Mo2men
amanvermagmat
a, b, c and N are four real numbers such that:
a/(b+c) = b/(c+a) = c/(a+b) = N. Also M is a positive number.
Is (M+N)/2 an odd integer?
(1) M = 3/2
(2) M = K/2, where K is an odd integer.
Hi
amanvermagmatAre you sure from the OA.
(1) M = 3/2
Let a=b=c=1...N= 1/2
M+N = 1/2 + 3/2= 4/2 = 2.........(M+N)/2 = 1..........Answer is Yes
Let a=1 b= 2 c=-3...N= -1
M+N = -1+ 3/2 = 1/2.........(M+N)/2 = 1/4................Answer is No
Insufficient
Hello
Thanks for your query. I have again edited the question in light of this new information.
Well -if three or more ratios are equal, say p/q = r/s = t/u = .... (and so on)
then each of these ratios is also equal to = (p+r+t+...)/(q+s+u+...) , i.e., each of these ratios is equal to the ratio of sum of numerators and the sum of denominators.
So in my question: a/(b+c) = b/(c+a) = c/(a+b) = N.. Here N is also = (a+b+c)/(b+c+c+a+a+b) = (a+b+c)/2(a+b+c) .
Since a+b+c is coming in denominator, we CANNOT take those values where a+b+c = 0.
In your second example: a=1 b= 2 c=-3... a+b+c = 0 which is not allowed.
This should have been mentioned in the question, so I am editing now. Thanks for this query.