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The two piles of coal have 1300kg. When the first pile burns 2 / 3 and the second pile burns 5 / 7, the remaining coal is the same. How many kg of each pile
Let the first pile x, then the second pile 1300-x
(1-2/3）x=(1-5/7)(1300-x)
x=600
1300-x=700
The first pile is 600kg, the second pile is 700kg
The distance between the two places is 480 km, and the passenger cars and freight cars depart from the two places at the same time. The passenger cars travel 68 km per hour, and the passenger cars 52 km per hour,
After a few hours, the two cars meet? (calculated by equation and arithmetic)
Arithmetic:
The total distance between the two cars is 480km
The total speed of the two vehicles is v = 68 + 52 = 120 km / h
It takes two hours to meet
Equation:
Let two cars meet after X hours
The equation 68x + 52X = 480
The solution is x = 4 hours
So after four hours, the two cars met
(68+52)*x=480 x=4h
480/(68+52)=4h
There are two freight cars, one is 96 meters long and runs 26 meters per second, the other is 104 meters long and runs 24 meters per second. How many seconds does it take for the two cars to run in opposite directions?
Relative displacement velocity after meeting: 26 + 24 = 50 m / S
Total length from front meeting to rear separation: 96 + 104 = 200m
So the time: 200 / 50 = 4 seconds
The total weight of the two piles of coal is 1980 kg. After 3 / 5 of the first pile of coal is used up and 2 / 3 of the second pile of coal is used up,
The weight of the remaining two piles of coal is equal. How many kilos did each pile of coal weigh?
Suppose that the first pile of coal weighs XKG, and the other pile weighs 1980-xkg. The first pile uses 3 / 5, leaving (1-3 / 5) x = 2x / 5kg. The second pile uses 2 / 3, leaving (1-2 / 3) (1980-x) = 660-x / 3kg
At the same time, the passenger and freight cars leave from a and B. the passenger cars travel 78 kilometers per hour and the freight cars 72 kilometers per hour. When they meet, the passenger cars travel 18 kilometers more than the freight cars
How many meters is the distance between the two places?
equation
If two cars travel for X hours, the distance between the two places is (78 + 72) x km
78x－72x＝18
x ＝18/6
x＝3
（78＋72）x ＝150x＝150*3=450
A car loaded 3 tons of goods, B car loaded 2 tons of goods, there are 13 tons of goods required to be loaded at a time and all the cars are full, how many cars each of a and B cars, using binary linear equation solution
Let a x, B y
3x+2y=13,x
There are the same two piles of coal, each pile weighs 10 tons. The first pile uses 3 / 5 of the coal, and the second pile uses 3 / 5 of the coal. Which pile uses more coal
First pile = 10 * 3 / 5 = 6 tons
Second pile = 3 / 5 tons
The first pile is much more used
Both passenger and freight cars leave from a and B at the same time. Passenger cars travel 78 kilometers per hour and freight cars 72 kilometers per hour. When they meet, passenger cars travel 18 kilometers more than freight cars
How many kilometers are there between a and B
Driving: 18 / (78-72) = 3 hours
The distance between the two places is (78 + 72) * 3 = 450 km
Let the time of two vehicles be t, then the distance is s
78t-72t=18
T=3
S = 78t + 72t = 78 × 3 + 72 × 3 = 450 km
It takes 6 hours for a batch of goods to be transported by car a alone, and 60 tons per hour by car B. now it's transported by car a and car B together. When it's finished, the weight ratio of car a and car B is 2:3
Since the traffic volume ratio of Party A and Party B is 2:3 at the same time, it can be concluded that the hourly traffic volume of Party A is 2 / 3 of that of Party B, that is, 40 tons per hour
If the joint transportation time of Party A and Party B is k, then
40K + 60K=40 X 6
The equation k = 2.4 can be solved once
Then the transportation capacity of B is 60x 2.4 = 144 tons
Although the book is not vehicle tonnage, but no one said that the weight of goods must reach the vehicle
There are two piles of coal with a total of 26 tons. Two thirds of the first pile of coal is as heavy as one fifth of the second pile of coal. How much is each pile of coal
Set the weight of the first pile of coal X
Then another pile of coal weighs 26-x
2X/3=(26-X)/5
5*2X=3(26-x)
10X=78-3X
13X=78
The weight of the first pile of coal is x = 6
Another pile of coal weighs 26-x = 20
Let x, the first pile of X tons, then the second pile of (26-x) tons 2 / 3x = 1 / 5 (26-x), then x is 6
Let the first pile be a, then the second pile be (26-a)
2/3*A=1/5*（26-A）
Then solve the equation
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