {1/(z)} }$

then we can notice that every fraction can be simplified to 3/5 and the powers also simplify: $(\frac{3}{5})^{ {1/(3x)} }$ < $(\frac{3}{5})^{ {1/(2y)} }$ < $(\frac{3}{5})^{ {1/(z)} }$

At this point, it becomes the comparison of fractions. We should know that when "x" is < 1 the larger exponent reduces "x" and when "x">1, the larger exponent increases it. So, we can rewrite the whole thing like this: $\frac{1}{3x} > \frac{1}{2y} > \frac{1}{z}$.

From here we know that to make $\frac{1}{3x}$ the largest, 3x has to be the smallest.
So, we write it again: 3x<2y<z. And we are left off with plugging in numbers. We only have 1,4,6,8,9 available because they are non prime. By plugging in values, we see only one combination that works 1,4, and 9. Thus, the answer is 14.

Bunuel

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(3/5)^(3/9x) < (3/5)^(2/4y) < (3/5)^(1/z)
(3/5)^(1/3x) < (3/5)^(1/2y) < (3/5)^(1/z)
Since, 0<(3/5)<1 , the function (3/5)^a decreases as “a” increases, so it means
1/3x > 1/2y > 1/z or 3x < 2y < z
Single digit non prime positives = 1,4,6,8,9
We need distinct values of x,y,z such that 3x<2y<z and z<=9;
If x=1.... Then 3(1)<2(4)<z=9 .....possible

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

B

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on simplifing the equation it becomes (0.6)^1/3x<(0.6)^1/2y<(0.6)^1/z

1/3x<1/2y<1/z

since x,y,z are non prime distinct single positive integer. the value can be 1,4,6,8,9

2y will be more than two digit if y=6,8,9 hence y can be 4 as 3x should be less than 1/2yand z should be 9 since 1/2y should be less than 1/z hence z should be 9
x+y+z=1+4+9=14
anser=14

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Deconstructing the Question
We are given distinct single-digit non-prime positive integers $x, y, z$.
The possible set of values for $x, y, z$ is derived from integers 1-9 excluding primes (2, 3, 5, 7).
Candidates: {1, 4, 6, 8, 9}.
We are given the inequality: $\sqrt[9x]{\frac{27}{125}} < \sqrt[4y]{\frac{9}{25}} < \sqrt[z]{\frac{3}{5}}$
Target: Find $x + y + z$.

Step 1: Simplify the Terms
Express all terms with the base $\frac{3}{5}$:
1. $\sqrt[9x]{\frac{27}{125}} = \left(\left(\frac{3}{5}\right)^3\right)^{\frac{1}{9x}} = \left(\frac{3}{5}\right)^{\frac{3}{9x}} = \left(\frac{3}{5}\right)^{\frac{1}{3x}}$
2. $\sqrt[4y]{\frac{9}{25}} = \left(\left(\frac{3}{5}\right)^2\right)^{\frac{1}{4y}} = \left(\frac{3}{5}\right)^{\frac{2}{4y}} = \left(\frac{3}{5}\right)^{\frac{1}{2y}}$
3. $\sqrt[z]{\frac{3}{5}} = \left(\frac{3}{5}\right)^{\frac{1}{z}}$

Step 2: Analyze the Exponents
The inequality is: $\left(\frac{3}{5}\right)^{\frac{1}{3x}} < \left(\frac{3}{5}\right)^{\frac{1}{2y}} < \left(\frac{3}{5}\right)^{\frac{1}{z}}$

Theory: Since the base $0 < \frac{3}{5} < 1$, the function is decreasing.
This means a smaller value implies a larger exponent. We must reverse the inequality signs for the exponents:

$\frac{1}{3x} > \frac{1}{2y} > \frac{1}{z}$

Since $x, y, z$ are positive, we can invert the fractions and reverse the signs again: $3x < 2y < z$

Step 3: Test Values
We need to pick distinct $x, y, z$ from $\{1, 4, 6, 8, 9\}$ to satisfy $3x < 2y < z$.

Case 1: Try the smallest possible value for $x$.
Let $x = 1$. Inequality becomes: $3(1) < 2y < z \implies 3 < 2y < z$.
We need a value for $y$ from $\{4, 6, 8, 9\}$.
* If $y = 4$, then $2y = 8$. We need $z > 8$. The only candidate is $z = 9$. Check: $3 < 8 < 9$. This works! Are they distinct? $\{1, 4, 9\}$. Yes.

* If $y = 6$, then $2y = 12$. We need $z > 12$. No single digit available.

So the only solution is $x = 1, y = 4, z = 9$.

Step 4: Calculate Sum $x + y + z = 1 + 4 + 9 = 14$.

Answer: B

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the answer should be 18 given 4,6,8 are single digit non- prime positive integers

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reducing the expressions to basic form, we get:
(3/5)^(1/3x) < (3/5)^(1/2y)<(3/5)^(1/z)

Since 3/5<1, we get
1/3x>1/2y>1/z
or
3x<2y<z
Since all x,y and z are single digit non prime positive integer, they cant be 2,3,5,7.
so x = 1, y=4, z=9 is the only possible solution
so x+y+z = 14

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Bunuel

Equate all the bases to 3/5

(3/5)^{3/9x}
(3/5)^{2/4y}
(3/5)^{1/z}

(3/5)^{1/3x}
(3/5)^{1/2y}
(3/5)^{1/z}

Since 3/5 is less than one, higher the power smaller is the number

Therefore we can conclude that 1/3x > 1/2y > 1/z

3x < 2y < z

x, y , z can be 1, 4, 6 , 8, 9

x = 1
y = 4
z= 9

x + y + z = 14

Option B

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After solving the exponential equation, we get
=> 3x<2y<z
=>So, single digit non-prime +ve values that satisy the above inequality are (Given)
=> x=1, y=4, z=9
=>thus x+y+z = 1+4+9=14 (B)

Bunuel

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distinct single digit non-prime +ve : {1, 4, 6, 8, 9}
bringing each to form 3/5 so as to compare comparable values
(3/5)^(3/9x) < (3/5)^(2/4y) < (3/5)^(1/z)
since 3/5 < 1
1/3x > 1/2y > 1/z
which gives, 2y > 3x and Z> 2y => z > 2y > 3x
plug and play with options: 1st 13 -> {1, 4, 8} -> 8(z) > 8(2y) > 3 (3x) not working
option 2: 14 -> {1, 4, 9} -> 9(z) > 8(2x) > 3 (3x) works --- skip checking further options to save time.

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We can rewrite the terms of the disequalities as (3/5) with different exponents.
The first term can be rewritten as (3/5)^(1/3x). (Because 27/125=(3/5)^3 and 3/9x=1/3x).
In a similar way we can rewrite the second term as (3/5)^(1/2y) and the third as (3/5)^(1/z).

Since 0 < 3/5 < 1, as the exponent grows, the result decreases. This allows us to rewrite the disequality as 1/3x > 1/2y > 1/z.
In a similar way, as the denominator grows, the number decreases, so we can invert the disequality: 3x < 2y < z.

x,y,z can be one of these values (non prime, single digit numbers): 1, 4, 6, 8, 9.

The only options that satisfy the disequality are x=1, y=4, z=9 -> 3*1 < 2*4 < 9.
1 + 4 + 9 = 14

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x, y & z could only be 1, 4, 6, 8, 9 as per the constraints. And as per the ans choices either, 1 + 4 + 8 = 13 works or 1 + 4 + 9 = 14 works.

Simplifying the inequality to powers of 3/5

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

When we have a number between 0 to 1, higher powers lead to smaller nos and vice versa. So fitting in our combinations here,

only x=1, y = 4 & z = 9 work and the ans is 14.

Why so? When we fit these we get the powers as 1/3, 1/8 & 1/9 (higher power 1/3, will lead to the smallest no, and 1/9 will lead to the biggest no)

If we fit x=1, y=4 & z=8, we get powers as 1/3, 1/8 & 1/8, which will not make the inequality work.

Ans (B) 14

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Bunuel

Here, firstly we will write the given equation in simpler form-
That is,
(3/5)^3/9x < (3/5)^2/4y < (3/5)^1/z
= (3/5)^1/3x < (3/5)^1/2y < (3/5)^1/z
Now, since the base is same, equating powers-
That is, 1/3x < 1/2y < 1/z
Multiplying the inequalities by their LCM- 6xyz,
will give, 2yz < 3xz < 6xy

Now, by trial and error check for values of x, y and z that might suit this inequality.

It is given that x, y and z values are positive single digit non primes. That is possible values- 1,4,6,8,9.
One possible combination is x=6, y=4 and z=1 which gives 8 < 18 < 24. But, it's sum is x + y + z = 11 which is not in options.
Another possible combination is x=9, y=4 and z=1 which gives 8 < 27 < 216. It's sum is x + y+ z = 14. This is option B.

Thus answer is option B: 14.

ikan444

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15 Dec 2025, 10:22

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<Please look at the attached file first>

After that, we get this inequality: z>2y>3x. (1)

The possible values of x,y,z can be (1,4,6,8,9) as per the condition stated in the question. (2)
Try out different combinations that fulfill both (1) and (2).

Start with the lowest possible number, the values satisfy both conditions:
x=1 => 3x=3
y=4 => 2y=8
z = 9

x+y+z = 1+4+9 = 14 (B)

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Firstly, x, y, z can only take 1, 4, 6, 8, 9. And all are distinct.

Now, simplifying the expression, we get
${\frac{3}{5}}^{\frac{1}{z}} > {\frac{3}{5}}^{\frac{1}{2y}} > {\frac{3}{5}}^{\frac{1}{3x}}$

Since 3/5 is 0.6, which lies between 0 and 1, in this range, as power decreases, the value increases.
So, which is why 1/z < 1/2y < 1/3x.

Taking the first two, we get z > 2y,
y can take either 1 or 4, since 2*6 is 12, and z can't be greater than 12.
If y takes 1, z can take any of 4, 6, 8, 9. If y takes 4, z can take only 9.

Next, we will take the next two, 1/2y < 1/3x, we get x < 2/3 * y.
If y is 1, x is < 2/3, there is no value for x. So, y has to be 4, which means x < 8/3, which means x = 1.
And since y is 4, z takes 9

1 + 4 + 9 = 14, option B.

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