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The total weight of two tons of coal is 2268 kg. Two fifths of coal a and one fourth of coal B are taken out, and the total weight is 708 kg?
Let a be x kg
X*2/5+(2268-X)*1/4=708
0.4X+567-X/4=708
0.15X=141
X=940
Pile B = 2268-940 = 1328Kg
Do not understand can ask, help please adopt, thank you!
Let a be x, B be y, x + y = 2268, 2 / 5x + 1 / 4Y = 708. Question: the process
The speed ratio of freight car to passenger car is 3:4. The two cars leave from station a and station B at the same time and meet at 18km from the midpoint of the two stations?
A: the distance between a and B is 252 km
Judge whether the two quantities in the following questions are in inverse proportion, and explain the reason. 1
The numerator of a fraction is fixed, and the value of the fraction is inversely proportional to the denominator
Because: according to "numerator / denominator = fractional value", denominator × fractional value = numerator (certain)
Therefore, the numerator of a fraction is fixed, and the value of the fraction is inversely proportional to the denominator
The total weight of the two piles of coal is 160 tons. If 2 / 5 more than 20 tons are transported from the first pile and 2 / 5 less than 6 tons are transported from the second pile, how many tons are transported from the two piles of coal?
Party A and Party B transported more than 2 / 5 of the total amount, 20-6 = 14 tons
Two piles of Coal Transported 160 × 2 / 5 + 14 = 78 tons
The speed ratio of a and B is 9:7. When they meet, a travels 72 kilometers more than B. A and B travel several thousand meters per hour?
The speed ratio of vehicle a and vehicle B is 9 ∶ 7, which means that vehicle a travels 2 / 7 more per hour than vehicle B
When meeting, car a travels 72 kilometers more than car B, then it can be seen that car B travels 72 △ 2 / 7 = 252 kilometers when meeting, so car B's speed is 252 △ 6 = 42 kilometers per hour
If the speed ratio of car a and car B is 9 ∶ 7, car a will travel 42 × 9 / 7 = 54 km per hour
Suppose the speed of car a and B 2 is 9x and 7x
So: 9x times 6h minus 7x times 6h = 72km
The results show that x = 24 V A = 216 v b = 168
72 △ 6 = 12 km
Speed of car a: 12 △ (9-7) × 9 = 54 km
Speed of vehicle B: 12 △ (9-7) × 7 = 42 km
72 △ 6 = 12 km, speed of vehicle a: 12 △ 9-7 × 9 = 54 km, speed of vehicle B: 12 △ 9-7 × 7 = 42 km
Determine the proportion of the two quantities in the following questions. 1. The load of the car is fixed, the number of times the goods are transported and the total mass of the goods
2. Build a road, the road that has been built and the road that has not been built. 3. The volume of the cylinder is fixed, and its bottom area and height are the same
4 copy a manuscript, the number of pages copied and the remaining pages 5 a car from a to B, car wheel circumference and car wheel rotation number 6 with a section of 40 cm long wire surrounded into a rectangle, the length and width of the rectangle
7 a map, distance on the map and actual distance 8 the number of copies of a book and the money used
The number of times of transportation is proportional to the total mass of the goods. The total mass of the goods is proportional to the number of times of transportation = the load of the car. 2. When building a road, the ratio or product of the repaired road and the unmodified road is not constant
The total weight of the two piles of coal is 160 tons. The first pile carries more than 20 tons of 2 / 5, and the second pile carries less than 6 tons of 2 / 5. How many tons are the two piles carrying
Let pile a have X tons and pile B have y tons
x+y=160.
Two piles transported = (2 / 5) x + 20 + (2 / 5) y-6
=(2/5)*(x+y)+14
=(2/5)*160+14
=78 tons
160 * (2 / 5) + 20 = 84 tons.
B is 160 * (2 / 5) - 6 = 60 tons.
The two piles are 144 tons in total
The speed ratio of the bus to the truck is 7:4. The two cars start from two places at the same time and run in opposite directions. They meet 18 kilometers away from the midpoint. How many kilometers did the bus travel?
The distance of the whole journey: 18 △ (74 + 7-12) = 132 (km), the kilometers of the bus: 132 × 711 = 84 (km)
Vehicle a carries a ton of goods. The weight of vehicle B is 0.5 tons more than 1 / 2 of vehicle a. the tonnage of vehicle B is expressed as () tons by the formula with letters
.
0.5+a/2
The total weight of the two piles of coal is 160 tons. If team a transports more than 20 tons of 2 / 5 and team B transports less than 6 tons of 2 / 5, how many tons of coal will the two teams transport?
Jiadui transport more than 2 / 5 20 tons: 160x2 / 5 + 20 = 64 + 20 = 84 (tons)
6 tons less than 2 / 5 transported by reactor B: 160x2 / 5-6 = 64-6 = 58 (tons)
Two piles of coal transported together: 84 + 58 = 142 (T)
(1) Coal transportation team a
Two
160 * --- + 20 = 84 (tons)
Five
(2) Coal transportation team B
Two
160 * --- 6 = 58 (tons)
Five
(3) The two teams transport coal together
84 + 58 = 142 (tons)
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