Party A and Party B face each other from a and B, and the speed ratio is 7:4. After the first meeting, they continue to move forward and return to the destination immediately. On the way, they meet for the second time. The second meeting is 10 meters away from A. how far is the distance between the two places?

A: the distance between the two places is 110 meters
Party A and Party B walk from AB to each other at the same time. For the first time, they meet at 40km away from A. after that, they still advance at the same speed and return to their destination,
If they meet again 20km away from a, what is the distance between a and B
When the two met for the second time, the sum of their total journey was three times the distance between AB and a, of which a's journey was
40 × 3 = 120 km
When they meet again, a's journey is 20 kilometers less than twice the distance of ab
Therefore, the distance between AB and ab is
(120 + 20) △ 2 = 70km
The two cars started from ab at the same time and went in opposite directions. When they first met, they were 900 kilometers away from A. after meeting, they continued to move forward at the same speed, and the two cars stopped
When you arrive at your destination, return immediately, and meet for the second time on the way. At this time, you are 300 kilometers away from B. how many kilometers is the distance between AB and B? How many kilometers is the distance between the two meeting places? If you solve the equation, there can only be one X
The distance between the two places
3*900=x+300
x=2700-300
x=2400
The first encounter is 900 km away from a
The second distance a 2400-300 = 2100km
Two encounters 2100-900 = 1200 km
A and B leave from ab at the same time, and their speed ratio is 7:11. After the first meeting, the two vehicles continue to move in the same direction and reach the terminal respectively
The title is incomplete
480KM
If the velocity of a is x, the velocity of B is 11 / 7X. The distance between two places is L
According to the same service time of the two vehicles, the equations are as follows:
(L+80)/X=(2L-80)/(11/7X)
3L=880+560
L = 480km
There is a batch of goods in the freight yard. If they are transported by 3 trucks, they can be transported in 4 days; if they are transported by 4 trucks, they can be transported in 5 days; if they are transported by 20 tractors, they can be transported in 6 days. Now there are 2 trucks, 3 trucks and 7 tractors. After they transport together for 2 days, the rest will be transported by tractors, and they must be transported in 2 days. How many tractors will they use in these two days?
[1 - (1 ﹣ 3 ﹣ 4 × 2 + 1 ﹣ 5 ﹣ 4 × 3 + 1 ﹣ 6 ﹣ 20 × 7) × 2] / (1 ﹣ 6 ﹣ 20 × 2), = [1 - (16 + 320 + 7120) × 2] / 160, = [1-45120 × 2] / 160, = [1-4560] / 160, = 1560 ﹣ 160, = 15 (tractors), a: 15 tractors will be used in these two days
How to determine group distance and group number in compiling group distance series
Group distance series is the abbreviation of group distance variable series. Group distance series is usually made up of discontinuous variables with large variation range of variables. Therefore, group distance series has both continuous variable series and discontinuous variable series. The difference between the maximum value and minimum value of each group in group distance series is called group distance
Two sixth grade math problems, two cars a and B leave AB at the same time, and meet 15 kilometers away from the destination
1. Two cars of a and B leave AB at the same time and meet 15 kilometers away from the destination. It is known that the speed ratio of a and B is 7:6. How many kilometers is the distance between AB and B?
2. Party A and Party B set out from a and B at the same time, facing each other. When they were 6 hours old, they met on the way. If the speed ratio of Party A and Party B was 2:3, how many hours would it take for Party B to reach a after they met?
It's better to use the proportion solution, if it doesn't work, the formula, thank you
It's the midpoint.
1. Should it be the midpoint?
The distance between the two places is x km
（½X+15）：（½X-15）=7:6
X=390
A: the distance between a and B is 390km
2. Set B's speed to 3x km / h
A's speed is 2x km / h
6（2X+3X)=390
X=13
B's speed is 39 km / h
390 / 39-6 = 4 hours
A: it will take four hours to get to a
1。 Let AB be x km away from each other, then according to the equal running time of two vehicles, we get: (X-15): 7 = 15:6, and the solution is x = 32.5Km
2. Wrong title, no solution
According to the speed ratio equal to the distance ratio, when it is 6 hours, Party A takes 2 parts of the whole distance and Party B takes 3 parts. First, find out how long it takes Party B to take: 6 △ 3 = 2 (hours), and the rest is 2 × 2 = 4 (hours)
Should the "end" of the first question be changed to "Midpoint"
1.（x-15）：7=15：6
x =32.5
A: AB is 32.5Km apart.
I don't know
A company wants to transport 159 tons of goods from warehouse A to warehouse B. each truck can transport 7.5 tons, and the freight of each truck is 130 yuan. Each truck can transport 3 tons, and the freight of each truck is 60 yuan. Please design the transportation scheme with the least freight. How much is the minimum freight
First, figure out which is the lower freight for large and small trucks,
It's 17.3 yuan per ton for big trucks,
And small card is 20 yuan per ton
In this way, there are 21 7.5 tons in 159 tons, and the remaining 1.5 tons,
Obviously, 21 large cards and 1 small card are not suitable,
So it's just right to use 20 large cards and 3 small cards
Freight: 20 × 130 + 3 × 60 = 2600 + 180 = 2780 yuan
159 △ 75 ≈ 21 (car)
21-1 = 20 (vehicle)
159÷20÷3
＝7.95÷3
About 3 (car)
20×130+60×3
＝2600+180
= 2780 yuan
A: it will cost at least 2780 yuan.
159 △ 75 ≈ 21 (car)
21-1 = 20 (vehicle)
159÷20÷3
＝7.95÷3
About 3 (car)
20×130+60×3
＝2600+180
= 2780 yuan
A: it will cost at least 2780 yuan.
130 / 7.5 * 159 = 2756 yuan
A: the minimum freight is 2756 yuan.
(Note: all trucks are used.)
Hello:
First, figure out which is the lower freight for large and small trucks,
It's 17.3 yuan per ton for big trucks,
The small card is 20 yuan per ton.
In this way, there are 21 7.5 tons in 159 tons, and the remaining 1.5 tons,
Obviously, 21 large cards and 1 small card are not suitable,
So it's just right to use 20 large cards and 3 small cards.
Freight: 20 × 130 + 3 × 60 = 2600 + 180 = 2780 yuan
Hello:
First, figure out which is the lower freight for large and small trucks,
It's 17.3 yuan per ton for big trucks,
The small card is 20 yuan per ton.
In this way, there are 21 7.5 tons in 159 tons, and the remaining 1.5 tons,
Obviously, 21 large cards and 1 small card are not suitable,
So it's just right to use 20 large cards and 3 small cards.
Freight: 20 × 130 + 3 × 60 = 2600 + 180 = 2780 yuan
159 / 75 = 21.2 (car)
21*130+60*1
=2730+60
=2790 yuan
It is judged that in compiling variable series, continuous variable can only adopt group distance form
correct
If the variable value changes a lot, the number of times, or continuous variables, use group distance grouping
The distance between a and B is 150 kilometers. A and B start from a and B respectively. It takes 10 hours for them to meet each other. The speed of a is 2 / 3 of that of B, so it takes? Hours for B to walk alone
Let B be x, then a be 2x / 3
Solving equation 10 * (x + 2x / 3) = 150
X=9
So B speed is 9 km / h
It takes 150 / 9 = 16.67 hours to walk alone
Sixteen
16 hours 40 minutes
According to the calculation of 10 hours, the total speed of the two is 15km / h. The speed of a is 2 / 3 of that of B. the calculation shows that a 6, B 9, time s = 150 / 9 = 16 hours and 40 minutes
(1 + 2 / 3) * 10 = 50 / 3 hours
If it takes x hours for B to walk alone, then the speed of B is 150 △ X and that of a is (150 △ x) × 2 △ 3
。 Two speeds have been set. Can you do it

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