Solution:
( 1).25×4= 100, 100÷ 1= 100;
(2). Assuming that the number of newly grown grasses per day is K and the total number of grasses at the beginning is A, there are:
(20×K + A)÷ 15 = 20
(10× k+a) ÷ 20 =10 = > solution: K= 10/ day, A= 100.
Number of grazing times per cow per day b: b = [(20× k+a) ÷ 15] ÷ 20]
= 1/ day
Therefore, the number of newly planted forage grass for cattle every day is 10÷ 1= 10.
(3). As mentioned above, assuming that the number of newly grown grasses per day is K and the total number of grasses at the beginning is A, there are:
(6×K + A)÷24 = 6
(8×K+A)÷2 1 = 8 == > solution: K= 12/ day, A=72.
The number of grazing times per cow per day b: B:B = [(6×K+A)÷24 ]÷6 24] ÷ 6.
= 1/ day
Suppose you need to eat for c days, there are:
(c× k+a) =16× b× c = > Substitution and solution: C= 18 days.
(4). As mentioned above, assuming that the number of newly grown grasses per day is K and the total number of grasses at the beginning is A, there are:
(20×K + A)÷5 = 20
(15×K+A)÷6 = 15 == > solution: K=2/ day, A=60.
Number of grazing times per cow per day b: b = [(20× k+a) ÷ 5] ÷ 20]
= 1/ day
Suppose you need d cows to eat for six days, and there are:
(6×K+A)= d×b×6 = = & gt； Alternatives and solutions: D= 12.
(5). Assuming that the number of grasses falling every day is K and the total number of grasses at the beginning is A, there are:
(A - 5×K)÷5 = 20
(A-6×K)÷6 = 16 == > solution: K=4/ day, A= 120.
The number of times each cow eats grass every day b: b = [(a-5× k) ÷ 5] ÷ 20]
= 1/ day
Suppose you need to eat for c days, there are:
(a-c× k) =11× b× c = > Substitution and solution: C=8 days.
(6). As above, assuming that the number of grasses falling every day is K and the total number of grasses at the beginning is A, there are:
(A - 8×K)÷8 = 30
(a-9× k) ÷ 9 = 25 = > solution: K= 15/ day, A=360.
The number of times each cow eats grass every day b: b = [(a-8× k) ÷ 8] ÷ 30].
= 1/ day
Suppose you need to eat for c days, there are:
(a-c× k) = 2/kloc-0 /× b× c = > Substitution and solution: C= 10 days.
(7). Assuming that the number of grasses falling every day is K and the total number of grasses at the beginning is A, there are:
(A - 5×K)÷5 = 20
(A-6×K)÷6 = 15 == > solution: K= 10/ day, A= 150.
The number of times each cow eats grass every day b: b = [(a-5× k) ÷ 5] ÷ 20]
= 1/ day
Suppose it takes d cows to eat for ten days, and there are:
(a-10× k) = d× b×10 = > alternatives and solutions: D=5 heads.
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Xiaoshengchu mathematics is incredible. The author has limited ability and can only solve equations. I hope I can understand.