The school assigned 540 tree planting tasks according to the number of students in three classes of grade 6. There were 44 students in class 1, 45 students in class 2 and 46 students in class 3. How many trees should be planted in each of the three classes?

The school assigned 540 tree planting tasks according to the number of students in three classes of grade 6. There were 44 students in class 1, 45 students in class 2 and 46 students in class 3. How many trees should be planted in each of the three classes?


44 + 45 + 46 = 135540 × 44135 = 176 (trees) 540 × 45135 = 180 (trees) 540 × 46135 = 184 (trees) answer: 176 trees should be planted in class one, 180 trees in class two and 184 trees in class three



Sixty trees are planted in three classes in grade six. There are forty students in one class, thirty-eight students in two classes, forty-two students in three classes. How many trees are there in three classes?


40 + 38 + 42 = 120 people
Class 1: 60 × 40 / 120 = 20 trees
Class 2: 60 × 38 / 120 = 19 trees
Class 3: 60 × 43 / 120 = 21 trees



The school assigned the task of planting 220 trees according to the number of students in two classes of grade six. There were 46 students in class one and 42 students in class two. How many trees were planted in each class
J: the way to solve the proportion. Thank you


Hello, ESC
Set up a class to plant x trees
X:(220-X)=46:42
42X=46×220-46X
88X=10120
X=10120÷88
X=115
115 trees are planted in class one, and 105 trees are planted in class two: 220-115



The school assigned 960 trees to each class according to the number of students in three classes of grade 6. There were 40 students in class 1, 38 students in class 2 and 42 students in class 3. How many trees were assigned to each class?


There are 320 trees in class one, 304 trees in class two and 336 trees in class three. Is this the problem of primary school students?



In the process of wine making, what components of grape become alcohol? Please explain in detail, urgently needed, thank you!


The process of wine making is the conversion of sugar into alcohol under the catalysis of yeast. When the alcohol concentration is greater than 15 degrees, yeast is killed, which is why the highest degree of wine is less than 15 degrees



Can 75% medical alcohol be used as liquid alcohol in hot pot
In addition, solid alcohol and liquid alcohol are good


Yes, because 75% alcohol can be burned. The difference between solid alcohol and liquid alcohol is that acetate is added into solid alcohol. The purpose is to make solid alcohol easier to use, safer and less easy to pour than liquid alcohol. Industrial alcohol can be used as fuel. General industrial alcohol contains methanol and other alcohol



There are 11 liters of pure alcohol in container a and 15 liters of water in container B. for the first time, pour part of pure alcohol in container a into container B to mix alcohol with water. For the second time, pour part of mixed liquid in container B into container A. in this way, the pure alcohol content in container a is 62.5%, and the alcohol content in container B is 25%. How many liters of mixed liquid are poured into container a from container B for the second time ?


For container B: because the alcohol: mixture = 25%, so the alcohol: water = 25%: (1-25%) = 1:3, that is, the poured alcohol is 15 △ 3 = 5 (L). For container a: the remaining alcohol is 11-5 = 6 (L). After setting, pour x L from container B, then (6 + 25% x) △ 6 + x) = 62.5%, the solution is x = 6. A: pour 6 L of mixture from container B for the second time



There are 11 liters of pure alcohol in container a and 15 liters of water in container B. for the first time, pour part of pure alcohol in container a into container B to mix alcohol with water. For the second time, pour part of mixed liquid in container B into container A. in this way, the pure alcohol content in container a is 62.5%, and the alcohol content in container B is 25%, How many liters of mixed liquid is poured from container B to container a for the second time? Two solutions


1. Because the alcohol content in container B is 25%, that is, water accounts for 75% 15 / (1-25%) = 20 liters (the amount of alcohol in container a after pouring a part of alcohol into container B) 20-15 = 5 liters (the amount of alcohol poured into container B) 11-5 = 6 liters (pure alcohol in container a) (100% - 62.5%): (62.5% - 25%) = 1:1, so



There are 11 liters of pure alcohol in container a, 11 liters of pure alcohol in container a and 15 liters of water in container B. pour part of pure alcohol in container a into container B for the first time
Then pour some of the shaken solution into container A. at this time, the alcohol concentration of container a is 62.5% and that of solution B is 25%. How many liters of mixed liquid is poured into container a?


To grasp the constant amount, pour a part of pure alcohol into container B for the first time, then the constant amount in container B is water, or the original 15 liters, and the alcohol concentration of solution B is 25%
15 (1-25%) = 20 (L) is the volume of solution in B after the first operation, then the volume poured from a is 20-15 = 5 (L), and there is 11-5 = 6 (L) left in a
For the second time, we need to use the knowledge of average number when pouring from container B into container A. for the second time, there are 6 liters of pure alcohol (100%) in container a, which is 62.5% higher than the total concentration after mixing
(100% - 62.5%) × 6 ÷ (62.5% - 25%) = 6 (liter)



There are 11 liters of pure alcohol in container a and 15 liters of water in container B. for the first time, pour part of pure alcohol in container a into container B to mix alcohol with water. For the second time, pour part of mixed liquid in container B into container A. in this way, the pure alcohol content in container a is 62.5%, and the alcohol content in container B is 25%. How many liters of mixed liquid are poured into container a from container B for the second time ?


For container B: because the alcohol: mixture = 25%, so the alcohol: water = 25%: (1-25%) = 1:3, that is, the poured alcohol is 15 △ 3 = 5 (L). For container a: the remaining alcohol is 11-5 = 6 (L). After setting, pour x L from container B, then (6 + 25% x) △ 6 + x) = 62.5%, the solution is x = 6. A: pour 6 L of mixture from container B for the second time