Data Sufficiency

Is n an integer?

1) sqrt(n) > 1
2) sqrt(n) < 2

Please explain thy answers. Thanks

IMO the answer should be (E)

By statement 1, sqrrt(n)>1 n could be any number be it integer or a fraction (decimal).
By statement 2, sqrrt(n)<2, again n could be an integer or a fraction (decimal)

Taking both statements together 1<sqrrt(n)<2, this again doesn't tell for sure whether n is an integer.

Thus E.

IMO C

1) NOT sufficient
2) NOT SUFF

combined
1<sqrt(n)<2
so the square root is not an integer. the number is not an integer.

tohellandback wrote:IMO C

1) NOT sufficient
2) NOT SUFF

combined
1<sqrt(n)<2
so the square root is not an integer. the number is not an integer.

tohellandback,

I think it wouldn't be advisable to assume that since the sqrrt of a number is not integer, the number itself is also not an integer.

As you would agree that the sqrrt(2) = 1.414, while 2 is an integer. But the sqrrt(2.5) = 1.581 which again lies in the range 1<sqrrt(n)<2 but n=2.5 is not an integer, so together also the two statements are not sufficient.

Cheers,
Vivek

Hi treker,

Could you please post the OA.


Cheers,
Vivek

vivekjaiswal wrote:

tohellandback wrote:IMO C

1) NOT sufficient
2) NOT SUFF

combined
1<sqrt(n)<2
so the square root is not an integer. the number is not an integer.

tohellandback,

I think it wouldn't be advisable to assume that since the sqrrt of a number is not integer, the number itself is also not an integer.

As you would agree that the sqrrt(2) = 1.414, while 2 is an integer. But the sqrrt(2.5) = 1.581 which again lies in the range 1<sqrrt(n)<2 but n=2.5 is not an integer, so together also the two statements are not sufficient.

Cheers,
Vivek

NO. I don't think so.
integers may have integer or non-integer square root.
for example 2- it is 1.414......(here 1.414 is just a convention. its non terminating and the square of 1.414 is NOT 2)
and for 9- it is 3

so you can be sure that if the square root is not an integer, then the number is not an integer

tohellandback wrote:

vivekjaiswal wrote:

tohellandback wrote:IMO C

1) NOT sufficient
2) NOT SUFF

combined
1<sqrt(n)<2
so the square root is not an integer. the number is not an integer.

tohellandback,

I think it wouldn't be advisable to assume that since the sqrrt of a number is not integer, the number itself is also not an integer.

As you would agree that the sqrrt(2) = 1.414, while 2 is an integer. But the sqrrt(2.5) = 1.581 which again lies in the range 1<sqrrt(n)<2 but n=2.5 is not an integer, so together also the two statements are not sufficient.

Cheers,
Vivek

NO. I don't think so.
integers may have integer or non-integer square root.
for example 2- it is 1.414......(here 1.414 is just a convention. its non terminating and the square of 1.414 is NOT 2)
and for 9- it is 3

so you can be sure that if the square root is not an integer, then the number is not an integer

agreed on your point about sqrrt(2).

But we are not really concerned about the square of a non-integer number, what the question is stating is that the sqrrt of n should lie in the range 1<sqrrt(n)<2. Which is true for both n=2 and n=2.5 (irrespective of the fact that whether the sqrrt(n) is a terminating decimal) while 'n' is integer in one of the case and it is not in another.

If the question asks "is 'n' an integer?" and to sufficiently derive that it is not, by following the statements is not possible because of the above mentioned points.

hope that clarifies my point

Cheers,
Vivek

vivekjaiswal wrote:

tohellandback wrote:

vivekjaiswal wrote:

tohellandback wrote:IMO C

1) NOT sufficient
2) NOT SUFF

combined
1<sqrt(n)<2
so the square root is not an integer. the number is not an integer.

tohellandback,

I think it wouldn't be advisable to assume that since the sqrrt of a number is not integer, the number itself is also not an integer.

As you would agree that the sqrrt(2) = 1.414, while 2 is an integer. But the sqrrt(2.5) = 1.581 which again lies in the range 1<sqrrt(n)<2 but n=2.5 is not an integer, so together also the two statements are not sufficient.

Cheers,
Vivek

NO. I don't think so.
integers may have integer or non-integer square root.
for example 2- it is 1.414......(here 1.414 is just a convention. its non terminating and the square of 1.414 is NOT 2)
and for 9- it is 3

so you can be sure that if the square root is not an integer, then the number is not an integer

agreed on your point about sqrrt(2).

But we are not really concerned about the square of a non-integer number, what the question is stating is that the sqrrt of n should lie in the range 1<sqrrt(n)<2. Which is true for both n=2 and n=2.5 (irrespective of the fact that whether the sqrrt(n) is a terminating decimal) while 'n' is integer in one of the case and it is not in another.

If the question asks "is 'n' an integer?" and to sufficiently derive that it is not, by following the statements is not possible because of the above mentioned points.

hope that clarifies my point

Cheers,
Vivek

let's tackle it like this
sqrt(1)=1
sqrt(4)=2

now 1<sqrt(n)<2
for n to be an integer, n should be wither 2 or 3
for any value between 1 and 2, the square of that value cannot be 2 or 3.
you can't say 1.414 or 1.732, they are just conventions. the values are non terminating and their squares are not 2 or 3

tohellandback,

I do agree with your point about 1.414 and 1.732 being conventional numbers used to represent sqrt of 2 and 3.

What I have been trying to emphasise is that the question is not concerned about getting an integer by squaring a non-integer that lies between 1 and 2.
It is just concerned whether we can be sure that n is an integer (or not) if its sqrt lies between 1 and 2.
Keeping that question in my mind; I have an integer 2, sqrt of which (although being a non-terminating decimal) lies between 1 and 2 and another non-integer 2.5 whose sqrt again lies between 1 and 2. Thus we cannot be sure about the property of n given that 1<sqrt(n)<2

I am not sure whether what I say is making any sense or not, but to answer the question by going the other way round, that is trying to find a number between 1 and 2 whose square would yeild an integer, is not the best way to approach this question.

Cheers,
Vivek

The answer should be E.

For sqrt(n) to be between 1 and 2, n can have many values such as 2,3, and even 2.5 or 2.7

sqrt(2) = 1.4
sqrt(3) = 1.73
sqrt(2.5) ~ between 1.4 and 1.7
sqrt(2.7) ~ "same as above".

So even with 2 statements we still dont know whether n is an integer or not.

A

Must be E.

can someone pls let me know if I am evaluating the question too simplistically and hence missing the point of the question (regardless of whether my ans is correct or not).

I chose E for following reasons:
1) sqrt(n) > 1 --> squaring both sides gives n>1 so number can be int or non-int.
2) sqrt(n) < 2 -->squaring both sides gives n<4 so number can be int or non-int

1) and 2) number is more than 1 but less than 4 so can be int or non-int.

many thanks.