shaderon wrote:
Asifpirlo wrote:
If a, b, and c are consecutive positive odd integers, not necessarily in that order, which of the following must be true?
I. a + b > c
II. bc > a
III. (a + c)^2 > b
A. I only
B. II only
C. III only
D. I and II only
E. I and III only
Can someone explain what the highlighted part means?
What are the possibilities that one would need to consider?
Hi Shaderon,
The same order would mean that
a < b < c or
being precise : b = a + 2
c = b + 2 = a + 4, where a is a positive odd integer
However as they are not consecutive in the same order so, above inequality a<b<c does not hold true.
i.e., any case is possible :
a < b < c or
a < c < b or
b < a < c or
b < c < a or
c < a < b or
c < b < aNow,definitely we should consider the boundary case values to find how the inequality behaves.
Boundary case : taking odd numbers 1,3,5
I. a + b > c
put a =1,b=3,c=5 ;does not work
II. bc > a
put b=1,c=3,a=5 ;does not work
III. (a + c)^2 > b
put a=1,c=3,b=5 ; will work
We can reject inequalities I and II.
It is pretty clear, only inequality III is only left which holds true as shown above.
Looking at the answer options,
correct option is
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