If you can use the solution of the equation, write the solution. If you can't use the solution of the equation, use the arithmetic method 1. The clothing factory used to make 2.5 meters of cloth for a suit of clothes. After improvement, 0.5 meters can be saved for each suit. How many sets of cloth can be made for 60 sets of clothes? 2. The speed sum of car a and car B is 110 km / h. car a travels 10 km more per hour than car B

If you can use the solution of the equation, write the solution. If you can't use the solution of the equation, use the arithmetic method 1. The clothing factory used to make 2.5 meters of cloth for a suit of clothes. After improvement, 0.5 meters can be saved for each suit. How many sets of cloth can be made for 60 sets of clothes? 2. The speed sum of car a and car B is 110 km / h. car a travels 10 km more per hour than car B

1. (60x2.5) / (2.5-0.5) = 150 / 2 = 75 sets
A: the cloth used to make 60 sets of clothes can now be made into 75 sets
2. Suppose that the speed of car B is X
x+(x+10)=110
x+x+10=110
2x=110-10
2x=100
x=50
A: the speed of car B is 50 km / h

Application of equations in solving urgent problems
Four hours later, the two cars meet at a distance of 15 meters from the destination. It is known that car a travels 41 kilometers per hour, and how many kilometers per hour does a car travel?

Let car B travel x kilometers per hour. When car B is fast, 4x-41 * 4 = 0.015 * 2 leads to x = 41.0075. When car B is slow, 41 * 4-4x = 0.015 * 2 leads to x = 40.9925. That is to say, car B travels 41.0075 kilometers or 40.9925 kilometers per hour