Does work time of a task increase during scheduling? (preemptive scheduling)

this is something I found while studying an excercise which the result to was already given but something seems a bit off from what a studied so far:

In this picture you can see the scheduling table of a preemptive scheduling on 2 CPUs with 6 tasks to do and of course each one of them was given the time needed to finish the task and priorities.

And now the main question:

Shouldn't the remaining work time of task #1 be 4 when its work is continued in CPU #2 at t=7 ? In the picture it says that 6 out of 6 task was done although 6 of the originally 10 work time was done at t=0 which means that only 4 task is left to do. Is the excercise doing it wrong or did I miss something you should know about multi procossed scheduling?

(At first I thought it is because switching processor but if I look at task #4 it doesn't seem like thats the case)

Would really appreciate your opinion Thanks.

So yeah it turned out to be a mistake in the sheet. so task #1 is supposed to have only 4 task left by the time t = 7