A and B have 300 tons of coal in total. A transports two fifths of the coal and B transports back two fifths of the coal. Now B has 20 tons more than A. how many tons each of a and B?

A and B have 300 tons of coal in total. A transports two fifths of the coal and B transports back two fifths of the coal. Now B has 20 tons more than A. how many tons each of a and B?

Let a be x tons and B be y tons
x+y=300
x(1-2/5)+20=y(1+2/5) 3x+100=7y
X = 200 (tons)
Y = 100 (ton)
A was 200 tons, B was 100 tons
A + B = 300
7 B / 5-3 A / 5 = 20
B = 100
A = 200
A passenger car and a freight car leave from Party A and Party B at the same time. The speed ratio of passenger to freight is 5:4. When they meet, the freight car is 15 minutes away from the midpoint, and they meet in a few hours
2 hours
One
A passenger car and a freight car leave from Party A and Party B at the same time. The speed ratio of passenger to freight is 5:4. When they meet, the freight car is 15 minutes away from the midpoint, and they meet in a few hours
Should it be 15 kilometers from the midpoint?
If the speed ratio is 5:4, the travel ratio is also 5:4
That is, the passenger car line: 5 / [5 + 4] = 5 / 9, the freight car line: 4 / 9
When we meet again, the bus runs more than the truck: 15 * 2 = 30 km
So the whole journey length is: 30 / (5 / 9-4 / 9) = 270 km
It's hard to find the meeting time. Less... More
A passenger car and a freight car leave from Party A and Party B at the same time. The speed ratio of passenger to freight is 5:4. When they meet, the freight car is 15 minutes away from the midpoint, and they meet in a few hours
Should it be 15 kilometers from the midpoint?
If the speed ratio is 5:4, the travel ratio is also 5:4
That is, the passenger car line: 5 / [5 + 4] = 5 / 9, the freight car line: 4 / 9
When we meet again, the bus runs more than the truck: 15 * 2 = 30 km
So the whole journey length is: 30 / (5 / 9-4 / 9) = 270 km
It's hard to find the meeting time. There are few conditions. Put it away
A, B, C three transport teams jointly transport a batch of goods, team a transported 14 of this batch of goods, team B transported a part of this batch of goods, team C transported 13 of this batch of goods, just all of them were transported. It is known that team a transported 10 tons less than team C, how many tons did Team B transport?
10 (13-14), = 10 (13-14), = 112, = 120 (tons); 120 × (1-13-14), = 120 × 512, = 50 (tons); answer: Team B transported 50 tons
Party A and Party B transport a batch of coal, Party B transport alone for 10 days, party a transport 50 cubic meters per day
How many cubic meters of coal does the two teams transport for six days at the same time?
Don't solve equations
[50 △ 1-6 △ 10)] × 6 = 750 M3
Second, it took 10 days to finish the single team and 6 days for both teams. That is to say, second team did 6 days and 6 / 10 of the whole team. Second team did 6 days and 6 / 10 of the whole team at the same time. That is to say, first team finished the remaining 4 / 10
[50 △ 1-6 △ 10)] × 6 = 750 M3
How simple
[50 △ 1-6 △ 10)] × 6 = 750 M3
Second, it took 10 days to finish the single team and 6 days for both teams. That is to say, second team did 6 days and 6 / 10 of the whole team. Second team did 6 days and 6 / 10 of the whole team at the same time. That is to say, first team finished the remaining 4 / 10
The passenger and freight cars leave from a and B at the same time and meet at a distance of 15 kilometers from the midpoint. It is known that the speed of the passenger car is 4 / 5 of that of the freight car. How many kilometers are there between a and B?
15×2÷（1-4/5）×（1+4/5）
=30÷1/5×9/5
=270 km
Answer
The speed ratio of passenger cars to freight cars is 4:5
The speed of two cars is equal to the distance of two cars
The bus takes 4 / (4 + 5) = 4 / 9 of the total length
15 meters is equivalent to 1 / 2-4 / 9 of the total length = 1 / 18
The distance between a and B is: 15 ÷ (1 / 18) = 270 km
There are six vehicles a and four vehicles B. A's daily transportation volume is 30t, and the cost is 0.9 yuan. B's daily transportation volume is 40t, and the cost is 1000. It undertakes at least 280t of business every day,
In order to minimize the daily transportation cost, how many vehicles do you need?
First, set X trucks of type A and Y trucks of type B, list the constraint conditions, then draw the feasible region according to the constraint conditions, set z = 0.9x + y, and then use Z
Analysis: X type a trucks and Y type B trucks are required
From the theme
30x+40y≥280
0≤x≤6
0≤y≤4
And X, y ∈ Z
Objective function of transportation cost z = 0.9x + y
When the objective function passes through a (4,4), the minimum Z is 7.6 thousand yuan
And need a truck and B truck each 4
In order to minimize the daily transportation cost, 4 vehicles of a are required
There are pictures on the Internet. You can search them
Two transport teams transport coal. Team a transports coal 6 hours later and team B transports coal. Because team B's work efficiency is 25% higher than team a, it completes the task 1 hour earlier than team a alone. If team a transports 15 tons and team B transports coal, it can complete the task 2 hours earlier than team a alone. How many tons does team a transport per hour? How many hours does team a transport alone?
The time ratio of Party A and Party B is (1 + 25%): 1 = 125%: 1 = 5:4, the transportation demand of Party A alone is: 1 △ 54-1) × 54 + 6 = 1 △ 14 × 54 + 6 = 5 + 6 = 11 (hours), the transportation demand of Party B transported by Party A alone is: 2 △ 54-1) × 54 = 2 △ 14 × 54 = 10 (hours), the transportation demand of Party A per hour is: 15 △ 11-10 = 15 (tons) a: the transportation demand of Party A per hour is 15 tons, the transportation demand of Party A alone is 11 hours
At the same time, the passenger and freight cars leave from a and B. when they meet, the distance ratio of the passenger and freight cars is 5:4; after meeting, the freight car is 18 kilometers faster than the passenger car per hour
As a result, the two cars arrived at the other side's departure station at the same time. It is known that the freight car has been running for 10 hours. How many kilometers is the distance between a and B?
Let the speed of the freight car be V goods, and the speed of the passenger car be V passengers. It is known that V goods = V passengers + 18. Since the movement time of the two cars is equal when they meet, so V passengers / V goods = 5:4. The speed of the freight car can be calculated from the above two formulas, then V goods * 10 is the distance between a and B
4÷5＝4/5
5÷4＝5/4
5/4÷4/5＝25/16
18 ÷ (25 / 16-1) = 32 km / h
32 × 10 = 320 km
The transport team carries a batch of goods, 20 tons a day. In six days, three fifths of the goods are lucky. How many tons are there in total? (equation)
Suppose that there are x tons in total,
3/5x=20×6
x=120×5/3
x=200
Suppose there are x tons in total.
X*3/5=20*6
X=200
There are x tons in total
3/5X=20*6
X=200
It takes 5.5 hours for Party A and Party B to complete the shipment______ What's the efficiency ratio______ .
7: 5 = 14:11; 17:15.5, = 5.5:7, = 11:14; a: the time ratio of a and B is 14:11, the work efficiency ratio is 11:14
The time ratio is: 7:5.5
The efficiency ratio is 5.5:7
14 to 11 and 11 to 14
Time ratio 7:5.5
The efficiency ratio is 1 / 7:1 / 5.5