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What is the value of x? (1) X3 is a 2-digit positive odd integer. (2) X4 is a 2-digit positive odd integer.

I don't know whether the answer is correct. I got a different one.

What is the value of x?

Note that we are not told that x is an integer

(1) x^3 is a 2-digit positive odd integer --> now, if x is an integer then x=3 as x^3=27 is the only odd 2-digit positive cube of an integer (1^3=1 and 5^3=125) but if x is not an integer then it can be cube root of any 2-digit positive odd integer, for example if x=\sqrt[3]{11} then x^3=11. Not sufficient.

(2) x^4 is a 2-digit positive odd integer --> basically the same here: if x is an integer then x=3 or x=-3 as x^4=81 is the only odd 2-digit positive integer which is in fourth power of an integer (1^4=1 and 5^4=625) (so even if x is an integer this statement is still insufficient as it gives two values for x: 3 and -3). x also can be non-integer as above: it can be fourth root from any 2-digit positive odd integer, for example if x=\sqrt[4]{11} then x^4=11. Not sufficient.

(1)+(2) x cannot be an irrational number (so that both x^3 and x^4 to be integers), so x must be 3. Sufficient.

Re: What is the value of x? [#permalink]
05 Jan 2013, 22:42

1

This post received KUDOS

Glad that helped.

Always watch out for ZIP trap (assuming Zero, Integer, Positive) -> (Make sure to check for 0, factions and negatives) Especially for inequalities, algebraic, number/fraction problems. _________________

(1)+(2) x cannot be an irrational number (so that both x^3 and x^4 to be integers), so x must be 3. Sufficient.

Answer: C.

Hi Bunuel: Just like others, I also have a hard time visualizing that there does not exist an irrational number whose 3rd and 4th power both result in an odd digit integer. I mean integer is a smaller set compared to irrational numbers and we still have 3 (an integer) whose 3rd and 4th power both result in an odd 2-digit integer. On the other hand in terms of irrational numbers we have tremendous possibilities even between two integers we have infinite irrational numbers and we cannot have such a number. It some how feels odd to me. I have no doubt what you are saying is right but I have hard time imagining it. Maybe my understanding of irrational numbers and their powers is still primordial.

Say x IS an irrational number and x*x*x=x^3=integer. In this case x*x*x*x=x^3*x=integer*irrational=irrational.

If x is an irrational number and x*x*x*x=x^4=integer, then x^3=x^4/x=integer/irrational=irrational.

So, as you can see if x is an irrational number, then both x^3 and x^4 cannot be rational.

What is the value of x? (1) X3 is a 2-digit positive odd integer. (2) X4 is a 2-digit positive odd integer.

I don't know whether the answer is correct. I got a different one.

What is the value of x?

Note that we are not told that x is an integer

(1) x^3 is a 2-digit positive odd integer --> now, if x is an integer then x=3 as x^3=27 is the only odd 2-digit positive cube of an integer (1^3=1 and 5^3=125) but if x is not an integer then it can be cube root of any 2-digit positive odd integer, for example if x=\sqrt[3]{11} then x^3=11. Not sufficient.

(2) x^4 is a 2-digit positive odd integer --> basically the same here: if x is an integer then x=3 or x=-3 as x^4=81 is the only odd 2-digit positive integer which is in fourth power of an integer (1^4=1 and 5^4=625) (so even if x is an integer this statement is still insufficient as it gives two values for x: 3 and -3). x also can be non-integer as above: it can be fourth root from any 2-digit positive odd integer, for example if x=\sqrt[4]{11} then x^4=11. Not sufficient.

(1)+(2) x can not be an irrational number (so that both x^3 and x^4 to be integers), so x must be 3. Sufficient.

Tricky one, I considered the integer constraint that didn't exist. Must take care with this.

Bunuel wrote:

shan123 wrote:

What is the value of x? (1) X3 is a 2-digit positive odd integer. (2) X4 is a 2-digit positive odd integer.

I don't know whether the answer is correct. I got a different one.

What is the value of x?

Note that we are not told that x is an integer

(1) x^3 is a 2-digit positive odd integer --> now, if x is an integer then x=3 as x^3=27 is the only odd 2-digit positive cube of an integer (1^3=1 and 5^3=125) but if x is not an integer then it can be cube root of any 2-digit positive odd integer, for example if x=\sqrt[3]{11} then x^3=11. Not sufficient.

(2) x^4 is a 2-digit positive odd integer --> basically the same here: if x is an integer then x=3 or x=-3 as x^4=81 is the only odd 2-digit positive integer which is in fourth power of an integer (1^4=1 and 5^4=625) (so even if x is an integer this statement is still insufficient as it gives two values for x: 3 and -3). x also can be non-integer as above: it can be fourth root from any 2-digit positive odd integer, for example if x=\sqrt[4]{11} then x^4=11. Not sufficient.

(1)+(2) x can not be an irrational number (so that both x^3 and x^4 to be integers), so x must be 3. Sufficient.

Re: What is the value of x? [#permalink]
16 Jun 2013, 02:59

Bunuel wrote:

carcass wrote:

basically 1) is insuff because we have to consider integers and non integers (so irrational numbers). Same for 2)

Bothe statements are suff because we have only 3 that mettes the criteria so we have to consider only the 3 (the integer). So sufficient

But why we C is sufficient ?' why we can not consider the irrational numbers ??

Thanks. Now I hope is more clear what I mean. I'm sorry if I have explained myself badly

If x is an irrational number then x^3 and x^4 cannot both be integers as given in the statements, so x can only be 3.

Hi Bunnel,

Still did not get this part: If x is an irrational number then x^3 and x^4 cannot both be integers as given in the statements, so x can only be 3

Irrational no cannot be expressed as p/q, where p and q are integers.

I made my understand it like this: Their is only 1 number possible whose cube is 27 and only one number has fourth power equal to 81. Which is integer 3.

Re: What is the value of x? [#permalink]
16 Jun 2013, 03:15

Expert's post

cumulonimbus wrote:

Bunuel wrote:

carcass wrote:

basically 1) is insuff because we have to consider integers and non integers (so irrational numbers). Same for 2)

Bothe statements are suff because we have only 3 that mettes the criteria so we have to consider only the 3 (the integer). So sufficient

But why we C is sufficient ?' why we can not consider the irrational numbers ??

Thanks. Now I hope is more clear what I mean. I'm sorry if I have explained myself badly

If x is an irrational number then x^3 and x^4 cannot both be integers as given in the statements, so x can only be 3.

Hi Bunnel,

Still did not get this part: If x is an irrational number then x^3 and x^4 cannot both be integers as given in the statements, so x can only be 3

Irrational no cannot be expressed as p/q, where p and q are integers.

I made my understand it like this: Their is only 1 number possible whose cube is 27 and only one number has fourth power equal to 81. Which is integer 3.

Please explain why have you mentioned it here.

I don't understand your question. Please elaborate. _________________

(1)+(2) x cannot be an irrational number (so that both x^3 and x^4 to be integers), so x must be 3. Sufficient.

Answer: C.

Hi Bunuel: Just like others, I also have a hard time visualizing that there does not exist an irrational number whose 3rd and 4th power both result in an odd digit integer. I mean integer is a smaller set compared to irrational numbers and we still have 3 (an integer) whose 3rd and 4th power both result in an odd 2-digit integer. On the other hand in terms of irrational numbers we have tremendous possibilities even between two integers we have infinite irrational numbers and we cannot have such a number. It some how feels odd to me. I have no doubt what you are saying is right but I have hard time imagining it. Maybe my understanding of irrational numbers and their powers is still primordial. _________________

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