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l/m+n=m/n+l

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l/m+n=m/n+l [#permalink] New post   17 Sep 2018, 20:27
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\(\frac{l}{m+n}=\frac{m}{n+l}=\frac{n}{l+m}=k\) , where k is a real number

Quantity A
Quantity B
k
\(\frac{1}{3}\)


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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A
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jarabhuiyan
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Re: l/m+n=m/n+l [#permalink] New post   23 Oct 2018, 14:35
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16. The given equation is l
m+ n
= m
n + l
= n
l + m
= k . Forming the three equations yields l = (m + n)k, m =
(n + l)k, n = (l + m)k. Summing these three equations yields
l + m + n = (m + n)k + (n + l)k + (l + m)k
= k[(m + n) + (n + l) + (l + m)]
= k(m + n + n + l + l + m)
= k(2m + 2n + 2l)
= 2k(m + n + l)
1 = 2k by canceling m + n + l from each side
1/2 = k
Hence, Column A equals 1/2. Since 1/2 is greater than 1/3, Column A is greater than Column B, and the
answer is (A).
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Re: l/m+n=m/n+l [#permalink] New post   23 Oct 2018, 09:46
Can you provide with explanation to this problem?
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Re: l/m+n=m/n+l [#permalink] New post   02 Nov 2018, 10:36
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jarabhuiyan wrote:
16. The given equation is l
m+ n
= m
n + l
= n
l + m
= k . Forming the three equations yields l = (m + n)k, m =
(n + l)k, n = (l + m)k. Summing these three equations yields
l + m + n = (m + n)k + (n + l)k + (l + m)k
= k[(m + n) + (n + l) + (l + m)]
= k(m + n + n + l + l + m)
= k(2m + 2n + 2l)
= 2k(m + n + l)
1 = 2k by canceling m + n + l from each side
1/2 = k
Hence, Column A equals 1/2. Since 1/2 is greater than 1/3, Column A is greater than Column B, and the
answer is (A).


You, are incredible!! Thanks!! This opened up my eyes to a whole new way of problem solving!!!
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l/m+n=m/n+l [#permalink] New post   04 Nov 2018, 02:20
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Carcass wrote:
\(\frac{l}{m+n}=\frac{m}{n+l}=\frac{n}{l+m}=k\) , where k is a real number

Quantity A
Quantity B
k
\(\frac{1}{3}\)


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.


\(\frac{l}{m+n}=\frac{m}{n+l}=\frac{n}{l+m}=k\)....
three equations can be formed :-
    \(\frac{l}{m+n}=k........l=(m+n)k\)
    \(\frac{m}{l+n}=k........m=(l+n)k\)
    \(\frac{n}{m+l}=k........n=(m+l)k\)

add the three equations...
\(l+m+n=(m+n)k+(l+n)k+(m+l)k...........l+m+n=(m+n+l+n+m+l)k..........
l+m+n=2(m+n+l)k.........2k=1.....k=\frac{1}{2}=A\) and \(B=\frac{1}{3}\) thus \(A>B\)

A
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Re: l/m+n=m/n+l [#permalink] New post   09 Nov 2018, 15:48
jarabhuiyan wrote:
16. The given equation is l
m+ n
= m
n + l
= n
l + m
= k . Forming the three equations yields l = (m + n)k, m =
(n + l)k, n = (l + m)k. Summing these three equations yields
l + m + n = (m + n)k + (n + l)k + (l + m)k
= k[(m + n) + (n + l) + (l + m)]
= k(m + n + n + l + l + m)
= k(2m + 2n + 2l)
= 2k(m + n + l)
1 = 2k by canceling m + n + l from each side
1/2 = k
Hence, Column A equals 1/2. Since 1/2 is greater than 1/3, Column A is greater than Column B, and the
answer is (A).


Thank you so much for this.
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kunalkmr62
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Re: l/m+n=m/n+l [#permalink] New post   27 Nov 2018, 10:57
I tried the following approach and got k = -1 , can someone explain where I went wrong.
l(n+l)=m(m+n) { cross multiplying the first 2 terms}
ln+lpow2 = mpow2 +mn
n(l-m)= (m+l)*(m-l)
n= -(m+l)

Substituting this value in the last equation, yields k=-1 , and hence B>A
@chetan2u
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chetan2u
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Re: l/m+n=m/n+l [#permalink] New post   27 Nov 2018, 18:08
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kunalkmr62 wrote:
I tried the following approach and got k = -1 , can someone explain where I went wrong.
l(n+l)=m(m+n) { cross multiplying the first 2 terms}
ln+lpow2 = mpow2 +mn
n(l-m)= (m+l)*(m-l)
n= -(m+l)

Substituting this value in the last equation, yields k=-1 , and hence B>A
@chetan2u



You cannot cancel out l-m from both sides..
n(l-m)= (m+l)*(m-l).....
n(l-m)-(m+l)(m-l)=0
n(l-m)+(m+l)(l-m)=0
(l-m)(n+m+l)=0
So either l=m or n=-(m+l)
Now you do not know which one is correct.
If l=m, then also (l-m)(n+m+l)=0 and then n+m+l is not necessarily equal to 0
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Re: l/m+n=m/n+l [#permalink] New post   19 Jul 2021, 10:00
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As nothing is mentioned about l,m,n , they can be fraction or integer or any real number which satisfies the equation
\(\frac{l}{m+n}=\frac{m}{n+l}=\frac{n}{l+m}\)

The above equation is satisfied only when l=m=n

Hence the least value either \(\frac{l}{m+n}\) or \(\frac{m}{n+l}\) or \(\frac{n}{l+m}\) can take is \(\frac{1}{2}\)

So Quantity A = \(\frac{1}{2}\) and Quantity B is \(\frac{1}{3}\)

Therefore the answer is Quantity A is greater
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Re: l/m+n=m/n+l [#permalink] New post   20 May 2022, 06:15
What about l = -4 and m = n = 2?

l / (m + n) = -4 / 4 = -1
m / (n + l) = 2 / (2 - 4) = -1
n / (m + l) = 2 / (2 - 4) = -1

what about this case?
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Re: l/m+n=m/n+l [#permalink] New post   20 May 2022, 06:16
santosh93 wrote:
As nothing is mentioned about l,m,n , they can be fraction or integer or any real number which satisfies the equation
\(\frac{l}{m+n}=\frac{m}{n+l}=\frac{n}{l+m}\)

The above equation is satisfied only when l=m=n

Therefore the answer is Quantity A is greater


What when l = -4 and m = n = 2?
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Re: l/m+n=m/n+l [#permalink] New post   20 May 2022, 07:18
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here the substitution method does not work best because could lead to a conflict answers

The best is to boil down the equality and see if k is > or < 1/3

regards
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l/m+n=m/n+l [#permalink] New post   20 May 2022, 17:22
Very simple.

Just use Componendo, Dividendo property, and you'll come up with l=m=n.
Hence K which is = l/(m+n) gets equal to 1/2.

Hence A is greater.
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