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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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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

Originally posted by smartmundu on 18 Jul 2010, 13:23.
Last edited by Bunuel on 20 Aug 2012, 11:37, edited 1 time in total.
Edited the question.
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Re: New Ques [#permalink]

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New post 18 Jul 2010, 13:57
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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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New post 19 Jul 2010, 08:57
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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New post 19 Jul 2010, 12:06
Bunuel as per you what level of question is this?
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Re: New Ques [#permalink]

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New post 19 Jul 2010, 12:18
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Re: New Ques [#permalink]

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New post 19 Jul 2010, 16:10
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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New post 20 Aug 2012, 11:30
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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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New post 26 Aug 2012, 12:02
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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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New post 05 Nov 2012, 06:06
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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New post 13 Jan 2017, 08:52
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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New post 03 Apr 2017, 22:45
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, 22:45
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