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# m05 #22

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Intern
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21 Sep 2008, 11:08
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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

Source: GMAT Club Tests - hardest GMAT questions

REVISED VERSION OF THIS QUESTION IS HERE: m05-70531-20.html#p1255538
CIO
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01 Apr 2009, 03:32
5
KUDOS
I've changed the OE a bit.

From S1 we know that $$X$$ is an integer divisible by 10. Not sufficient by itself.
From S2 we know that $$X$$ might be divisible by 15 but not in all cases. Consider $$X=\sqrt{30}$$ (not divisible by 15) vs. $$X=30$$ (divisible by 15).
From S1+S2 we know that $$X$$ is an integer and that its square is divisible by 30. This is only possible if $$X$$ is divisible by 30.

Hope this helps.

klb15 wrote:
The explanation does not seem correct...
My ANS is E.

Can anyone explain further?

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01 Apr 2009, 05:04
C.
From 1 ) X is an integer =>NSF
From 2) X has atleast one 2,5 and 3 as factors.

1 + 2. X is divisible by 15.
CIO
Joined: 02 Oct 2007
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01 Apr 2009, 05:08
Not necessarily. $$X$$ can equal $$\sqrt{30}$$ or $$\sqrt{60}$$.

Economist wrote:
C.
From 1 ) X is an integer =>NSF
From 2) X has atleast one 2,5 and 3 as factors.

1 + 2. X is divisible by 15.

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27 Jul 2010, 07:53
3
KUDOS
Hi,

This is Data Sufficiency (DS) GMAT quant question. DS questions have 2 statements (1) and (2). Here are the options for DS questions:

(A) Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
(B) Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
(D) EACH statement ALONE is sufficient
(E) Statements (1) and (2) TOGETHER are NOT sufficient

Welcome to the Forum!

kalyanit wrote:
I am new to the forum. Pls tell me What r the options your considering ? What is the meaning of Option C ?

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27 Jul 2010, 17:34
Came to same conclusion as many, C.

1. means x must be an integer.
2. + 1 means that Sqrt of n*30 must be an integer

2 + 1 mean that x must be divisible by 30. i.e. for Sqrt of n*30 to be an integer n must equals to at least (2*5*3).

Good question!
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28 Jul 2010, 03:25
1
KUDOS
Is X divisible by 15?

1. When X is divided by 10, the result is an integer -- X=K*2*5--- NOt sufficient
2. X^2 is a multiple of 30 -- X^2= K1*2*3*5 since X might not be integer..Not sufficeint

Combining both .. sufficient Ans C
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05 Apr 2011, 20:55
4
KUDOS
I'd take any division questions the following way:

From 1, X=10k for some positive integer k. Since this value could be ANy integer and nt merely a multiple of 3, I'd consider this insufficient.

From 2, X^2 = 30m for some positive integer m. Now this means that X = 30^1/2*m^1/2 which could result in an irrational number also apart from the regular integers. This appears insufficient too.

Let both clauses be combined now:

We have X^2 = 100(k^2) = 30m for some k and m. This means k^2/m is some 3/10. So K^2 is proportional to 3 and this is impossible for integer values of K unless K is a factor of some multiple of 3 and hence is itself divisible by 3. (Reductio ad absurdum in simpler terms). So X=10k now reduces to X=30n for some natural number n. Ergo, X is a multiple of 30 and hence 15.

Regards
Rahul
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Rahul

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29 Jul 2011, 07:07
With B you can have X = square root (30) also as a possibility, which is not divisible by 15.
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29 Jul 2011, 08:02
HAi retro ,

I got your point up to k*k/m=3/10. After that we said k is proportional to 3 and so on. I did not understand that. Can you explain in detail. I got almost wrong these type of questions. Your explanation will help me a lot

tomB
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31 Jul 2011, 11:39
I thought the answer is 'B'
second statement says
2. X^2 is a multiple of 30

I guessed this would mean any multiple of 30 which is a square number would satisfy the equation x^2=30n.
for example 900 or 3600.
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Joined: 28 Jul 2011
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31 Jul 2011, 12:12
I agree hemadri. That's the same doubt I've. Cause every multiple of 30 is surely divisible by 15.

Posted from my mobile device
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31 Jul 2011, 23:59
1
KUDOS
hey,
S1: 10 i.e is not divisible but 30 is. 20 isn't and 60 is NS
S2: you said that x^2=30n so:
x^2 can be 30*30 which means that x=30 hence divisible by 15, but

x^2 cab be 30*2 which means that x=root 60 , hence not divisible by 15. NS

combining:

x^2=30A
x=10B

X^2/X=30A/10B
X= 3*(A/B)
since x has 3 in it and from S1 we know that A/B must be 10(0r 2*5) or else it wont be an integer

x=3* 2something*5 something. which means 30;60;90 etc.. allways divisible by 15
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10 Aug 2011, 22:55
st 1 : X divisible by 10
Insuff

st 2 : X^2 divisible by 30 . As pointed out by dzyubam, x could be root 30 (30^1/2)
Insuff

My doubt is combining both, we get X could be 10 root 30 . Which is not divisible by 15. Pls explain
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11 Aug 2011, 00:54
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GMATmay wrote:
st 1 : X divisible by 10
Insuff

st 2 : X^2 divisible by 30 . As pointed out by dzyubam, x could be root 30 (30^1/2)
Insuff

My doubt is combining both, we get X could be 10 root 30 . Which is not divisible by 15. Pls explain

$$10\sqrt{30}$$ doesn't satisfy statement 1. Per statement 1 X has to be ....-20,-10,0,10,20,30... A multiple of 10. $$10\sqrt{30}$$ is NOT a multiple of 10.
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11 Aug 2011, 01:21
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KUDOS
s1- x can be 10- no or 30-yes (divisible by 15). so x=10a
S2- x^2=30b. b can be 2 so not divisible or b=30 then x divisible.

combining: x^2=30b and x=10a

x^2/x=30b/10a so x= 3(a/b).
from S1 we know that x is an integer and a multiple of 10 so x^2 must conclude at least one 2 one 3 and one 5 to be an integer (30=2*5*3)
so 3*(a/b)= 3*(2*5/2*5*3) and it must be an integer.
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02 Aug 2012, 05:45
Bunuel, please throw some light here,
(c) is right as per many of us but why not (E)?
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02 Aug 2012, 06:10
Statement 1 is clearly insufficient.

Now coming to Statement 2, we have x^2 = 30*k for some positive integer k. So now, x^2 = 30*k is a quadratic equation (degree 2) which will have 2 roots. Hence, x = +sqrt(30*k) and x = -sqrt(30*k). Nowhere in the question it is given that x is positive. So we by ourselves cannot take x = +sqrt(30*k). Therefore, the answer has to be E.
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02 Aug 2012, 06:39
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thevenus wrote:
Bunuel, please throw some light here,
(c) is right as per many of us but why not (E)?

Is $$x$$ divisible by 15?

(1) When $$x$$ is divided by 10, the result is an integer --> $$\frac{x}{10}=integer$$ --> $$x=10*integer$$. Now, if $$x=0$$ (in case $$integer=0$$), then the answer is YES but if $$x=10$$ (in case $$integer=1$$), then the answer is NO. Not sufficient.

From this statement though we can deduce that $$x$$ is an integer (since $$x=10*integer=integer$$).

(2) $$x^2$$ is a multiple of 30 --> if $$x=0$$, then the answer is YES but if $$x=\sqrt{30}$$, then the answer is NO. Not sufficient.

(1)+(2) Since from (1) $$x=integer$$ then $$x^2=integer$$, and in order $$x^2$$ to be divisible by 30=2*3*5, $$x$$ must be divisible by 30 ($$x$$ must be a multiple of 2, 3 and 5, else how can this primes appear in $$x^2$$?), hence $$x$$ is divisible by 15 too. Sufficient.

Notice that $$x$$ can be positive, negative or even zero, but in any case it'll be divisible by 30.

Next, every GMAT divisibility question will tell you in advance that any unknowns represent positive integers (ALL GMAT divisibility questions are limited to positive integers only). So, we edited this question and in the new GMAT Club tests this question reads:

If x is a positive integer, is x divisible by 15?

(1) x is a multiple of 10 --> if $$x=10$$, then the answer is NO but if $$x=30$$, then the answer is YES. Not sufficient

(2) x^2 is a multiple of 12 --> since $$x$$ is an integer, then $$x^2$$ is a perfect square. The least perfect square which is a multiple of 12 is 36. Hence, the least value of $$x$$ is 6 and in this case the answer is NO, but if for example $$x=12*15$$ then the answer is YES. Not sufficient.

Notice that from this statement we can deduce that $$x$$ must be a multiple of 3 (else how can this prime appear in $$x^2$$?).

(1)+(2) $$x$$ is a multiple of both 10 and 3, hence it's a multiple of 30, so $$x$$ must be divisible by 15. Sufficient.

Hope it's clear.
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02 Aug 2012, 07:07
Bunuel wrote:
thevenus wrote:
Bunuel, please throw some light here,
(c) is right as per many of us but why not (E)?

Is $$x$$ divisible by 15?

(1) When $$x$$ is divided by 10, the result is an integer --> $$\frac{x}{10}=integer$$ --> $$x=10*integer$$. Now, if $$x=0$$ (in case $$integer=0$$), then the answer is YES but if $$x=10$$ (in case $$integer=1$$), then the answer is NO. Not sufficient.

From this statement though we can deduce that $$x$$ is an integer (since $$x=10*integer=integer$$).

(2) $$x^2$$ is a multiple of 30 --> if $$x=0$$, then the answer is YES but if $$x=\sqrt{30}$$, then the answer is NO. Not sufficient.

(1)+(2) Since from (1) $$x=integer$$ then $$x^2=integer$$, and in order $$x^2$$ to be divisible by 30=2*3*5, $$x$$ must be divisible by 30 ($$x$$ must be a multiple of 2, 3 and 5, else how can this primes appear in $$x^2$$?), hence $$x$$ is divisible by 15 too. Sufficient.

Notice that $$x$$ can be positive, negative or even zero, but in any case it'll be divisible by 30.

Next, every GMAT divisibility question will tell you in advance that any unknowns represent positive integers (ALL GMAT divisibility questions are limited to positive integers only). So, we edited this question and in the new GMAT Club tests this question reads:

If x is a positive integer, is x divisible by 15?

(1) x is a multiple of 10 --> if $$x=10$$, then the answer is NO but if $$x=30$$, then the answer is YES. Not sufficient

(2) x^2 is a multiple of 12 --> since $$x$$ is an integer, then $$x^2$$ is a perfect square. The least perfect square which is a multiple of 12 is 36. Hence, the least value of $$x$$ is 6 and in this case the answer is NO, but if for example $$x=12*15$$ then the answer is YES. Not sufficient.

Notice that from this statement we can deduce that $$x$$ must be a multiple of 3 (else how can this prime appear in $$x^2$$?).

(1)+(2) $$x$$ is a multiple of both 10 and 3, hence it's a multiple of 30, so $$x$$ must be divisible by 15. Sufficient.

Hope it's clear.

Thanks a lot, Kudos for you +1
Wasn't the previous one was tougher? why did you changed / modified ? Now the GMAT can't put such an option (of choosing irrational no. at least if not negative numbers?)
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# m05 #22

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