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The least common multiplier of A and B is 120, the ratio of [#permalink]
 09 Jan 2009, 20:52
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57% (02:00) correct
42% (02:39) wrong based on 3 sessions
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
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I did not get how a = 120 x 3 and b = 120 x4. Pmenon, did you get it ? If yes, can you explain please?
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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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krishan wrote: 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. you got it.. I thought something else..... I edited/deleted my post..
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anyone want to take a stab at this ? 
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krishan wrote: 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. I agree with you. We have one 3 as divisor in 120 and one 4 as divisor in 120 => So, A=30, B=40. GCD=10
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How do you get x= 4 and y=3 from the two equations? ( as the least possible values ?)
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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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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 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 = 10Hence the greatest common divisor is 10
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cicerone wrote: 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 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 = 10Hence the greatest common divisor is 10 Yup, I got there the same way.
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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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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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thank you ... that was helpful
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So just a question Greatest Common divisor is equal to common factor in A and B???
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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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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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peraspera wrote: cicerone wrote: 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 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 = 10Hence the greatest common divisor is 10 Yup, I got there the same way. Thanks for a smart and efficient way!
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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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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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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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