Zeal Test Series 2019: Operating System - Process Schedule

Question

What should be less time consuming approach for such type of questions

Comments

Suppose burst times of A, B and C are $T_A,T_B,T_C$ respectively.
According to ques.-
$T_A = 6+x$. (Here $x$ is A's remaining time .)
A ran for 6-time units and got preempted and B got CPU)
This is possible only when $x > T_B$
Now,
$T_B = 2+y$. (Here $y$ is B's remaining time .)
B ran for 2-time units and got preempted and C got CPU)
This is possible only when $y > T_C$

So we have-
$T_C = 4\\T_B=2+y\\y >T_C \ so \ y >4\\T_A=6+x\\
x >T_B$
So minimum value of y can be 5 so minimum value of $T_B=2+5=7.$
 Minimum value of x can be 8 so minimum value of $T_A=6+8=14$

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If D is the correct option then T_A-6 =5 and T_B=7 (given) i.e. remaining time of A is less than B's BT. Then how could B have pre empted A?

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yes Non of the options are correct