Please solve q 31?

Question

1107 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-15T16:00:15

The answer is $"option (d)"$

$"Question (31)"$

Let

$a/(b+c)=k$ , $=>$ , $a=(b+c)k$

$b/(c+a)=k$ , $=>$ , $b=(a+c)k$

$c/(a+b)=k$ , $=>$ , $c=(a+b)k$

Therefore,

$(a+b+c)=(b+c)k+(a+c)k+(a+b)k$

$(a+b+c)=(bk+ck)+(ak+ck)+(ak+bk)$

$(a+b+c)=2k(a+b+c))$

$k=1/2$

Similarly if $k'=1/k$

$k'=2$

The answer is $"option (d)"$