Question

If $A + B > C + D$ and $B + C > A + D$ , then it is definite that
A.$D > B$
B.$C > D$
C.$A > D$
D.$B > D$

Answer 1

Hint: For solving this particular question, we have to take the given expressions then add the expressions. After adding and simplifying we get the required result. While adding we have to keep in mind that the left side of the expression must be added with the left side of the other expression and right with the right side only.

Complete step by step solution:
In this particular question , it is given that ,
$
  A + B > C + D, \\
  B + C > A + D. \\
 $
let us denote the given expressions as ,
$
  A + B > C + D...........(1) \\
  B + C > A + D.......(2) \\
 $
Now , add equation one with equation two . we get ,
$ \Rightarrow A + 2B + C > A + C + 2D$
Now , simplify the above expression.
$ \Rightarrow 2B > 2D$
Now , divide the above expression by two. We get,
$ \Rightarrow B > D$

Therefore , we can say that option $D$ is the correct option for this particular question.

Note:
If it is given that $A > B$ , then when we add the same number let say $C$ both the sides of inequality as $A + C > B + C$ , then there is no change in the direction of the symbol, it will remain the same.
Similarly , if it is given that $A > B$ , then when we subtract the same number let say $C$ both the sides of inequality as $A - C > B - C$ , then there is no change in the direction of the symbol, it will remain the same.
Similarly , if it is given that $A > B$ , then when we multiple positive number let say $C$ both the side of inequality as $A * C > B * C$ , then there is no change in the direction of the symbol, it will remain the same.