First, the issue of chartering.

1, the problem of chartering runs through the mathematical method of assuming first and then adjusting. The most economical way to rent a boat is: first, try to rent a seat with a more favorable unit price, and second, try not to have an empty seat.

2. Judging whether it is cheaper to rent a big boat or a small boat, according to common sense of life, it is generally cheaper to rent a big boat; Try to rent a big boat. How many big ships are there? 32 ÷ 6 = 5 big ships, and there are 2 people left.

3, learn to adjust, try not to empty seats, one boat leaves two people, but two seats are still empty, which is not the most economical. If you adjust to five big boats, there will be eight people left, just taking two boats.

4. However, this scheme is not absolute. For example, a teacher takes 48 students to go boating in the park, and the boat is limited to 5 people. The rent for each boat is 30 yuan, and the boat is limited to three people. The rent per ship is 2 1 yuan. How to rent a boat is the most economical, according to the two most economical principles.

We should rent a boat like this: a big boat with eight people and a small boat with three people, just enough for 49 people. The cost is 8×30+3×2 1=303 yuan. But the most economical plan is to rent a large ship of 10, which can take 50 people and keep two seats. The cost is 10×30=300 yuan.

6. Applying formulas is not the fundamental method to solve problems, but a universal law. When practicing, let the students experience the mathematical method of assuming first and then adjusting, and let the students realize that the mathematical method is more important and can't apply the formula without thinking.

7. What students tend to overlook in this problem is that there are not only 326 students but also 14 teachers by bus. The total number of passengers is 326+65,438+04 = 340,340 ÷ 40 = 8 carts ... 20 people, 20÷20 = 1 car.

Second, the issue of buying tickets.

1, 6 adults and 4 children, which scheme is cost-effective? For this kind of problem, two schemes are generally formulated and then compared.

2. Option 1 purchase:150× 6+60× 4 =140 (yuan) Option 2 purchase: (6+4) × 100 = 1000 (yuan).

There are 4 adults and 6 children. Which scheme is cost-effective? Similarly, the first option is to buy 150× 4+60× 6 = 960 (yuan), and the second option is to buy (6+4) × 100 = 1000 (yuan). 4 adults and 6 children, children tickets are cheap. choose

However, it should be noted that sometimes the topic is special, and it will be cheaper to mix the two schemes. For example, for the second question above, there is a more economical scheme. Five people can buy group tickets, and four adults plus 1 child make up five group tickets.

5,5×100 = 500 yuan, at this time, there are 6- 1 = 5 children. Buy a child ticket: 60× 5 = 300 yuan, 500+300 = 800 yuan, in fact, the cheapest ticket is only 800 yuan.