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In the fraction x/y, where x and y are positive integers, wh [#permalink]
 29 Jan 2012, 16:58
Question Stats:
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63% (00:52) wrong based on 2 sessions
In the fraction x/y, where x and y are positive integers, what is the value of y? (1) The least common denominator of x/y and 1/3 is 6 (2) x=1 For me the answer is C.
Statement 1 is clearly insufficient because x can take any value. Value of y has to be 6 though for LCD to be 6.
Statement 2 gives us the value of x.
Combining the two the value of y should be 6. Am I right? _________________
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Last edited by Bunuel on 23 Dec 2012, 05:45, edited 2 times in total.
Added the OA
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Re: In the fraction x/y, where x and y are positive integers, wh [#permalink]
05 Jan 2013, 02:32
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Sachin9 wrote: Bunuel wrote: Sachin9 wrote: So how did you derive the following?
The least common denominator of x/y and 1/3 is 6 --> LCM of y and 3 is is 6 The least common denominator of x/y and 1/3 is 6 --> the least common multiple of y and 3 is is 6. I am still not understanding how you derived the following : The least common denominator of x/y and 1/3 is 6 --> the least common multiple of y and 3 is is 6. Please elaborate.. Responding to a pm: It's pretty simple really and I don't know what to add to Bunuel's post. Are you comfortable working with fractions? If yes, what do you do to add two fractions say 1/2 and 1/3? First, you give them a common denominator (6 in this case). Why 6? Because it is a common multiple of 2 and 3. Similarly, what if the two fractions are 1/4 and 1/6? What common denominator will you take? You can take the LCM of 4 and 6 which is 12 or you can even multiply 4 and 6 to get 24. 1/4 + 1/6 = 3/12 + 2/12 = 5/12 or 1/4 + 1/6 = 6/24 + 4/24 = 10/24 = 5/12 If you need to find the LOWEST common denominator, it will be the LCM. So in the question above, the least common denominator of x/y and 1/3 is 6 implies that the least common multiple of y and 3 is 6.
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Re: Least Common Denominator [#permalink]
29 Jan 2012, 17:02
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Re: In the fraction x/y, where x and y are positive integers, wh [#permalink]
29 Jan 2012, 17:07
Sorry - careless mistake at my part. Thanks for putting it right.
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Re: In the fraction x/y, where x and y are positive integers, wh [#permalink]
29 Jan 2012, 17:08
enigma123 wrote: In the fraction x/y, where x and y are positive integers, what is the value of y?
(1) The least common denominator of x/y and 1/3 is 6
(2) x=1
For me the answer is C.
Statement 1 is clearly insufficient because x can take any value. Value of y has to be 6 though for LCD to be 6.
Statement 2 gives us the value of x.
Combining the two the value of y should be 6. Am I right? I don't quite follow your logic. If you thought that, from Statement 1 alone, you could be certain that y = 6, you should have selected the answer A, since the question asks for the value of y. It doesn't ask for the value of x. The answer is not A, however, since using both statements, x/y could be equal to 1/2 or it could be equal to 1/6, so y could be 2 or 6, and the answer is E.
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Re: In the fraction x/y, where x and y are positive integers, wh [#permalink]
29 Jan 2012, 21:33
enigma123 wrote: In the fraction x/y, where x and y are positive integers, what is the value of y?
(1) The least common denominator of x/y and 1/3 is 6
(2) x=1
For me the answer is C.
Statement 1 is clearly insufficient because x can take any value. Value of y has to be 6 though for LCD to be 6.
Statement 2 gives us the value of x.
Combining the two the value of y should be 6. Am I right? I could not get the question here... does here the statement least common denominator of x/y and 1/3 means lcm(x,y)/hcf(x,y) (if f1 = num1/den1 and f2 = num2/den2 are both fractions in their simplest form, then LCM(f1,f2) = LCM(num1.num2)/GCD(den1,den2)).
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Re: In the fraction x/y, where x and y are positive integers, wh [#permalink]
30 Jan 2012, 02:39
subhajeet wrote: enigma123 wrote: In the fraction x/y, where x and y are positive integers, what is the value of y?
(1) The least common denominator of x/y and 1/3 is 6
(2) x=1
For me the answer is C.
Statement 1 is clearly insufficient because x can take any value. Value of y has to be 6 though for LCD to be 6.
Statement 2 gives us the value of x.
Combining the two the value of y should be 6. Am I right? I could not get the question here... does here the statement least common denominator of x/y and 1/3 means lcm(x,y)/hcf(x,y) (if f1 = num1/den1 and f2 = num2/den2 are both fractions in their simplest form, then LCM(f1,f2) = LCM(num1.num2)/GCD(den1,den2)). The least common denominator of x/y and 1/3 is 6 means the least common multiple of y and 3 is 6. Hope it's clear.
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Re: In the fraction x/y, where x and y are positive integers, wh [#permalink]
22 Dec 2012, 22:54
Bunuel, how to find the LCM and GCD of fractions?
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Re: In the fraction x/y, where x and y are positive integers, wh [#permalink]
23 Dec 2012, 05:48
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Re: In the fraction x/y, where x and y are positive integers, wh [#permalink]
23 Dec 2012, 09:53
So how did you derive the following? The least common denominator of x/y and 1/3 is 6 --> LCM of y and 3 is is 6
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Re: In the fraction x/y, where x and y are positive integers, wh [#permalink]
24 Dec 2012, 00:46
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Re: In the fraction x/y, where x and y are positive integers, wh [#permalink]
04 Jan 2013, 05:28
Bunuel wrote: Sachin9 wrote: So how did you derive the following?
The least common denominator of x/y and 1/3 is 6 --> LCM of y and 3 is is 6 The least common denominator of x/y and 1/3 is 6 --> the least common multiple of y and 3 is is 6. I am still not understanding how you derived the following : The least common denominator of x/y and 1/3 is 6 --> the least common multiple of y and 3 is is 6. Please elaborate..
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Re: In the fraction x/y, where x and y are positive integers, wh [#permalink]
04 Jan 2013, 05:34
Sachin9 wrote: Bunuel wrote: Sachin9 wrote: So how did you derive the following?
The least common denominator of x/y and 1/3 is 6 --> LCM of y and 3 is is 6 The least common denominator of x/y and 1/3 is 6 --> the least common multiple of y and 3 is is 6. I am still not understanding how you derived the following : The least common denominator of x/y and 1/3 is 6 --> the least common multiple of y and 3 is is 6. Please elaborate.. Not sure how to elaborate this further. We are told that the least common denominator of x/y and 1/3 is 6. The denominators of these fractions are y and 3, so we are basically told that the least common multiple of y and 3 is 6. It's quite straightforward.
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Re: In the fraction x/y, where x and y are positive integers, wh [#permalink]
04 Jan 2013, 05:56
Bunuel wrote: Not sure how to elaborate this further.
We are told that the least common denominator of x/y and 1/3 is 6. The denominators of these fractions are y and 3, so we are basically told that the least common multiple of y and 3 is 6.
It's quite straightforward. I am new here, but I will try to explain my way of thinking. Sorry if I get some terms wrong, English is not my primary language. Remember the way to add up 2 fractions with different denominators: you should look for the smallest (least) common denominators (and muliply the nominators). x/y + p/q = (x*q + p*y)/(y*p) Also you would show y and p in factors (prefferaby prime numbers) to find the least denominator, but because 3 is a prime number and den(y*3) = 6 = 3*2 you can skip this step. The rest of the question is explained  Y can be 2 or 6...
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Re: In the fraction x/y, where x and y are positive integers, wh [#permalink]
05 Jan 2013, 02:46
I get it now. Thanks a lot Karishma
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In the fraction x/y, where x and y are positive integers, wh [#permalink]
22 Jan 2013, 00:14
1. LCD of x/y and 1/3 is 6
LCD(3,2) = 6
LCD(3,6) = 6
y could be 2 or 6.
INSUFFICIENT.
2. x = 1 ... This doesn't give us anything much. x/y could be 1/7 or 1/8
INSUFFICIENT.
Together: 1/y and 1/3 has an LCD of 6
LCD(1/2,1/3) is 6
LCD(1/6,1/3) is 6
INSUFFICIENT.
Answer: E
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In the fraction x/y, where x and y are positive integers, what i [#permalink]
10 Feb 2013, 03:31
It is E. Statement 1: Since the least common denominator for 1/y and 1/3 is 6, there are two possible solutions. y can be 2 or 6. INSUFFICIENT Statement 2: x=1, doesn't say anything about y. INSUFFICIENT Together statements are insufficient and do not reveal any new information.
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In the fraction x/y, where x and y are positive integers, what i
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