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# If x,y and z are positive integers, and 30x=35y=42z, then

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If x,y and z are positive integers, and 30x=35y=42z, then [#permalink]

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18 Jul 2010, 12:23
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If x,y and z are positive integers, and 30x=35y=42z, then which of the following must be divisible by 3?

I. x
II. y
III. z

(a) I
(b) II
(c) III
(d) I & III
(e) I, II & III

PS: oops posted in the wrong forum, Moderator can you please move this to PS, will take care next time. Thanks
[Reveal] Spoiler: OA

Last edited by Bunuel on 20 Aug 2012, 10:37, edited 1 time in total.
Edited the question.

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Re: New Ques [#permalink]

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18 Jul 2010, 12:57
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smartmundu wrote:
If x,y and z are positive integers, and 30x=35y=42z, then which of the following must be divisible by 3?
I) x
II) y
III) z

(a) I
(b) II
(c) III
(d) I & III
(e) I, II & III

PS: oops posted in the wrong forum, Moderator can you please move this to PS, will take care next time. Thanks

$$3*10x=35y=3*14z=some \ multiple \ of \ 3$$ --> so the result is some multiple of 3 --> $$x$$ and $$z$$ may or may not be multiples of 3 ($$30x$$ or $$42z$$ already are multiples of 3 thus it's not necessary for $$x$$ or $$z$$ to be multiples of 3). But $$35y$$ to be multiple of 3, $$y$$ must be multiple of 3 (as 35 is not).

Answer: B.

With the same logic: $$z$$ must be multiple of 5, $$y$$ must be multiple of 2 (as it also must be multiple of 3 thus it must be multiple of 2*3=6) and $$x$$ must be multiple of 7.

Hope it's clear.
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Re: New Ques [#permalink]

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19 Jul 2010, 07:57
I did it wrong, i thought its x and Y which are divisible by 3.

I forgot about this

Bunuel wrote:

$$x$$ and $$z$$ may or may not be multiples of 3 ($$30x$$ or $$42z$$ already are multiples of 3 thus it's not necessary for $$x$$ or $$z$$ to be multiples of 3). But $$35y$$ to be multiple of 3, $$y$$ must be multiple of 3 (as 35 is not).

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Re: New Ques [#permalink]

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19 Jul 2010, 11:06
Bunuel as per you what level of question is this?

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Re: New Ques [#permalink]

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19 Jul 2010, 11:18
I'd say 600 level, not too hard.
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Re: New Ques [#permalink]

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19 Jul 2010, 15:10
here is my explanation:Is this approach correct?

30X=35y=42z

x=35y/30---now check whether x is divisible by 3--35y/30 is not divisible by 3
y=30X/35---divisible by 3
z=35y/42----not divisible by 3

hence y is only divisible by 3.

Ans:B

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Re: New Ques [#permalink]

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20 Aug 2012, 10:30
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Bunuel wrote:
smartmundu wrote:
If x,y and z are positive integers, and 30x=35y=42z, then which of the following must be divisible by 3?
I) x
II) y
III) z

(a) I
(b) II
(c) III
(d) I & III
(e) I, II & III

PS: oops posted in the wrong forum, Moderator can you please move this to PS, will take care next time. Thanks

$$3*10x=35y=3*14z=some \ multiple \ of \ 3$$ --> so the result is some multiple of 3 --> $$x$$ and $$z$$ may or may not be multiples of 3 ($$30x$$ or $$42z$$ already are multiples of 3 thus it's not necessary for $$x$$ or $$z$$ to be multiples of 3). But $$35y$$ to be multiple of 3, $$y$$ must be multiple of 3 (as 35 is not).

Answer: B.

With the same logic: $$z$$ must be multiple of 5, $$y$$ must be multiple of 2 (as it also must be multiple of 3 thus it must be multiple of 2*3=6) and $$x$$ must be multiple of 7.

Hope it's clear.

another way

$$30x= 35y = 42z$$

now from $$30x= 35y \Rightarrow x =\frac {35}{30}{y} \Rightarrow x = \frac{7}{6}{y}$$ $$(i)$$
and from $$30x = 42z \Rightarrow x = \frac{42}{30}{z} \Rightarrow x=\frac{7}{5}{z}$$ $$(ii)$$

hence whatever the values of Y and Z ( Both must be integers and must be chosen such so that x is also an integer ) , from $$(i)$$ and $$(ii)$$ we can see that X will always be a multiple of 7.

similarly for y ( $$Given 30x= 35y = 42z$$ )

$$35y = 42z \Rightarrow y = \frac 6 5z$$
and $$35y = 30x \Rightarrow y = \frac 6 7x$$

so for both the cases whatever the values of z and x( Both Integers and must be chosen such so that y is also an integer) y will always be a multiple of 6 and whatever is a multiple of 6 ( 0, 6,12,18....)
is also a multiple of 3, so y will always be a multiple of 3( and 6) no matter what values z and x take.( required answer)

now lets check z ( Given $$30x= 35y = 42z$$ )

we know $$30x= 42z$$
$$z= \frac 5 7x$$

and

$$42z = 35y$$
$$z=\frac 5 6y$$

so for both the cases above whatever the values of x and y ( Both integers and must be such so that z is also an integer) z must be a multiple of 5.
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Re: New Ques [#permalink]

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26 Aug 2012, 11:02
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Bunuel wrote:
smartmundu wrote:
If x,y and z are positive integers, and 30x=35y=42z, then which of the following must be divisible by 3?
I) x
II) y
III) z

(a) I
(b) II
(c) III
(d) I & III
(e) I, II & III

PS: oops posted in the wrong forum, Moderator can you please move this to PS, will take care next time. Thanks

$$3*10x=35y=3*14z=some \ multiple \ of \ 3$$ --> so the result is some multiple of 3 --> $$x$$ and $$z$$ may or may not be multiples of 3 ($$30x$$ or $$42z$$ already are multiples of 3 thus it's not necessary for $$x$$ or $$z$$ to be multiples of 3). But $$35y$$ to be multiple of 3, $$y$$ must be multiple of 3 (as 35 is not).

Answer: B.

With the same logic: $$z$$ must be multiple of 5, $$y$$ must be multiple of 2 (as it also must be multiple of 3 thus it must be multiple of 2*3=6) and $$x$$ must be multiple of 7.

Hope it's clear.

Hi bunuel,

I have slightly different and bit simpler way to solve this problem. Kindly have a look.
LCM of 30, 35, & 42 is 2.3.5.7
As 30x = 35y = 42z
Product of these parts is equal
In other words x has to be 7, y=6 & z =5
From this its very clear that y is a multiple of 3

I hope this method is simpler than the other methods.
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Re: If x,y and z are positive integers, and 30x=35y=42z, then [#permalink]

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05 Nov 2012, 05:06
30x = 35y = 42z

30 = 3*5*2

35 = 7*5

42 = 7*2*3

In order to sustain this equation, it is obvious that Y needs to hold a 3!

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Re: If x,y and z are positive integers, and 30x=35y=42z, then [#permalink]

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13 Sep 2014, 10:37
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Re: If x,y and z are positive integers, and 30x=35y=42z, then [#permalink]

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Re: If x,y and z are positive integers, and 30x=35y=42z, then [#permalink]

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13 Jan 2017, 05:24
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Re: If x,y and z are positive integers, and 30x=35y=42z, then [#permalink]

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13 Jan 2017, 07:52
smartmundu wrote:
If x,y and z are positive integers, and 30x=35y=42z, then which of the following must be divisible by 3?

I. x
II. y
III. z

(a) I
(b) II
(c) III
(d) I & III
(e) I, II & III

PS: oops posted in the wrong forum, Moderator can you please move this to PS, will take care next time. Thanks

$$30x = 35y = 42z = 210$$

So, $$x = 7$$ , $$y = 6$$ & $$z = 5$$

Check , only y is divisible by 3

Hence, answer will be (b) II....
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Re: If x,y and z are positive integers, and 30x=35y=42z, then [#permalink]

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03 Apr 2017, 21:45
30x=35y=42z
2*3*5*x= 5*7*y= 2*3*7*z

As these are equal, then the prime factors must be the same.
x must have 7 as a prime factor.
y must have 3 as a prime factor.
z must have 5 as a prime factor.

And Bunuel the powers of these prime factors would also be same. Right?

For example,
12= 3*4= 2^2*3
12=2*6= 2^2*3
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Re: If x,y and z are positive integers, and 30x=35y=42z, then   [#permalink] 03 Apr 2017, 21:45
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