Three vehicles of a, B and C transport a pile of coal. Vehicle a carries 40% of the total amount of coal, vehicle B carries 60% of taxis, and vehicle C carries 28 tons more than vehicle B, a total of several tons

Three vehicles of a, B and C transport a pile of coal. Vehicle a carries 40% of the total amount of coal, vehicle B carries 60% of taxis, and vehicle C carries 28 tons more than vehicle B, a total of several tons

B and C transported 1-40% of the total amount = 60%, of which B transported 60 / (100 + 60) = 3 / 8, that is, B transported 50% of the total amount
60% X3 / 8 = 9 / 40, then a is 40% - 9 / 40 = 7 / 40 more than B, so
This pile of coal has a total of 28 (7 / 40) = 160 tons

Party A, Party B and Party C transport 3500 tons of goods. It is known that Party A transports 40% of the total. The ratio of goods transported by Party B and Party C is 4:3. How many tons does Party A, Party B and Party C transport

A: 3500 * 0.40 = 140 tons;
Ethylene and propylene: 3500-140 = 3360 ((ton));
B: 3360 * 4 / 7 = 1920 tons;
C: 3360 * 3 / 7 = 1440 (tons) or 3500-140-1920 = 1440 (tons)

It costs 80 yuan to buy 5 pieces of a, 3 pieces of B and 4 pieces of C. It costs 250 yuan to buy 6 pieces of a, 8 pieces of B and 7 pieces of C. how much does it cost to buy one piece of a, B and C

Let the price of a, B and C be x, y and Z yuan respectively
Then: 5x + 3Y + 4Z = 80
6x+8y+7z=250
The sum of the two formulas is: 11x + 11y + 11z = 330
So x + y + Z = 30
That is to say, it costs 30 yuan to buy a piece of a, B and C

A batch of goods was allocated to Party A and Party B according to the ratio of 5:6. Party A has transported 80 kg, which is just 4 / 5 of the team's task. If the rest is completed by Party B, how many kg should Party B transport?

Team a shipped the goods: 80 ／ 4 / 5 = 100 kg
The weight of the goods transported by Party B: 100 △ 5 × 6 = 120 kg
This batch of goods is: 100 + 120 = 220 kg
B: 220-80 = 140 kg