Quote:
Originally Posted by sandeep jha
Hi puys,
can anyone solve the following problem for me....cudn't make much sense of the solution provided.....
Quant Question # 143 ---------------------------------------------------------
If a, b, c are real numbers such that a < b < c and a + b + c = 6, ab + bc + ca = 9, then which among the following is definitely true?
(a) 0 < a < 1 (b) 1 < b < 3 (c) 3 < c < 4 (d) All of them
Thanks...
|
trying anothwer method..
a= 6-B-C
so
bc + a(6-a)= 9
a^2 -6a +9 -bc=0
a to be real , 36 - 36 + 4bc >=0 or 4bc > 0 and same thing will work for b and c
but a = 3 - sqrt(bc)<b and c,
so (a) rejected. so (d) rejected.
since c can be more than 4 so (c) also rejected.
So (b) is the only soln.
I think this is the righ t one