Detailed explanation of the eighth Hope Cup in Grade 6

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1, original title:

? Analysis: Compared with the calculation problem given in Training Hundred Questions, this should be a simple calculation problem. The knowledge points used are mainly the fractional part of the cycle. If the fractional part of this infinite loop with a loop segment of "1" is one ninth, the correct answer should be worked out.

? 2. Original title:

? Analysis: This question is a slight revision of the ninth question in the "Training Hundred Questions".

? Then, the solution is naturally the same. Through the conditions given in the question, the following equation can be obtained:

? 3a+2=4b+3=5c+3

By: 4b+3=5c+3, and they are all natural numbers less than 10.

? It is easy for us to come to such a conclusion. B=5, c=4, further, a=7

So: (2a+b)/c=(2*7+5)/4=4.75.

? 3. Original question: If an operation is represented by "*" and the following relationship is satisfied:

( 1) 1* 1= 1; ? (2)(n+ 1)* 1 = 3×(n * 1)。

So, 5* 1-2* 1=? .

? Analysis: This is a question of "defining a new operation". "Training Hundred Questions" 2 1 came out after the title was changed.

? The way to do this kind of problem is to calculate step by step in strict accordance with the operation rules given in the problem.

? Specific to this question is:

5* 1-2* 1

? =3×(4* 1)-3×( 1* 1)

? =3×3×(3* 1)-3

? =3×3×3×(2* 1)-3

? =3×3×3×3( 1* 1)-3

? =3×3×3×3× 1-3

? =8 1-3

? =78

4. Original question: A fraction is equal to 2/3 after the numerator minus 1 and 1/2 after the numerator minus 2. What is the score? .

? Analysis: This question has no shadow on the "Hundred Training Questions", but it is a question with high frequency in junior high school mathematics. The problem itself is not difficult. Even those students who haven't studied Olympic Mathematics in primary school should see this problem in the synchronous practice of textbooks. Even if you can't find a method, you can make a trial calculation. The answer is: 5/6

5. Original question: Fill the eight numbers of 2, 3, 4, 5, 6, 7, 8 and 9 in the following eight boxes respectively (cannot be repeated), and many different subtraction formulas can be formed. In order to minimize the calculation result and make it a natural number, the calculation result is:

□□□□-□□□□

Analysis: This is a maximum problem. This problem has a prototype in many materials.

? "Use 1, 2, 3, 4, 5, 6, 7 and 8 once each to form two four-digit numbers. To minimize the difference between these two four digits, what are those two four digits and what are their differences? "

If you want to minimize the difference between two four-digit numbers, you should make them as close as possible.

First of all, the difference between the highest digits cannot exceed "1", that is, it can only be "1"

? Secondly, the three digits after the big number should be the minimum, and the three digits after the small number should be the maximum.

? The specific question is: 6234-5987=247?

? The answer to the prototype question is: 5 123-4876=247.

? Interested students can try it themselves:

? 9234-8765=

? 8234-7965=

? 7234-6985=

? 5236-4987=

? 4256-3987=

? 6. Original question: How many balls are there in a box? For the first time, Mr. Wang took out half the balls from the box and put them in 1 ball. The second time, he still took out half the ball from the box and put it in 1 ball. Was there a ball in the box before ....................................................................................... took it out? A.

Analysis: This is a very old problem. It often appears in many books and materials about children's intellectual training and development.

We can use reverse deduction to see what this problem is like.

Finally, there are two balls in the box. One of the two balls was just put in. If you don't put the ball, it's just a ball; And this ball is the other half left after taking half. If you don't take that half, there should be two balls. And if you take out half of the two balls and put one in, it's the opposite.

Therefore, we can say with certainty that there were two small balls in the box before the ball was taken out.

7. Original question: During the Chinese New Year, students should make some handicrafts for the elderly in nursing homes. At first, the students in the art group worked for one day, and later, the students who added 15 worked with them for another two days, just finished. Assuming that each student's work efficiency is the same, it takes 60 days for a student to complete it alone. Then, who are the students in the art group? A little.

Analysis: this is the 74 th question on the "Hundred Training Questions", but it is just a different statement.

We can assume that a classmate can only make one handicraft a day, so it is 60 handicrafts.

Because the additional 15 students worked for two days, then this 15 student completed 15*2=30 (pieces) of handicrafts, so the other 30 artworks were all completed by students of the art group. You can know that the art group has 1 until the students of the art group have worked for three days.

8. Original question: A supermarket has an average of 60 people queuing for payment every hour, and each cashier can handle 80 people per hour. One day, there was only one cashier in the supermarket, and there were no customers waiting in line for four hours after the payment began. If there were two cashiers working at that time, then the payment began? No one lined up after work.

Analysis: the 78 th question on "Training Hundred Questions" was copied as it is.

Obviously, this is a problem of "cows eating grass". We can change it into a model of "cows eating grass" first. That is, the grass on a meadow grows very fast, increasing by 60 servings per week, and a Niu Yi can eat 80 servings per week; If a cow is allowed to eat on this grass, it can last for four weeks. If you let two cows eat, how many weeks can you eat?

The amount of grassland grass is: 4*80-4*60=80 (parts)

In a week's time, after two cows have treated 60 newborn grasses, there are still 2*80-60= 100 (servings) to treat the original grasses. That is to say, the efficiency of these two cows in specially treating the original grasses is 100/week.

80/ 100=0.8 (weeks)

Specific to this problem, it is 0.8 hours.

? When this problem was solved here, I suddenly remembered the last question in the sixth grade test of Hope Cup and the last question in the preliminary test of China Cup (primary school group) a few days ago. Let's think about whether these problems have the same effect.

9. Original question: The following four figures are all composed of six identical squares, among which what can't be folded into a cube? .

Analysis: This question can be regarded as a question of giving points. The answer is "A".

? The replica of 64 questions in this topic "Training Hundred Questions".

10. Original question: The side lengths of the four squares shown in the figure below are all 1, and the shaded areas in the figure are indicated by S 1, S2, S3 and S4 in turn. What is the order of S 1, S2, S3 and S4 from small to large? .

Analysis: In this set of papers, this question should be regarded as a relatively difficult one. But judging from the students' answers, most of them got this question right. Of course, in this pair, "Meng" played a great role. If you really want to conduct strict argumentation and reasoning, I am afraid that few people can really answer it. Fortunately, this problem only depends on the results, not the process. This is naturally to be given. Here I will talk about my understanding of this problem.

Since we have to queue up from small to large, we must accurately calculate the shadow area in each picture.

The areas of Figure (1), Figure (2) and Figure (3) are all easy to find, which are 0.57, 0.2 15 and 0.5 respectively, while the area of Figure (4) is not so easy to find. It is obviously impossible to use the knowledge of primary school.

Here we can review the 60 th question of "Hundred Questions Training", which is also a question of comparing the size of the area. Under the conditions given by that question, it is impossible to directly find the area of the shadow part. But the answer given in the question skillfully adopts the method of cutting and filling, which easily solves the problem. Here we can get some enlightenment from it, and we can also solve this problem by cutting and filling.

As can be seen from figure 1, the area of the upper and lower red triangles is half that of the square.

From Figure 2, we can see that the area of the green part is not equal to the area of the yellow part. If the area of the green part is cut to the yellow area, it can be seen that the part representing the shadow area is smaller than the areas of the two red triangles in 1, that is, the area of the original shadow part is smaller than 0.5, but closer to 0.5.

From this, we can draw the conclusion that S2

Supplement: Regarding the fourth figure in the question 10, the shaded part can be combined into the sum of the red and cyan parts in the figure below by cutting and filling.

? The area of the red part is 0.2 15, which is exactly equal to the area of the second figure, and the cyan part is just the part where the fourth figure is more than the second figure. So? S4 area is larger than S2 area.

1 1. The original title was Question 72 of Hundred Questions Training, and the word has not been changed. I won't copy the original question here.

? Analysis: The key to solve this problem is that the length of two iron bars in water is equal. From this, it can be easily concluded that the ratio of the lengths of the two rods is 5: 6, and it is further concluded that the difference of the lengths of the two rods is 3 cm.

? 80% of the students did this problem correctly, which can be regarded as a sub-problem.

? In addition, I want to say that the second question of the China Cup preliminary test the day before was similar to this. Could it be that the teacher is giving a question?

? 12, A, B and C go fishing together. They put the fish they caught in the fish basket and lay down to rest. They are all asleep. A woke up first. He divided the fish in the fish basket into three parts equally. When he found another fish, he threw the extra fish back into the river and took one home. B woke up, divided the existing fish in the fish basket into three parts, and found one more fish. He also threw the extra fish back into the river and took one home. C finally woke up, and he also divided the fish in the fish basket into three parts. At this time, there was another fish. Require these three people to catch at least? A fish.

? Analysis: This problem can be solved by reverse calculation.

? Since it is the minimum, let's assume that there are only four fish left after C wakes up. Therefore, B should see seven fish after waking up, which is inconsistent with reality, because A threw a fish back into the river, which means A divided the fish according to the number of fish. In other words, the remaining two copies should add up to an even number. And 7 is not an even number;

? Then let's assume that C wakes up and sees seven fish. With the above example, nature is inconsistent with reality.

If C wakes up and sees 10 fish, B sees 16 fish, and A sees 25 fish before dividing the fish, then the answer is 25.

13, in winter, the white rabbit only stored 180 carrots, and the little gray rabbit only stored 120 Chinese cabbage. In order to have carrots to eat in winter, the little gray rabbit exchanged a dozen Chinese cabbages for some white rabbits' carrots, and at this time they stored the same amount of food. Then a Chinese cabbage can be exchanged for a carrot.

? Analysis: This issue should be considered as a whole first. Their total food is 180+ 120=300, so when they are equal, each rabbit should have 300/2= 150 (one, one).

? The original 120 was replaced by 150, an increase of 30.

? That is, the little gray rabbit took out more than a dozen, and then changed back to a number 30 more than these dozen.

? We can calculate that the possible situation is:

? The little gray rabbit took out 1 1 cabbage and replaced it with 4 1 carrot.

? The little gray rabbit took out 12 cabbage and changed 42 carrots.

? The little gray rabbit took out 13 cabbage and changed 43 carrots.

? Little gray rabbit took out 14 cabbage and changed 44 carrots.

? Little gray rabbit took out 15 cabbage and changed 45 carrots.

? The little gray rabbit took out 16 cabbage and changed 46 carrots.

? The little gray rabbit took out 17 cabbage and changed 47 carrots.

? The little gray rabbit took out 18 cabbage and changed 48 carrots.

? Little gray rabbit took out 19 Chinese cabbage and changed 49 carrots.

? In contrast, in these nine cases, the answer that best fits the meaning of the question is "? The little gray rabbit took out 15 cabbage and changed 45 carrots. "

? So, our answer is "3".

In this question, the answer given by some students is "4", which may be that ten trees are regarded as more than a dozen trees, and 10 trees are just taken out, and 40 trees are replaced, and the number has just increased by 30. But without further calculation, in fact, 15 tree is a better and more reasonable number.

? 14, Wang Yu plays balloon shooting. The game has two levels, and the number of balloons in the two levels is the same. If the number of balloons shot by Wang Yu in the first pass is four times more than the number of balloons missed, there will be two more; The number of balloons shot in the second level is 8 more than that in the first level, which is exactly 6 times the number of missed balloons, so there are 8 balloons in each level in the game.

? Analysis: this question is the same as the 43 rd question of "Training Hundred Questions", but the scene and quantity have been changed, which is essentially the same.

? It is easier to solve this problem with equations.

? Let the number of balls missed in the first pass be x, then the number of balloons hit in the first pass is 4x+2;

The number of leaking balls in the second pass is X-8, and the number of balloons in the second pass is 4X+2+8.

According to the conditions given in the question, (X-8)*6=4X+2+8.

? Solution: X=29

Therefore, the number of balloons in each gap is 29*(4+ 1)+2= 147 (only).

? 15, the original question: It is known that Xiaoming's parents are different in age, and the difference is no more than 10 years old. If last year, this year, and next year, the ages of mom and dad are all integer multiples of Xiaoming's age, what about Xiaoming this year? Years old.

? Analysis: This question evolved from the 4 1 question in the "Training Hundred Questions".

? Because age is calculated by integers, last year, this year and next year are three consecutive natural numbers, and one of these three consecutive natural numbers must be a multiple of 3.

? Because the age of both parents for three consecutive years is an integer multiple of Xiaoming's age, it is conceivable that Xiaoming's age will not exceed 4 years old.

? It is also known that the age difference between father and mother does not exceed 10, and the conditions are further narrowed. It is known that Xiao Ming's age in these three years can only be 1, 2 or 3 years old.

? However, the corresponding ages of their parents can only be: father: 3 1, 32, 33; Female: 25, 26, 27.

Or: father: 37,38,39, mother: 3 1, 32,33.

? If there is no age difference between parents in this question,

? Xiaoming's age may also be 2, 3 or 4 years old.

? And dad's age corresponds to: 38, 39, 40,

Mother's age corresponds to: 26, 27, 28.

16. Observing the subtraction formula shown in figure 1, it is found that the numerical order of the obtained number 175 and the minuend 57 1 is opposite. Then, after subtracting 396, there are three numbers * * * that make this number in the opposite order to the number being subtracted.

? Analysis: This is a question about "numbers and digits". It is the most common question type of Hope Cup, and it is a required question type. The 80th issue of Training Questions has given a detailed answer to this question. Here we use the method of digital riddle to analyze the problem.

? Let's look at Figure 2. This is a subtraction formula. If you subtract three digits from three digits, this number is still three digits. It means that a and c are definitely not zero.

? Look at the tenth number. B MINUS 9, B appears in the number again, indicating that B borrowed it when subtracting 9.

? Look at the hundreds again. After A is borrowed from "1", subtract 3 to get "C", which means A is a number that is 4 larger than C.

? From this we can determine that a and c may be:

5, 1;

6,2;

7,3;

8,4;

9, 5, * * * There are five groups of situations.

? When b is any single digit (including 0), there are 10 * * cases.

? All the formulas listed in Figure 2 are valid. 5* 10=50 (piece)

? 17, original title: two garment factories, A and B, produce the same kind of clothing. Factory A produces 2,700 sets of clothing every month, and the time ratio for producing tops and trousers is 2:1; Factory B produces 3600 sets of clothes every month, and the time ratio of producing tops and trousers is 3: 2. If the two factories cooperate for one month, what can they produce at most? Settings.

? According to the known conditions, a factory can produce 135 shirts and 270 pairs of trousers every day.

? Factory B can produce 200 coats and 300 pairs of trousers every day.

? By comparison, we can see that the efficiency of jacket production in factory B is much higher than that in factory A, but there is not much difference in the production of trousers between the two factories.

Because it's very troublesome to produce coats, we arranged for Factory B, which has the most advantages in this field, to spend all our time producing coats.

Then factory B can produce 200*30=6000 coats a month (30 days);

And let a factory also specialize in producing pants to match the tops produced by B factory at the beginning. While A only needs 6000/270=200/9 (days) to produce 6000 pairs of pants.

A factory still has 30-200/9=70/9 (days) to produce tops and trousers in proportion;

During these 70/9 days, a factory can also produce garments in sets: (70/9)/(30*2700)=700 (sets).

Plus 6000 sets of cooperative production, the maximum output can be: 6000+700=6700 sets.

? 18, original title: A cashier found that the cash was less than the book record when checking the accounts before leaving work 153 yuan. She knows that the actual collection can't be wrong, but the decimal point is wrong when bookkeeping. So how much cash did the wrong account actually receive? Yuan.

? Analysis: As a cashier, check whether the cash received matches the receipt before going to work every day.

? Of course, "the actual collection can't be wrong, but the cash and book records are missing 153 yuan", indicating that there is something wrong with bookkeeping.

? "There is an error in the decimal point of one of the counting points" and there are too many records, indicating that the decimal point has moved by one place, which has enlarged the original number by 10 times, that is, recorded 9 times more than the original number. Divided by 9, the extra 153 yuan is the actual cash received. 153/9= 17 (yuan).

? This question examines students' understanding of decimal points. Although it is a knowledge point in the fourth grade, in the junior high school exam, the frequency is very high, and the answers to such questions are also very simple. As long as the decimal point is moved by one place, the original number will be expanded to 10 times or reduced to one tenth.

? 19, there are 4 pieces of A with 5 tons, 6 pieces of B with 4 tons, 1/kloc-0 with 3 tons, and 7 pieces of D1ton. If you want to transport all the parts at once, you need a car with a load of at least 6 tons? Cars.

? Analysis: This is a global problem. Even if it appears on the examination paper of the second grade students in primary school, it can't be considered as super-class. But now it appears in the competition paper of grade six, occupying a very special position. Under normal circumstances, this position is the finale. This seems a little incredible, but it is for this reason that we see the cleverness of the teacher in charge of the examination paper. Because when grading, we found that more than half of the students lost points on this road. Is this even more incredible?

? In fact, this question is very simple. Just put the picture on the draft paper and put it together.

? 5? 1 5? 1 5? 1 5? 1

? 4 4 4 4 ? 4? 1? 1 4 1 1

? 3? 3 ? 3? 3 3? 3 3? 3 3? 3 3

Look at how many groups there are. Just arrange a few cars.

20. Original question: Both Party A and Party B start from A and B at the same time, and they go in opposite directions. When they set off, their speed ratio was 3: 2. After meeting, the speed of A increased by 20%, and the speed of B increased by 1/3, so that when A has been to B, B is still 4 1 km away from A, so what is the distance between A and B? kilometre (km)

? Analysis: "No fish, no table", the problem of travel has always been an indispensable and important dish in all primary school comprehensive examination papers. But putting this question here seems not to be the finale, but to complement the figures. In fact, this is a wonderful question, which comes from the 52nd question in "Hundred Training Questions". Although only two pictures have been changed, they have become the finishing touch, so that many students "look very simple and familiar, but they are not doing it right."

? Drawing line segments is the most commonly used and practical tool to solve travel problems. Because of the time, I won't draw it here.

? Because they walk in opposite directions at the same time, the speed ratio of A and B is 3: 2, so the ratio of their trips when they meet must be 3: 2, that is to say, A walked three-fifths of the whole journey and B walked two-fifths of the whole journey;

? After they met, they sped up separately, and the speed ratio changed from 3: 2 to 27: 20.

? A still walks fast, only two-fifths of the way to B, while B is still slow, three-fifths of the way to A, so when A arrives at B, B must still be on the way to A;

According to their speed ratio, it is easy to find that in the same time, when A walks the remaining two-fifths of the whole course, B can walk a small part of the whole course accordingly. That is, when A arrived at B, B walked 8/27 of the whole journey;

Then, from B to A, there is still 3/5-8/27 = 4 1135. Here, we will see a bright number "4 1", because it just corresponds to "B is still 41km away from A", so

Generally speaking, this set of papers is very level. Moreover, most of the questions come from "training 100 questions", which gives more "hope" to the participating students.

It is suggested that students who take the second exam should work harder on "training 100 questions", because we found that many weighty questions in "training 100 questions" did not appear in this set of papers and should be reserved for the second exam.

Everyone should pay more attention to counting, graph theory, combination and number theory.

We still remember the last two questions in the second exam of grade five and six last year, and that's where we really showed our level.