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If x,y and z are positive integers, and 30x=35y=42z, then
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Updated on: 20 Aug 2012, 11:37
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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
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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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18 Jul 2010, 13:57
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
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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
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19 Jul 2010, 12:06
Bunuel as per you what level of question is this?



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Re: New Ques
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19 Jul 2010, 12:18



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Re: New Ques
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19 Jul 2010, 16:10
here is my explanation:Is this approach correct?
30X=35y=42z
x=35y/30now check whether x is divisible by 335y/30 is not divisible by 3 y=30X/35divisible by 3 z=35y/42not divisible by 3
hence y is only divisible by 3.
Ans:B



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Re: New Ques
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20 Aug 2012, 11:30
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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26 Aug 2012, 12:02
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
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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
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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
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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 &nbs
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