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gmat club PS

by bblast » Fri May 27, 2011 11:37 pm
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

C
Last edited by bblast on Sat May 28, 2011 7:36 am, edited 1 time in total.
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by sjmit4 » Sat May 28, 2011 12:39 am
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.
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by cans » Sat May 28, 2011 4:45 am
First i thought B, but that was based on assumption that x is an integer.
So i guess C is the correct answer..
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by sameerballani » Sat May 28, 2011 4:57 am
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

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by manpsingh87 » Sat May 28, 2011 5:38 am
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
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by sameerballani » Sat May 28, 2011 6:04 am
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.
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by bblast » Sat May 28, 2011 7:39 am
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
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