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Bunuel
If x and y are both positive perfect cubes, what is x+y?

(1) The cube root of x plus the cube root of y equals 6
(2) 50 < x+y < 75

Kudos for a correct solution.

x and y are positive perfect cubes. Assume that x=\(k^3\) and y=\(z^3\)

statement 1: you end up with k+z=6. From which three different scenarios unfold:

case_1: if k=3 and z=3 --> y=x=27 and x+y=54
case_2: if k=4 and z=2 --> x=64 and y=8 --> x+y=72
case_3: if k=5 and z=1 --> x=125 and y=1 --> x+y=126.

statement 2: if k=1 and z=4 --> x=1 and y=64 --> x+y=65. If k=2 and z=4 then x=8 and y=64 --> x+y=72

1+2) case 3 from statement one is out. Both case_2 and case_1 hold.

Answer E.

I would assume that X and Y are two different positive cubes. In that case C should be the answer.
Just use the facts the gmat delivers to you. No one said they should be different or the same, but they COULD be. or they could not be. since you do not know, you can not give an answer. its E

But it would start with the second statement:

since x+y should be between 50 and 75, and the cube of 5 is 125, we can only consider the cube of 1,2,3,4
so we will get this pairs:
x/y or y/x
1-4
2-4
3-3


its not suff. then we can take a look at the first statement, and we see they pairs should sum up to 6, we see there are 2 possible values for x/y in the range of 50 and 75. then we can automatically assume that there are enough other pairs in a broader range of numbers.

so we do not have to do any math for the first statement.

we can take E
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X and Y both are positive integers.

Statement#1:
Cube root of x + cube root of y = 6

Taking x=64,y=8
X+Y = 72
Taking x=27,y=27
X+Y = 54

Insufficient

Statement#2
50
Provides no information for x and y
X and Y can be 27 and 27 or 8 and 64

Insufficient

#1 together

Considering x=27 y=27 satisfies both the statements and sum is 54

Considering x=64 y=8 satisfies both the statements and sum is 72

Insufficient

So answer is E.
What is the OA?

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Bunuel
If x and y are both positive perfect cubes, what is x+y?

(1) The cube root of x plus the cube root of y equals 6
(2) 50 < x+y < 75

Kudos for a correct solution.

Statement 1:
x=1
y=125
sum=126
OR
x=27
y=27
sum=54
Insufficient

Statement 2:
x=27
y=27
sum=54
OR
x=64
y=8
sum=72
Insufficient

Combined, The cube root of x plus the cube root of y equals 6 AND 50 < x+y < 75
x=27
y=27
sum=54
OR
x=64
y=8
sum=72
still proves insufficiency

Answer: E
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Bunuel
If x and y are both positive perfect cubes, what is x+y?

(1) The cube root of x plus the cube root of y equals 6
(2) 50 < x+y < 75

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION

Solution: E

A positive perfect cube is the result obtained when a positive integer is raised to the third power: the first few positive perfect cubes, for instance, are 1, 8, 27, 64, and 125, or 1^3, 2^3, 3^3, 4^3, and 5^3. If the sum of the two cube roots is 6, then either the roots are either 1 and 5, 2 and 4, or 3 and 3, all of which give us different values for x+y; statement (1) is INSUFFICIENT. Statement (2) is tricky. If x and y are 8 and 64, then x+y is in that range, but if x and y are both 27, x+y is in that range too; INSUFFICIENT. Together both options from statement (2) remain; (E)
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