A, B, C three students origami flowers. A 2 hours fold 42, B 3 hours fold 51, C 1.5 hours fold 33, whose work efficiency is high? Write the formula!

A, B, C three students origami flowers. A 2 hours fold 42, B 3 hours fold 51, C 1.5 hours fold 33, whose work efficiency is high? Write the formula!

Efficiency of a: 42 △ 2 = 21 (per hour)
Efficiency of B: 51 △ 3 = 17
The efficiency of C: 33 △ 1.5 = 22
Therefore, the efficiency of C is the highest
For processing the same parts, worker a processes 49 parts per hour, while worker B processes 51 parts per hour. What is the working efficiency of both workers?
The working efficiency of a is 49 / 1 = 49 / hour
The working efficiency of Party B is 51 / 1 = 51 / hour
Work efficiency of Party A 1 / 49, work efficiency of Party B 1 / 51 / equivalent to division sign work efficiency = work time divided by total work
There are 51 workers in a workshop of a factory. Each worker can process 16 A-type parts or 21 B-type parts every day, while one worker can not
If you don't complete the title, you can't try to solve it for you
The number ratio of the two engineering teams is 7:3. If team a sends 30 people to team B, the number ratio of the two teams is 3:2
There are X members in team B and 7 / 3x members in team a
7/3x-30=3/2x
(7/3-3/2)x=30
5/6x=30
x=30/(5/6)
x=36
7/3x=36*(7/3)=84
It is known that the sum of a, B and C is 35, and the difference between a and B is 7. If B is three times of C, then a = b = C=
Let a be x, B be y and C be Z
Then x + y + Z = 35
x-y=7
y=3z
If we change all three numbers to Z, we have 3Z + 7 + 3Z + Z = 35
The solution is Z = 4, y = 12, x = 19
If 30 people are transferred from team a to team B, the number of the two teams is equal; if 10 people are transferred from each team, the remaining number of team B is 25% of the remaining number of team A. how many people are there in each team?
The number of team a: (30 × 2) / (1-25%) + 10, = 60 ／ 34 + 10, = 60 × 43 + 10, = 80 + 10, = 90 (people); the number of team B: 90-30 × 2 = 30 (people). A: there were 90 people and 30 people in each team
A, B and C have saved a total of 350000 yuan. The amount of money saved by a is 3 / 4 of that of B, and the amount of money saved by B is 6 / 7 of that of C. how many yuan are saved by a, B and C?
It's about the strategy of solving problems in the second semester of sixth grade. It's better not to use equations
3/4×6/7=9/14
9/14+6/7+1=35/14
C 35 △ 35 / 14 = 140000 yuan
B 14 × 6 / 7 = 120000 yuan
A 12 × 3 / 4 = 90000 yuan
A: 9
B: 12
C: 14
Two pass marks, X3 for each, and the answer comes out. The pass marks of grade 6 are all right... Little p-boy, old boy, don't use the general score. You just say you've got to play with you, or you won't give a score. Look how good the one downstairs is. What grade are you in? Then tell me the grade I'm a sophomore???? If you want to be more detailed, just copy the following one. I'm only answering questions for the sake of wealth
A: 9
B: 12
C: 14
Two pass marks, X3 for each, and the answer comes out. The pass marks of grade 6 are all right... Little P child asked: old boy, don't use general score
The ratio of the original number of the two engineering teams is 7:5. If 30 people are transferred from team a to team B, the ratio of the two teams is 3:2?
Suitable for primary school students to solve problems
3/2-7/5=1/10
30 * 10 = 300 people
300 / (7 + 5) = 25
A 25 * 7 = 175
B 25 * 5 = 125
A: 175 in team a and 125 in team B
The sum of the three numbers of a, B and C is 35. The number of a is twice that of B by 5. One half of B is equal to one third of C?
Let a be x, B be y, C be Z. according to the conditional equations, we get x + y + Z = 35 (1) 2x-y = 5 (2) 1 / 2Y = 1 / 3Z (3) to solve the ternary linear equations as follows: x = (5 + y) / 2 from (2); Z = 3 / 2Y from (3). Substituting the above two equations into (1), we get the linear equation about y: (5 + y) / 2 + y + 3 / 2Y = 35
x+y+z=35
2x-y=5 ——> x=(y+5)/2
y/2=z/3 ——> z=3y/2
——> (y+5)/2 + y + 3y/2=35
——> y+5+2y+3y=70
——>6y=65 y=65/6 z=65/4 x=95/12
Let a be x, B be y and C be Z;
be
x+y+z=35
2x-y=5
y/2=z/3
There are two workshops a and B, and the number of workshop a accounts for five eighths of the total number of the two workshops,
If 90 people are transferred from workshop a to workshop 1, the number ratio of workshop a and workshop B is 2:3?
Later, the number ratio of workshop a and workshop B was 2:3, that is, the number of workshop a accounted for 2 / 5 of the total number of the two workshops. Originally, the number of people in workshop a accounted for 5 / 8 of the total number of people in the two workshops. In this way, workshop a reduced the total number of people by 5 / 8-2 / 5 = 9 / 40, that is, 90 people. So the total number of people = 90 / (9 / 40) = 400
The original number of people in workshop a = 400x5 / 8 = 250, the number of people in workshop B = 400-250 = 150 (the second method)
There are x people in workshop a and y people in workshop B
X=5/8（X+Y）
（X-90）：Y=2：3
The solution of the equations is as follows
X=150
Y=90
Let a be x and B be y, and let them be shown in two classes. The number of a is 90 and 150