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chetan2u I am not sure you can assume:
(m-n)/(n-m) =-1
If both m & n are negative then the division would lead to +1.

If such scenario is possible, then, there is more than one result possible and A can't be right.
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chetan2u I am not sure you can assume:
(m-n)/(n-m) =-1
If both m & n are negative then the division would lead to +1.

If such scenario is possible, then, there is more than one result possible and A can't be right.


\(\frac{(m-n)}{(n-m)}=\frac{-(n-m)}{(n-m)}=-1\)

so what ever be the values of m and n, ans will be -1.... ofcourse the denominator cannot be 0 that is \(n-m\neq{0}..n\neq{m}\)

take for example n=-3 and m=-2...
\(\frac{(-2-(-3))}{(-3-(-2))}=\frac{-2+3}{-3+2}=\frac{1}{-1}=-1\)
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Binggm14
As we have 2 variable and 2 equations..it's c

study mode

chetan2u
Binggm14
As we have 2 variable and 2 equations..it's c

study mode


Don't go with assumption always..
work on the equation..

Quote:
What is the numerical value of the expression \(\frac{(3m - n)}{(n - m)}\)?
\(\frac{(3m - n)}{(n - m)}=\frac{2m+m-n}{n-m}=\frac{2m}{n-m}+\frac{m-n}{n-m}=\frac{2m}{n-m}-1\)

(1) \(\frac{2m}{(n - m)} = \frac{7}{3}\)..
exactly what we are looking for..
\(\frac{2m}{n-m}-1=\frac{7}{3}-1=\frac{4}{3}\)
suff

(2) n – m = 6
nothing much
insuff

A
Okay thanks ..yeah ..it's A

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