bblast wrote:Is x divisible by 15?

1>When x is divided by 10, the result is an integer
2>x^2 is a multiple of 30

E

I think option A and B together give you the answer.
1> It proves that x is a integer
2> It proves x has factors 15=3* 5
As x^2 has factor 30= 2 x (3 x 5)hence x should also be a multiple of 30.

Let me know if I am making any mistake.

First i thought B, but that was based on assumption that x is an integer.
So i guess C is the correct answer..

bblast wrote:Is x divisible by 15?

1>When x is divided by 10, the result is an integer
2>x^2 is a multiple of 30

E

Considering 1) x is divisible by 10 - that means we are sure that x surely has 2 & 5 as factors. BUT for a number to be divisible by 15 we need a combination of 5 & 3. In this case, we are not sure about the presence of 3 so can't say. Shorter way : think of two numbers: 30 and 40(1 is divisible other is not) ----> NOT SUFFICIENT

Considering 2) x^2 is multiple of 30. This means when x^2 is divided by 30, we get an integer. This means x^2 surely has 2,3, and 5 as factor. ADDING some basic knowledge. We know that a perfect square will have two instances of each factor. Hence we can be sure that x is atleast divisible by 2,3, and 5 ie by 30 and 30 is divisible by 15. SO x is divisible by 15. I hope i have not confused you.
(What i am trying to say is
x^2= 2*3*5 * k[integer], but for x^2 to be a perfect square it needs to be
x^2 = 2*2*3*3*5*5*k[perfect square] Now taking root
x= 15 root(k[perfect square])---> divisible by 15) ---> Hence sufficient

Answer B

Thanks

bblast wrote:Is x divisible by 15?

1>When x is divided by 10, the result is an integer
2>x^2 is a multiple of 30

E

1) consider x=20; when 20 is divided by 10 result is an integer but 20 is not divisible by 15;
now consider x=30; when 30 is divided by 10 result is an integer and also 30 is divisible by 15;

hence 1 alone is not sufficient to answer the question.

2) x^2 is a multiple of 30;
now consider x to be sqrt(30); x^2=30 a multiple of 30; sqrt(30) is not divisible by 15
now consider x to be 30; x^2= 900 a multiple of 30; 30 is divisible by 15;

hence 2 alone is not sufficient to answer the question.

combining 1 and 2 we have; x should be integer so that x/10 results in integer;
and also x^2 should be a multiple of 30;
and the values for which it is possible are -30, 30,-60,60...etc...!!!! i.e. of the form (30k, where k can have both positive as well as negative values) all of which are divisible by 15... hence answer should be C

sameerballani wrote:

bblast wrote:Is x divisible by 15?

1>When x is divided by 10, the result is an integer
2>x^2 is a multiple of 30

E

Considering 1) x is divisible by 10 - that means we are sure that x surely has 2 & 5 as factors. BUT for a number to be divisible by 15 we need a combination of 5 & 3. In this case, we are not sure about the presence of 3 so can't say. Shorter way : think of two numbers: 30 and 40(1 is divisible other is not) ----> NOT SUFFICIENT

Considering 2) x^2 is multiple of 30. This means when x^2 is divided by 30, we get an integer. This means x^2 surely has 2,3, and 5 as factor. ADDING some basic knowledge. We know that a perfect square will have two instances of each factor. Hence we can be sure that x is atleast divisible by 2,3, and 5 ie by 30 and 30 is divisible by 15. SO x is divisible by 15. I hope i have not confused you.
(What i am trying to say is
x^2= 2*3*5 * k[integer], but for x^2 to be a perfect square it needs to be
x^2 = 2*2*3*3*5*5*k[perfect square] Now taking root
x= 15 root(k[perfect square])---> divisible by 15) ---> Hence sufficient

Answer B

Thanks

I seem to be completely mistaken here. When assuming that x^2 necessarily needs to b a perfect square and should not have any root in it. SO thinking again about the question it seems to me that the answer is C.

yup,,,manpsingh solution mirrors the official solution, I also erred on this and went for B.as 30 has factors 2,3 and 5. But key was that x can be root 30 as well. Even an OG ds question tests this logic.

correct answer is C