Master's work 15-5 = 10 days.

What the master did on 12- 10-2 is equivalent to what the apprentice did on 15- 12 = 3 days.

Therefore, the master completed the demand of the apprentice 15 days.

15 ÷ 3× 2 = 10 days,

Master needs to do it alone 10+ 10 = 20 days.

The second question:

A man's day is a working day.

The whole project needs 50× 30 = 1500 person-days.

24× 10 = 240 working days.

The task has 1500-240 = 1260 working days.

The time is 30- 10 = 20 days.

The number of workers needed is 1260 ÷ 20 = 63.

The third question:

If both piles are transported 1/5, the remaining 760 ÷ (1- 1/5) = 608 tons.

Only 608-592 = 16 tons is left.

It was caused by the fact that pile A was transported more than1/4-1/5 =1/20.

So reactor A is 16 ÷ 1/20 = 320 tons.

Pile B 760-320 = 440 tons.

The fourth question:

Team b completed 12 days = 3/5.

Team A finished 1-3/5 = 2/5.

Team A did 15× 2/5 = 6 days.

Shutdown 12-6 = 6 days.

The fifth question:

If both boys and girls lose 1/5, then this semester should be

750× (1- 1/5) = 600 people.

The difference of 7 10-600 = 1 10 is caused by the number of boys.

Equivalent to1/5+1/6 =11/30 of the number of boys last semester.

So the number of boys last semester was110 ÷11/30 = 300.

The number of boys this semester is 300× (1+ 1/6) = 350.

The number of girls this semester is 7 10-350 = 360.

The sixth question:

A 20 days is twice as long as B 20-5 = 15 days.

Then A working for 20 days is equivalent to B working 15× 2 = 30 days.

Therefore, the efficiency ratio of Party A and Party B is 30: 20 = 3: 2.

So A makes 3 ÷ (3-2) × 3 = 9 pieces every day.

B make 9 ÷ 3× 2 = 6 pieces every day.

So this batch has 9× 20+6× 15 = 270 parts.