Solution: Let each sluice discharge 1 share every hour.
1x30=30 (copies)
2x 10=20 (copies)
The upper reaches of the river come every hour;
(30-20)÷(30- 10)=0.5
The original water is:
30-30x0.5= 15 (copies)
20- 10x0.5= 15 (copies)
Need to discharge 15 parts of water within 5.5 hours;
15÷5.5+0.5=7 1/22>3
So at least four gates should be opened at the same time.
Or: let the flood discharge speed of each gate be a cubic meter/hour and the flood increase speed be b cubic meters/hour.
The flood that has exceeded the safety line is k cubic meters.
According to the meaning of the question:
(a-b)30=k
(2a-b) 10=k
The solution is a = k/ 15 and b = k/30.
Setting up for 5.5 hours requires x valves, so there are
(ax-b)5.5=k
Substitute a = k/ 15 and b = k/30.
(xk/ 15-k/30)*5.5=k
That is (x/15-1/30) * 5.5 =1.
The solution is x=35.5/ 1 1.
X=4 because x must be at least greater than this number.
Arithmetic method:
The hourly flood discharge of a gate is 1 unit,
The 30-hour flood discharge of single sluice is 30* 1=30, and that of double sluice 10 * 1 * 2 = 20.
The inflow per hour is (30-20)/(30- 10)=0.5 unit,
The original part beyond the safety line is 30-0.5*30= 15 or 20-0.5* 10= 15.
The total water volume in 5.5 hours is 15+0.5*5.5= 17.75.
The demand is 17.75÷5.5÷ 1=3.2, which means at least four gates should be opened.
Please adopt!