mathematical problem

How many days can several cows eat the same grass on different days?

The basic quantitative relationship of this kind of problem is:

The daily growth of grass = (the number of cattle multiplied by the number of days to eat more-the number of cattle multiplied by the number of days to eat less) divided by the number of days difference.

The original amount of grass = the number of cattle multiplied by the number of days to eat-the daily growth of grass multiplied by the number of days to eat.

1. With the cold weather, the grass on the pasture is decreasing at a constant speed every day. After calculation, the grass on the pasture can feed 20 cows for 5 days, or 16 cows for 6 days. So, 1 1 how many days can a cow eat?

1 solution: let 1 head cow eat grass 1 day as 1 unit.

Grass on the pasture naturally decreases (20×5- 16×6)÷(6-5)=4 (unit) every day;

It turns out that the pasture has grass (20+4)×5= 120 (unit);

Can feed 1 1 first cow120 ÷ (11+4) = 8 (days).

2. A pool with an open water outlet at the bottom. Water can be pumped by 5 pumps 20 and 8 pumps 15. How long will it take to drain the water if it only comes out of the outlet?

Solution: The total water quantity increases with the extension of water leakage, so the total water quantity is a variable. The increase of water leaking into the ship per unit time is constant. The original water quantity in the ship (that is, the existing water quantity in the ship when the water leakage is found) is also a constant.

If the amount of scouring water per person per hour is "1 unit", the sum of the original amount of water in the ship and the total water leakage in three hours is equal to the amount of scouring water per person per hour × time× number of people, that is,/kloc-0 /×××10 = 30.

The sum of the original water volume in the ship and the 8-hour water leakage is 1×5×8=40.

The water leakage per hour is equal to the difference between the total water volume of 8 hours and the total water volume of 3 hours ÷ time difference, that is, (40-30)÷(8-3)=2 (that is, the water leakage per hour is 2 units, which is equivalent to the scouring water volume of 2 people per hour).

The original water quantity in the ship is equal to the total water quantity fished out by 10 people in 3 hours-3 hours of water leakage. The water leakage in 3 hours is equivalent to 1 hour 3×2=6 people, so the original water quantity in the ship is 30-(2×3)=24.

If the water (24 units) takes 2 hours to get out, it needs 24 ÷ 2 = 12 (person), but at the same time it needs two people to get out every hour, so * * needs 12+2 = 14 (person).

3. There are three grasslands, with an area of 4 hectares, 8 hectares and 10 hectares respectively. The grass on the grass is as thick and grows as fast. The first grassland can feed 24 cows for 6 weeks, and the second grassland can feed 36 cows 12 weeks. Q: How many weeks can the third meadow feed 50 cows?

Solution: 4 hectares of grass can feed 24 cows for 6 weeks, and we can introduce 8 hectares of grass to feed 48 cows for 6 weeks. Let's assume that 1 cow eats one unit of grass in 1 week, then the grassland growth in (12-6) week is 36* 12-48*6= 144 unit, then 1 44 unit. It can be inferred that there are 36 *12-24 *12 =144 units of grass in 8 hectares of grassland. In this way, 10 hectare has144 * (10/8) =180 units of grass, and the grassland increment is 24 *( 10/8)= 30. To sum up, 10 hectare grassland can feed 30 cows. The original grass on the grassland can be used to raise cattle for rest. For food 180/(50-30)=9 weeks.

Several students went boating. They rented some boats. If there are four people on each boat, there will be five more. If there are five people on each boat, there are four seats on the boat. Q: How many students are there? How many ships?

Solution: This is profit and loss problem.

One profit and one loss (commonly used formula): (profit plus loss) divided by (twice distribution difference) = number of people.

Ship: (5+4)/(5-4)=9 (articles)

There are 4*9+5=4 1 student or 5*9-4=4 1 student.

Spiders have eight legs, dragonflies have six legs and two pairs of wings, and cicadas have six legs and 1 pairs of wings. At present, the worms of these three species have 16, * * * has 16 legs and 14 pairs of wings. Q: How many bugs are there in each kind?

Number of chickens = (number of feet per rabbit × total number of rabbits-actual number of feet) ÷ (number of feet per rabbit-number of feet per chicken)

Number of rabbits = total number of chickens and rabbits-number of chickens

Solution: ① Suppose that spiders also have six legs. How many legs do three animals have?

6× 18= 108 (pieces)

② How many spiders are there?

(118-108) ÷ (8-6) = 5 (only)

(3) How many dragonflies and cicadas are there?

18-5= 13 (only)

(4) Assuming that dragonflies are also a pair of wings, how many pairs of wings does * * * have? 1× 13= 13 (right)

⑤ How many dragonflies are there?

(20- 13)÷ 2- 1)= 7 (only)

There are seven dragonflies.