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Proportional valve mathematics
7 1/22 is greater than 3, so at least four gates should be opened at the same time.

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!