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The least common multiplier of A and B is 120, the ratio of
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09 Jan 2009, 20:52
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76% (00:47) correct 24% (02:09) wrong based on 74 sessions
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The least common multiplier of A and B is 120, the ratio of A and B is 3:4, what is the largest common divisor? Can someone break this down step by step for me ? I find that Im having some difficulty in understanding the concept behind these types of questions. Thanks in advance == Message from the GMAT Club Team == THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION. This discussion does not meet community quality standards. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.



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Re: PS: LCM
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09 Jan 2009, 22:53
Tiger, I think your solution is not correct. The problem says that LCM of a and b = 120 and from your calculations, a = 360 , b = 480. How can the numbers be greater than their Least Common Multiple? Here is my solution: a x X = 120 (1) b x Y = 120 (2) Also, a/b = 3/4 Doing (1) / (2) X/Y x 3/4 = 1 > 3X = 4Y > X = 4 and Y = 3 (I am taking the least possible values for X and Y) From (1) , a x 4 = 120 > a = 30 From (2), b x 3 = 120 > b = 40 Therefore HCF of a and b = 10.
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Re: PS: LCM
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19 Jan 2009, 07:41
How do you get x= 4 and y=3 from the two equations? ( as the least possible values ?)



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Re: PS: LCM
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28 Mar 2009, 21:08
Economist wrote: How do you get x= 4 and y=3 from the two equations? ( as the least possible values ?) 3X = 4Y .. The equation holds good when X = 4 and Y =3 . With these values: LHS = 3 x 4 =12 RHS = 4 x 3 = 12 > LHS = RHS for X =4 and Y = 3. Hope it is clear now.
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Re: PS: LCM
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01 Dec 2009, 01:11
Not sure if i did this correctly, but I got 10 as well. 120 broken down into its prime factors is 2^3 x 5 x 3 If the ratio is 3:4, the number must be divisible by 3, 4(2^2) That leaves 2^1 x 5 in the prime factorization = 10 Any thoughts on this approach?
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Re: PS: LCM
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05 Apr 2010, 11:59
Hey Mustafaj, Your approach is very correct and kind of logical too compared to other apporaches. Thanks, mustafaj wrote: Not sure if i did this correctly, but I got 10 as well.
120 broken down into its prime factors is 2^3 x 5 x 3
If the ratio is 3:4, the number must be divisible by 3, 4(2^2) That leaves 2^1 x 5 in the prime factorization = 10
Any thoughts on this approach?
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Re: PS: LCM
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17 Aug 2010, 16:33
Very simple approach. No equations involved. all the prime factors of 120: 2,3,4,5. A:B=3:4 means that 2 and 5 are factors of both A and B So, factors of A are 3,2,5 anf factors of B are 4,2,5. looking at both, it is clear that 2 and 5 are factors of both, so LCD=5x2=10 E
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Re: PS: LCM
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17 Nov 2010, 10:58
Hello there everyone,
I'm not sure if anybody's still interested in this question, but I found the solutions posted earlier a bit tedious to think up in under two minutes. This got me thinking a bit more deeper into what the question's actually telling; and here's how I reasoned an answer:
Given: LCM (A,B) = 120 That bit's telling that for the two numbers A and B:
The [Common multiples between (A,B)] x [Uncommon multiples between them] = 120 That's the wordy definition of LCM right? Take out all that's common, and multiply them with whatever uncommon remains and you'll get the LCM between A and B
Now, what's worth noting in this equation is that the first part of it [Common multiples (A,B)] is simply their GCF; So, 120 = GCF (A,B) x [Uncommon Factors of (A,B)]
I got this far in my thought process, and then gave up, because I thought there wasn't any information on what's uncommon between A,B.. But wait! The question says that A/B = 3/4 ... that's as good as saying that 3 and 4 are the only factors that would remain if I divided A and B  3 and 4 are the Uncommon factors between A and B
So, using this bit of information, now you can solve for the GCF:
120 = GCF (A,B) x 3 x 4 120 = GCF x 12 GCF = 10
From this I realized something really simple but which was not obvious to me:
LCM / GCF = product of uncommon factors; if you are given the ratio between the numbers, then each value in the ratio is an uncommon factor belongs to one of the numbers [i.e if two numbers are in the ratio 15:16, then their GCFx15x16 would give you their LCM]
Hope that helps! Raj
ps: excuse me if you find any errors/ if i am not clear, but as you may notice below my username, this is my first post!



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Re: PS: LCM
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20 Nov 2010, 15:44
Werewolf wrote: peraspera wrote: cicerone wrote: The least common multiplier of A and B is 120, the ratio of A and B is 3:4, what is the largest common divisor?
Can someone break this down step by step for me ? I find that Im having some difficulty in understanding the concept behind these types of questions. Thanks in advance
Given LCM = 120 and the ratio of the numbers = 3:4. We need to find the GCD.
Let the numbers be 3x and 4x. So it is clear that their GCD = x.
We have,
\(LCM \times GCD = Product \quad of \quad the \quad two \quad numbers\)
\(\Rightarrow 120 \times x = 3x \times 4x\)
\(\Rightarrow x = 10\)
Hence the greatest common divisor is 10 Yup, I got there the same way. Thanks for a smart and efficient way! +1 , Thats indeed a smart way . A concept can be used to solve this problem When you say LCM of A and B (for example A=3 and B=4 ) then we know for sure that the LCM of A and B (which is 120 here) will be divisible by both A and B ,that give us no1= 120/A=120/3= 40 & no2= 120/B=120/4 =30 GCF(40,30)= 10 .
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Re: PS: LCM
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22 May 2011, 07:24
pmenon wrote: The least common multiplier of A and B is 120, the ratio of A and B is 3:4, what is the largest common divisor?
Can someone break this down step by step for me ? I find that Im having some difficulty in understanding the concept behind these types of questions. Thanks in advance This can be solved using the principal of LCM X HCF = Product of the numbers. LCM = 120 Ratio is 3:4. So let the numbers be 3x and 4x. Now, when we multiply numerator and denominator with x , the variable x becomes the HCF, thinking logically. Because cancelling the greatest factor we turn up to 3:4 which cannot be simplified further. So finally we have 3x*4x = x*120 which gives x = 10. Ans : 10
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Re: PS: LCM
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23 May 2011, 14:01
Let X be the highest common factor
So one number be 3X and the other is 4X
If there are two numbers , say 12 (3*4) and 28 (7*4) whose highest common factor/divisor is 4 , then the LCM is 3*7*4 = 84
Lets say another example , if there are two numbers , say 10 (5*2) and 14 (7*2) Then the LCM is 5*7*2 = 70
Thus following the same logic in this question , 3 * 4 * X = 120
So X = 10
This could have been also solved by the concept of interpreting HCF and LCM using Venn Diagrams but i am unable to depict the diagram here
Hope this clears the problem



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Re: PS: LCM
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23 May 2011, 22:54
A/B = 3/4 3 * GCD * 4 * GCD = 120 * GCD => GCD = 120/12 = 10
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Re: PS: LCM
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24 May 2011, 02:15
a=3*x b= 4(2^2)*y a*b= 3*2^3*5(120)*x*y (hence the factor will be less than x*y we can ignore the xy)
a has the 3 from 120 and b has the 2^2 from 120=hencewhat's left from 120 is 2*5=10



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Re: PS: LCM
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18 Jun 2011, 09:25
hope i got it right: 120x=4x*3x
120 is 2*2*2*3*5 the numbers ar 3/4 means 3/2*2 hence 2*5 will be the solution



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Re: The least common multiplier of A and B is 120, the ratio of
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12 Nov 2014, 02:50
Ratio = \(\frac{3}{4}\) LCM = 120 = 3 * 4 * 10Just observe: Ignore the numbers in ratio (3 & 4) They are already in LCM. What remains is 10 Answer = 10
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