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The least common multiplier of A and B is 120, the ratio of

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The least common multiplier of A and B is 120, the ratio of [#permalink] New post 09 Jan 2009, 19:52
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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
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Re: PS: LCM [#permalink] New post 09 Jan 2009, 21:53
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 [#permalink] New post 19 Jan 2009, 06:41
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 [#permalink] New post 28 Mar 2009, 20:08
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 [#permalink] New post 29 Mar 2009, 05:28
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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 = 10

Hence the greatest common divisor is 10
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Re: PS: LCM [#permalink] New post 01 Dec 2009, 00:11
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 [#permalink] New post 05 Apr 2010, 10:59
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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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Re: PS: LCM [#permalink] New post 17 Aug 2010, 15:33
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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

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Re: PS: LCM [#permalink] New post 17 Nov 2010, 09:58
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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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Re: PS: LCM [#permalink] New post 20 Nov 2010, 14:44
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 [#permalink] New post 22 May 2011, 06:24
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 [#permalink] New post 23 May 2011, 13:01
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 [#permalink] New post 23 May 2011, 21:54
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A/B = 3/4

3 * GCD * 4 * GCD = 120 * GCD

=> GCD = 120/12 = 10
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Re: PS: LCM [#permalink] New post 24 May 2011, 01:15
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=hence-what's left from 120 is 2*5=10
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Re: PS: LCM [#permalink] New post 18 Jun 2011, 08:25
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 [#permalink] New post 03 Oct 2013, 18:29
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Re: The least common multiplier of A and B is 120, the ratio of   [#permalink] 03 Oct 2013, 18:29
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