The carrying capacity of a certain ship is 260 tons, and the volume of its cargo hold is 1000 cubic meters. There are two kinds of goods (A and b) to be transported, of which the volume of a cargo is 8 cubic meters per ton, and that of B cargo is 2 cubic meters per ton. If we want to make full use of the carrying capacity and cargo hold volume of the ship, how many goods should be loaded in each of them?

The carrying capacity of a certain ship is 260 tons, and the volume of its cargo hold is 1000 cubic meters. There are two kinds of goods (A and b) to be transported, of which the volume of a cargo is 8 cubic meters per ton, and that of B cargo is 2 cubic meters per ton. If we want to make full use of the carrying capacity and cargo hold volume of the ship, how many goods should be loaded in each of them?

Suppose that the mass of armored goods is x tons and that of type B goods is y tons. Solve the equations x + y = 2608x + 2Y = 1000 and get x = 80y = 180. Answer: the mass of armored goods is 80 tons and that of type B goods is 180 tons

A ship has a capacity of 300 tons and a volume of 1000 cubic meters. There are two kinds of cargo to be loaded, namely, 7 cubic meters for class a cargo and 2 cubic meters for class B cargo
How to arrange the cargo transportation in order to make full use of the carrying capacity and volume of the ship?

x+y=300
7x+2y=1000
x=80
y=220

There is a barge with a carrying capacity of 800 tons and a volume of 795 cubic meters. Now it is carrying pig iron and cotton. The volume of pig iron per ton is 0.3 cubic meters, and that of cotton per ton is 4 cubic meters. How many tons of pig iron and cotton can we make full use of the carrying capacity and capacity of the barge?

Suppose the pig iron transports x tons, then the cotton transports (800-x) tons. From the meaning of the question, we can get: 0.3x + 4 (800-x) = 795, the solution is: x = 650800-650 = 150 (tons). Answer: the pig iron transports 650 tons, and the cotton transports 150 tons