The total number of plants is 900+ 1250 = 2 150, and 24+30+32 = 86 plants can be planted every day.
The planting days are 2 150 ÷ 86 = 25 days.
24× 25 = 600 trees will be completed in 25 days.
Then B will finish 900-600=300 trees before helping C.
That is, after 300 ÷ 30 = 10 days, I transferred from A to B on 1 1 day.
2. There are three grasslands with an area of 5 15 and 24 mu respectively. The grass on the grassland is as thick and grows as fast. The first grassland can feed 10 cows for 30 days, and the second grassland can feed 28 cows for 45 days. How many cows can eat on the third grass for 80 days?
This is a complicated problem of herding cattle.
The grass eaten by each cow every day is 1 serving.
Because the first piece of grassland with an area of 5 mu+grassland with an area of 5 mu = 10× 30 = 300 copies for 30 days.
Therefore, the amount of grass per mu and the amount of grass per mu for 30 days are 300 ÷ 5 = 60.
Because the original grass quantity of the second grassland with an area of 15 mu+the grass quantity of 65438 with an area of 45 days = 28× 45 = 1260.
Therefore, the amount of raw grass per mu and the amount of grass in 45 days per mu are 1260 ÷ 15 = 84 copies.
So 45-30 = 15 days, and the area per mu is 84-60 = 24.
Therefore, the area per mu is 24/ 15 = 1.6 parts/day.
Therefore, the amount of grass per mu is 60-30× 1.6 = 12.
The third plot covers an area of 24 mu, and needs to grow 1.6× 24 = 38.4 pieces every day, and the original grass has 24× 12 = 288 pieces.
Every day, 38.4 cows need to eat the newly grown cows, and the remaining cows eat the original grass every day, so the original grass is enough for 80 days, so 288 ÷ 80 = 3.6 cows.
So a * * * needs 38.4+3.6 = 42 cows to eat.
Two solutions:
Solution 1:
Assume that the daily grazing amount of each cow is 1, and the total grass amount per mu for 30 days is10 * 30/5 = 60; The total grass yield per mu in 45 days is: 28*45/ 15=84, so the new grass yield per mu per day is (84-60)/(45-30)= 1.6, and the original grass yield per mu is 60-1.6 *.
Scheme 2: 10 cows eat 5 mu in 30 days, and 30 cows eat 5 mu in 30 days 15 mu. According to 28 cows eating15mu for 45 days, it can be deduced that15mu of new grass (28 * 45-30 * 30)/(45-30) = original grass amount15mu:1260-24 *. 15mu cattle required for 80 days 180/80+24 (head) 24mu: (180/80+24) * (24/15) = 42 heads.
3. A project, contracted by both parties, can be completed in 2.4 days and needs to be paid 1800 yuan; Contracted by Team B and Team C, it can be completed in 3+3/4 days, and it needs to be paid 1500 yuan; Contracted by two teams, Party A, Party B and Party C, it can be completed in 2+6/7 days at a cost of 1.600 yuan. On the premise of ensuring the completion within one week, which team will spend the least?
The cooperation between Party A and Party B is completed in one day 1 ÷ 2.4 = 5/ 12, and the payment 1800 ÷ 2.4 = 750 yuan.
The cooperation between ethylene, propylene and Fang Yitian is 1 ÷ (3+3/4) = 4/ 15, and the payment is 1500× 4/ 15 = 400 yuan.
The cooperation between Party A, Party C and Fang Yitian is 1÷ (2+6/7) = 7/20, and the payment is 1600× 7/20 = 560 yuan.
Three people cooperate in one day (5/12+4/15+7/20) ÷ 2 = 31/60,
Three people cooperate to pay (750+400+560) ÷ 2 = 855 yuan a day.
Party A alone completes 31/60-4/15 =1/4 every day, and pays 855-400 = 455 yuan.
Party B alone completes 3 1/60-7/20 = 1/6 every day, and pays 855-560 = 295 yuan.
Party C alone completes 31/60-5/12 =110 every day and pays 855-750 = 105 yuan.
So by contrast,
Choose b with 1 ÷ 1/6 = 6 days, only 295× 6 = 1770 yuan.
There is a rectangular iron block in the cylindrical container. Now turn on the tap and pour the water into the container. In 3 minutes, the water surface is just above the top of the cuboid. /kloc-After 0/8 minutes, the container has been filled with water. It is known that the height of a container is 50 cm and the height of a cuboid is 20 cm. Find the ratio of the bottom area of a cuboid to the bottom area of a container.
Divide this container into upper and lower parts. According to the time relationship, it can be found that the volume of water in the upper part is 18 ÷ 3 = 6 times that of the lower part.
The height ratio of the upper half and the lower half is (50-20): 20 = 3: 2.
So the bottom area of the upper part is 6 ÷ 3× 2 = 4 times the bottom area of the lower part filled with water.
Therefore, the ratio of the bottom area of the cuboid to the bottom area of the container is (4- 1): 4 = 3: 4.
Unique solution:
(50-20): 20 = 3: 2, and when there is no cuboid, it takes 18*2/3= 12 (minutes) to fill 20 cm.
So the volume of a cuboid is 12-3=9 minutes of water, because the height is the same.
So the volume ratio is equal to the bottom area ratio, 9: 12 = 3: 4.
Two bosses, A and B, bought a fashion at the same price, and B bought 1/5 sets more than A, and then they sold them at profit margins of 80% and 50% respectively. After both of them were sold out, A still made more profits than B, just enough for him to buy 65,438+00 sets of this fashion, which A bought at the beginning.
Consider that the number of groups of A is 5 and the number of groups of B is 6.
The profit earned by A is 80% × 5 = 4, and the profit earned by B is 50% × 6 = 3.
A is 4-3 more than B = 1 copy, and this1copy is 10 set.
So, A initially purchased 10× 5 = 50 sets.
6. There are two water pipes A and B, and water is injected into two pools with the same size at the same time. At the same time, the water injection ratio of A and B is 7: 5. After 2+ 1/3 hours, the sum of the water injected into A and B is just a pool. At this time, the water injection rate of pipe A is increased by 25%, while pipe B remains unchanged.
Think of a pool of water as 1.
Because a pool of water was injected after 7/3 hours, 7/ 12 was injected into the A pipe and 5/ 12 was injected into the B pipe.
The water injection rate of a pipe is 7/ 12 ÷ 7/3 = 1/4, and that of b pipe is 1/4× 5/7 = 5/28.
The later water injection of a pipeline is1/4× (1+25%) = 5/16.
The time spent is 5/ 12 ÷ 5/ 16 = 4/3 hours.
It takes 1 ÷ 5/28 = 5.6 hours for tube B to fill the pool.
Water injection needs 5.6-7/3-4/3 = 29/ 15 hours.
That is 1 hour and 56 minutes.
Continue to do another method:
According to the original water injection rate, it takes 7/3 ÷ 7/ 12 = 4 hours to fill the pool with pipes.
The filling time of tube B is 7/3 ÷ 5/ 12 = 5.6 hours.
Time difference is 5.6-4 = 1.6 hours.
Later, the speed of nailing pipe increased, so the time was less and the time difference was more.
After the speed of A is increased, it takes 7/3× 5/7 = 5/3 hours.
The shortened time is equivalent to1-1÷ (1+25%) =1/5.
So the time is shortened by 5/3× 1/5 = 1/3.
So the second pipe needs1.6+1/3 = 29/15 hours.
Do it another way:
(1) The remaining nail tubes need time.
7/3× 5/7 ÷ (1+25%) = 4/3 hours.
(2) The time required to find the remaining B tubes.
7/3× 7/5 = 49/ 15 hours
(3) When the tube A is full, the tube B is evacuated.
49/ 15-4/3 = 29/ 15 hours If you want to enter a key junior high school, I suggest you review your math! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !