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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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New post Updated on: 20 Aug 2012, 10:37
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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

Originally posted by smartmundu on 18 Jul 2010, 12:23.
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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New post 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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New post 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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New post 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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New post 19 Jul 2010, 11:18
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Re: New Ques  [#permalink]

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New post 19 Jul 2010, 15:10
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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, 10:30
1
1
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, 11:02
4
1
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, 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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New post 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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New post 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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If x,y and z are positive integers, and 30x=35y=42z, then  [#permalink]

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New post 13 Dec 2018, 21:01
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




First take out LCM of 30,35 and 42....which is equal to 210
Next divide all three by 210, i.e.,
\(\frac{30x}{210}\)=\(\frac{35y}{210}\)=\(\frac{42z}{210}\)
=> \(\frac{x}{7}\)=\(\frac{y}{6}\)=\(\frac{z}{5}\)

Now lets equate these with a constant, i.e.,
let, \(\frac{x}{7}\)=\(\frac{y}{6}\)=\(\frac{z}{5}\)=K
=> x=7K , y=6K and z=5K

As K is a constant, we can say that y must be divisible by 3.
Hope it helps.
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If x,y and z are positive integers, and 30x=35y=42z, then &nbs [#permalink] 13 Dec 2018, 21:01
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