Analysis of one-day cooperation between Party A and Party B: 1 ÷ 2.4 = 5/ 12, 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.
Question: 200 grams of 60% alcohol solution and 300 grams of 30% alcohol solution are mixed, and the concentration of the obtained alcohol solution is ().
Analysis:
Solution mass = solute mass+solvent mass
Solute mass = solution mass × concentration
Concentration = solute mass ÷ solution mass
Solution mass = solute mass/concentration
To require the concentration of mixed solution, we must find out the total mass of mixed solution and the mass of pure alcohol.
The total mass of the mixed solution is the sum of the mass of the first two solutions:
200+300=500 grams.
The content of pure alcohol after mixing is equal to the sum of pure alcohol in the two solutions before mixing:
200× 60%+300× 30% =120+90 = 210 (g)
The concentration of the mixed alcohol solution is:
2 10÷500=42%
A: The concentration of mixed alcohol solution is 42%.
Jin Dian: When two solutions with different concentrations are mixed, the total amount of solution and solute remains unchanged.
Questions A, B and C plant trees in plots A and B, with 900 trees in plot A and 0/250 trees in plot B. It is known that Party A, Party B and Party C can plant 24, 30 and 32 trees each day. Party A plants trees in Site A, Party C plants trees in Site B, Party B plants trees in Site A first, and then goes to Site B ... The two plots start and end at the same time. On what day should B be transferred from A to B?
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, 300 ÷ 30 = 10 days later.
That is, on 1 1 day, it was transferred from a place to b place.
There are three grasslands in the experiment, with an area of 5 15 and 24 mu respectively. The grass on the grass is as thick and grows as fast. The first meadow can feed 10 cows for 30 days, the second meadow can feed 28 cows for 45 days, and how many cows can the third meadow feed for 80 days?
This is a problem of raising cattle, which is a complicated problem.
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+65438 with an area of 45 days+grass quantity = 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.
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. In another 18 minutes, the water has filled the container. 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.
The container is divided 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.
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 water injected into pools A and B is exactly one pool. At this time, the water injection rate of pipe A is increased by 25%, while the water injection rate of pipe B is unchanged. So, how many hours does it take for pipeline A to fill pool A and pipeline B to fill pool B?
The analysis regards 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