The ratio of copper to zinc in the alloy is 2∶3. Now add 6 grams of zinc to get 36 grams of new alloy. What is the ratio of copper to zinc in the new alloy?

Analysis: The ratio of copper and zinc in the new alloy is required, and the respective weights of copper and zinc in the new alloy must be calculated separately. It should be noted that when the ratio of copper to zinc is 2∶3, the weight of the alloy is (36-6) grams instead of 36 grams. The weight of copper has never changed.

Solution: When the ratio of copper to zinc is 2∶3, the alloy weight is:

36-6 = 30 (grams).

Weight of copper:

The weight of zinc in the new alloy;

36- 12 = 24 (g).

The ratio of copper to zinc in the new alloy;

12∶24= 1∶2.

A: The ratio of copper to zinc in the new alloy is 1: 2. The biggest problem.

The lecture hall has 10 rows of seats, and each row has 16 seats. When there are 65,438+050 people, some rows have the same number of people. We hope that the fewer rows with the same number of people, the better, at least there are rows with the same number of people.

Solution: There are at least 4 lines.

If the number of people in the platoon is different, the maximum number of people in 10 platoon is 16, 15, 14, 13, ... 7, and the maximum number of people is16+/kloc-0.

If there are at most two rows with the same number of people, the maximum number of seats is (16+15+14+13+12) × 2 =140 (person);

If there are at most three rows with the same number of people, the maximum number of people is (16+15+14) × 3+13 =148 (people);

If there are at most four rows with the same number of people, the maximum number of seats is (16+15) × 4+14× 2 =152 (people).

Due to 148

Some students in A and B schools know each other. It is known that none of the students in A school can know all the students in B school, and any two students in B school have a friend in A school. Q: Can we find two students A and B in A school and three students C, D and E in B school, so that A can know C and D but not E? Explain why. Understanding is mutual, that is, when A knows B, B also knows A..

Analysis: If you choose any two students in school B as C and D, then there are people in school A who know C and D, so let's call them A, because A can't know all the students in school B, and there are students in school B, so A doesn't know E. At this time, A knows C and D, but doesn't know E. According to this idea, it is a bit troublesome to consider choosing B. Although there are students in school A for D and students in school B for E, it is known that there are students in school A. The reason for this dilemma is that C and D in B school are too "arbitrary". Checking C and D in B may result in failing to check B in the last A. It seems that we should choose a special person.

Because all the students in school A don't recognize all the students in school B, one person knows the most students in school B (or one of the students who knows the most in school B). Choose him as A. Because A can't recognize all the students in B school, so choose a student from B school who A doesn't know as E, and let the students from B school who A knows as D..

For D and E, there is one person in a school. Think of it as B. B knows D and E. Because B knows E and A doesn't know E, A and B are not the same person.

Among the students in school B that A knows, there must be people that B doesn't know. Otherwise, when any student in school B whom A knows knows knows B, B certainly knows at least one person E than A, which contradicts the assumption that "A is the one who knows the most students in school A". In school B, student C knows A but not B, so there are:

A knows C and D, but doesn't know E.B. knows D and E, but doesn't know C. regularity.

The station delivered 2000 boxes of glass to a factory. According to the contract, the money was delivered to 5 yuan intact. If a box is damaged, it will compensate 40 yuan, not the freight. After the glass was delivered, the station * * * received the payment of 9 190 yuan. How many cases of glass have been damaged?

Solution: ① Arithmetic solution: If all 2000 cases of glass are delivered without damage, the payment should be 2000×5 = 10000 yuan.

Difference between actual freight and payment:

10000-9190 = 810 (yuan).

Now, let's exchange a box of good glass for a box of bad glass. The total number of boxes remains unchanged at 2,000, but the freight per box has decreased:

40+5 = 45 (yuan)

So how many boxes were changed, and the payment was only reduced by 8 10 yuan? Do division:

8 10 ÷ 45 = 18 (box).

Answer: * * * change 18 cases.

② Algebraic solution:

If the X box is damaged, the X box is not damaged.

Establish an equation according to the meaning of the question.

5(2000-x)-40x=9 190

45x= 10000-9 190

45x=8 10

x= 18。

Answer: 18 box is damaged.

collect coins

1 cent, 2 cents, 5 cents, how many different ways to play a dollar?

Analysis: Using 1 min, 2 points and 5 points to make up one yuan is the same as using 2 points and 5 points to make up less than one yuan. So, this question is transformed into: "How many different ways to make up for less than one yuan?"

Solution: Count according to the number of nickels in 2 1 category;

If there are 20 nickels, obviously there is only one way to put them together;

If there are 65,438+09 nickels, the value of two nickels will not exceed 65,438+000-5× 65,438+09 = 5 (cents), so there are three different ways to make two nickels.

If there are 18 nickels, the value of a 2-cent coin does not exceed100-5×18 =10 (cents), so there are 6 different nickels.

... in this way, we can get different ways to get together. * * * Yes:

1+3+6+8+ 1 1+ 13+ 16+ 18+2 1+…+48+5 1

=5×( 1+3+6+8)+4×( 10+20+30+40)+5 1

=90+400+5 1

=54 1 (species).