CAT 1997 — QA Question 28

This question can be done by assuming some values for x, y and z, e.g. 

Let x = 4, y = 3 and z = 1.
la(4, 3, 1) = min(7, 4) = 4;
le(4, 3, 1) = max(1, 2) = 2; 
ma(x, y, z) = $\frac{1}{2}$ (4 + 2) = 3. 

Hence, we can reject options (a) & (c).

Let x = 10, y = 3 and z = 1.
la(10, 3, 1) = min(13, 4) = 4;
le(10, 3, 1) = max(7, 2) = 7; 
ma(x, y, z) = $\frac{1}{2}$ (4 + 7) = 5.5. 

Hence, we can reject option (b) also.

Hence, option (d).