Test questions and detailed answers of the sixth grade mathematics competition in primary school

Calculate the following questions first and write a short operation flow (*** 15, 5 points for each small question).

Fill in the blanks (***40 points, 5 points for each small question)

1. Fill in the appropriate operation symbol in the following "□" to make the equation hold:

( 1□9□9□2)×( 1□9□9□2)×( 19□9□2)= 1992

2. The length of an isosceles trapezoid with three sides is 55cm, 25cm and 15cm respectively, and its base is the longest side. Therefore, the circumference of this isosceles trapezoid is _ _ cm.

There are 90 seats on a row of benches, and some seats have been taken. At this moment, another man came and sat on this bench. Interestingly, no matter where he sits, he is adjacent to the person who has already sat. It turns out that at least _ _ people have been seated.

4. 1992 divided by the natural number a, the quotient is 46, and the remainder is R. a=_ _, r=_ _.

On the Double Ninth Festival, 25 old people came to Yanling Teahouse for tea. Their ages are exactly 25 consecutive natural numbers. Two years later, the sum of the ages of these 25 old people is exactly 2000. Among them, the oldest old man is _ _ _ years old this year.

6. The school bought several books on history, literature and art, and popular science, and each student borrowed two books at will. Then, at least two of these _ _ _ students must have borrowed similar books.

7. In a math contest, five players scored 404 points, each player scored unequally, and the player with the highest score scored 90 points. Then the player with the least score will get at least _ _ _ _ points and at most _ _ _ _ points. (Each player's score is an integer)

8. To saw high-quality copper tubes with a length of 1 meter into small copper tubes with a length of 38 mm and a length of 90 mm, each saw consumes 1 mm copper tubes. Then, only when the sawed 38 mm copper tube is _ _ _ _ _ section and the sawed 90 mm copper tube is _ _ _ _ _ _ section, the copper tube loss can be minimized.

Third, answer the following application questions (write a column solution process. When making a formula, you can do it step by step, you can also do a comprehensive formula, and you can also do an equation (***20 points, 5 points for each small question).

1. Two engineering teams, Party A and Party B, jointly built a 4200m-long highway, and Team B built more than Team A every day 100m. Now it will be repaired by engineering team A for 3 days. The rest of the road sections were jointly repaired by Team A and Team B, and it took only 6 days to complete. Q: How many meters do the A and B engineering teams build roads every day?

2. A person rides a bicycle from the county seat to the township to set up a factory. He started from the county seat by bike and completed half the journey in 30 minutes. At this time, he sped up, driving 50 meters more every minute than before. After riding for another 20 minutes, he knew from the mileage card on the side of the road that it would take another 2 kilometers to set up a factory in the township and seek the total distance from the county to the township.

3. The width and height of a cuboid are equal, both equal to half the length (as shown in figure 12). Cut this cuboid into 12 small cuboids, and the sum of the surface areas of these small cuboids is 600 square decimeters. Find the volume of this big cuboid.

Three workers in a bookbinding workshop will pack a batch of books and send them to the post office (each package is required).

35 more copies. The second time, they brought all the rest of the books, plus the fraction of the first time, which just filled 1 1. How many books are there in this batch?

Four. Questions and answers (***35 points)

1. There are 1992 buttons. Two people take turns to take a few buttons from them, but each person takes at least 1 buttons and at most 4 buttons. Whoever takes the last button loses. Q: What are the countermeasures to ensure victory? (5 points)

There is a square thick paper with a side length of 24 cm. If you cut a small square at each of its four corners, you can make a carton without a lid. Now, in order to maximize the size of the carton, how many centimeters should the side length of the cut small square be? (6 points)

3. Jin Shifu of individual iron works needs two shapes of iron blanks (A) and (B) as shown in figure 13 to process some iron products. At present, there are two pieces of scrap iron A and B (as shown in figure 14 and figure 15). The small squares in figure 13, figure 14 and figure 15 are all equilateral squares. Jin Shifu wants to choose one piece from them, so the selected iron sheet is just suitable for processing the complete set of this iron sheet product ("complete set" means that there are as many iron sheets as (a) and (b)) without wasting any materials.

Q: (1) Which scrap iron should Jin Shifu choose from? (3 points)

(2) How to cut the selected scraps? (Please draw a cut line mark on the drawing or use a shadow to indicate the blank of a shape) (5 points)

4. Just modify one digit of 2 1475, and the modified number can be divisible by 225. How to modify it? (6 points)

5.( 1) How to divide nine identical chocolates equally among four children (each chocolate can only be cut into two parts at most)? (5 points)

(2) If the word "four" in the above (1) is changed to "seven", will it be divided? If so, how to do it? If not, why? (5 points)

Detailed explanation and explanation

First, the calculation problem

Note: In order to get a simple algorithm, we must first make a comprehensive analysis of each number and operation symbol in the problem.

It should be known immediately that it can become 3.6; However, 3.6 is only one decimal point away from 36, so it is easy to think of transforming "654.3×36" into "6543×3.6". When this step is completed, it will be positive.

"By the same means, this skill has been introduced many times in this newspaper.

Note: The solution to this problem can be observed from different angles. The first solution is to observe and compare the factors of each addend (continued product) in the molecular part, and find out the multiple relationship before and after, so that "1×3×24" is mentioned as the common factor, and the denominator part is similarly deformed. The second solution is to focus on the whole complex fraction, look at the denominator from the numerator and find the left, middle and right multiplication of the numerator.

The sum of the three products in brackets in the molecular part has been removed. This question is adapted from Example 5 on page 2 of Friends of Mathematics (7).

3. Solution 1:

Solution 2:

Note: The first solution is a general method to find the sum of the first n terms of geometric series, which is introduced in the column of "Good Partner Email" in the first edition of 2 17. Because the last addend in this problem is always half of the previous addend, as long as a minimum addend is added, it can make up "2 times", that is, the previous addend, so it is not difficult to think of scheme 2.

Second, fill in the blanks

1. Solution: (/kloc-0 /× 9× 9+2 )× (1+9+2 )× (19-9-2).

=83×3×8

= 1992

Or (1× 9× 9+2 )× (/kloc-0 /× 9 ÷ 9× 2 )× (19-9+2).

=83×2× 12

= 1992

The answer to this question is not unique, as long as the symbols filled in can make the equation hold, they are all correct.

Note: Fill in three operation symbols between four numbers to make the calculation result a known number, which is a "formula puzzle" that players are familiar with. However, this problem is not easy to judge what the calculation results should be in brackets at once, so it is necessary to decompose 1992 into the product of three numbers, 1992=83×3×2×2×2, because there are many combinations of the products of 83, 3, 2, 2 and 2, which are1.

2. Solution: 55+ 15+25×2= 120 (cm)

Note: To calculate the circumference, you need to know how long the upper sole, the lower sole and the two waists are. Easy to judge: the longest bottom should be 55 cm. The key is to judge the waist length. If the waist length is 15cm, 15x2+25 = 55, it means that the sum of the lengths of the upper sole and the two waists is exactly equal to the length of the lower sole, and the four sides cannot form a trapezoid, so the waist length can only be 25cm. Readers should start with "Can any three sticks form a triangle?" In the third edition of this newspaper 190.

3. Solution: At least

Description: According to the meaning of the question, it can be inferred that there are two empty seats between any two adjacent people who are already sitting on this bench. But only from this result, we can't be sure how many seats there are on the bench, because the leftmost (rightmost) who has sat can sit in the first seat on the left (right) or the second seat on the left (as shown in figure 16, "●" means the person who has sat, "○").

But when asked how many people are "at least" seated in the topic, the second case should be chosen, with three people in each group (○● ○) and one person in each group already seated.

( 1)●○○●○○●……

(2)○●○○●○○●○……

Figure 16

4. Solution 1: from 1992 ÷ 46 = 43... 14.

Study immediately: a=43, r= 14.

Solution 2: According to the basic relationship between division and remainder, there are

1992=46a+r(0≤r 4 = 77 (points)

Explanation: To solve this problem, we need to consider two extreme situations:

(1) In order to make the player with the lowest score as few as possible, it is necessary to make the top four score ahead of the fifth as much as possible under the condition that the total score of the five players is fixed. The score of the first place is known (90 points), which requires the scores of the second, third and fourth places to be as close to 90 points as possible and not equal to each other. Only the second, third and fourth place scored 89 points, 88 points and 87 points in turn, and the fifth place scored the least.

(2) To make the player with the lowest score get the most points, the scores of the second, third, fourth and fifth players should be as close as possible under the condition that the total score and the first score are certain. Considering that their scores are not equal to each other, this can only be done when the scores of the second, third, fourth and fifth places are four consecutive natural numbers, and the maximum score of the fifth place can be calculated by the method of "leveling".

This question is adapted from "Friends of Mathematics" (7) page 46 13.

8. Solution: Let's set up 38mm and 90mm copper pipes to saw X and Y parts respectively. Well, according to the meaning of the question, there are

38X+90Y+(X+Y- 1)= 1000

39X+9 1Y= 100 1

In order to reduce the loss, we should saw as many 90 mm long copper pipes as possible, that is to say, X in the above formula should be as small as possible and Y should be as large as possible. Since both X and Y must be natural numbers, it is not difficult to infer that X=7 and Y=8. That is, when 38 mm copper tube saw 7 sections and 90 mm copper tube saw 8 sections, the loss was the least.

Note: After reading the question, the contestants can immediately think that in order to minimize the loss, they should saw as many 90 mm long copper pipes as possible, but they must meet the two conditions of "both kinds of copper pipes are available" and "the sum of the lengths of the two kinds of copper pipes plus the length of the lost part should be equal to 1 m", which is not so simple to calculate. This kind of problem is easier to reason with the help of equivalence relation. However, don't forget the loss of 1mm when listing the equations. The loss of several "1mm" can't be miscalculated, but it should be "total number of segments-1".

After listing the equations, there are two points to note: (1) deformation should be reasonable; (2) Choose a simple algorithm. As in the above solution, 100 1 is written as 7×1×13, 39 as 3× 13, 9 1 as 7×/kloc-0.

This question is the original question of exercise 6 on page 5 1 of Friends of Mathematics (7).

Third, the application problem.

1. Scheme 1: Assuming that Team B repairs as many roads as Team A every day, the roads repaired by the two teams will be 600 meters less and 4,200 meters less, which is equivalent to the roads completed by Team A 15 = 3+6× 2. The formula is as follows:

(4200-600)÷(3+6×2)

= 360015 = 240m

240+100 = 340m

Option 2: If Team A builds roads x meters every day, then Team B builds roads "X+ 100" meters every day. Make an equation according to the meaning of the question.

3X+6×(X+X+ 100)=4200

The solution is X=240.

So X+ 100=340 (m)

A: Team A builds 240 meters of roads every day, while Team B builds 340 meters of roads every day.

Explanation: "Hypothesis" is an arithmetic method that we often use to solve application problems. It embodies wit and agility, and can get answers quickly. This question is adapted from the example in the "Question and Answer" column of the second edition of the 234th issue of this newspaper.

2. solution: from the topic, we can see that half of the total distance was completed in the first 30 minutes, and the other half was not completed in the last 20 minutes, which is 2 kilometers less than half of the total distance. In other words, the last 20 minutes are 2000 meters less than the previous 30 minutes. Why is it missing? There are two reasons: (1) The last 20 minutes are less than the previous 30 minutes 10 minutes; (2) The last 20 minutes are 50 meters more than the previous 30 minutes. In this way, it is easy to infer that the distance traveled every 10 minute in the first 30 minutes is 20×50+2000=3000 (meters). 3000÷ 10= 300 meters per minute for the first 30 minutes. The total distance is

300×30×2

= 18000 (m)

A: The total distance from the county seat to the township factory is 18km.

Description: The key to solve this problem is: (1) By comparison, know how many miles this person walked in the first 30 minutes compared with the last 20 minutes; (2) Find out why the first 30 minutes are 2000 meters more than the last 20 minutes. For details, please refer to the article "Grasping Contradictions and Finding Reasons" in issue 209 of this newspaper.

3. Solution 1: If the left (right) surface area of a large cuboid is x square decimeter, then the surface area of this large cuboid is 10X. After cutting into 12 small cuboids, the newly increased surface area is

(3X+2×2X)×2= 14X

/kloc-the sum of the surface areas of 0/2 small cuboids is

10X+ 14X=600

X=25

V=25× 10=250 (cubic decimeter)

Solution 2: Take a large cuboid with a surface area of-"1"and cut it into 12 small cuboids.

V=25×5×2=250 (cubic decimeter)

The volume of this big cuboid is 250 cubic decimeters.

Note: this problem is relatively simple, as long as you understand the relationship that "the newly increased surface area is equal to twice the cross-sectional area" after cutting a geometric figure into two parts, but when calculating the newly increased surface area, you will make mistakes carelessly. This topic is adapted from the example in the column "Teach you to think" in the first edition of the 226th issue of this newspaper.

Because 10 package +25 books +35 books ←→ 1 1 package.

So 1 package ←→60 copies.

(14+11) × 60 =1500 (Ben)

Solution 2: (Solution of the column equation)

There are 7X= 14Y+35 (1).

5X= 1 1Y-35 (2)

(1)-(2) to obtain zx-3y+70 (3)

(1)+(2), and 12X=25Y (4) is obtained.

(3)×6， 12X= 18Y+420 (5)

Comparing (4) and (5), there are

25Y= 18Y+420

The answer is Y=60.

12X=25×60= 1500 (Ben)

A: There are 1500 books in this batch.

Note: the quantitative relationship in this problem is actually easy to see, and the result of solving 1 is almost calculated in your mind. So, don't think too complicated. Option 2 is easy to think of, but setting the "unknown" is also very particular. If there are X copies in this batch, the deformation will be more troublesome.

Fourth, question and answer.

1. A: The countermeasure to ensure a certain victory is: (1) Take 1 buttons first, and then there are 199 1 buttons left. (2) It's the opponent's turn to take it. If the opponent takes n (1≤n≤4), take "5-n" by himself. After 398 cycles, he took out 398×5= 1990 buttons, leaving 1 buttons.

Note: In this question, only the word "win" in "Example 1" in the column of "Olympic School" in the 233rd issue of this newspaper is changed to "lose". A word difference, the countermeasures have changed. We know that there is a basic idea to solve the countermeasure problem: leave the possibility of failure (loss) to the opponent. In this question, whoever gets the last button loses, so if you want to win, you must get 199 1. With this in mind, it is easy to find countermeasures to ensure victory.

2. Answer: The side length of the cut small square should be 4 cm.

Note: To answer this question, you can compare it with a table first. By comparison, it is easy to know that when the side length of the cut small square is several centimeters, the volume of the carton is the largest.

You can see the results immediately from the above table.

It can also be assumed that the side length of the cut small square (the height of the carton) is H, so the side length of the bottom surface of the carton is 24-2h. Its volume is

Because 24-2h+24-2h+4h=48 (fixed number), according to the conclusion introduced by Friends of Mathematics (7) on page 23, when 24-2h=4h, the product of (24-2h)×(24-2h)×4h is the largest. That is to say, when h=4, v is the largest.

3. Answer: (1) Metal sheet should be selected.

(2) The cutting method is shown in figure 17.

Note: The title requires that a piece of iron sheet material is suitable for making "complete sets" of iron sheet products, which requires that the selected iron sheet material contains as many as (a) and (b) blanks; Because we can't waste materials, we just need to count (how many small squares are there in two materials A and B) to see if the total number of small squares in A (or B) can be divisible by (10+7= 17).

When answering the question (2), two blanks (a) and (b) can be spliced into a figure 18, and then the four shapes of the figure 18 can be obtained by cutting from the center in four directions according to the above calculation results. If you carefully observe the figure 17, it is easy to find the symmetrical beauty in the figure, and this beauty can also inspire you to find a way to cut the iron sheet.

Answer: You can change "1" to "0", "4" to "3", "1" to "9" and "2" to 1 ".

Note: There are four answers to this question, depending on whether you have considered the question comprehensively. Because 225=25×9, the number to be modified can be divisible by 225, that is, by both 25 and 9. Dividing by 25 is not a problem. The last two digits of 75 need not be modified, just look at the first three digits. There are 2+1+4+7+5 =19 =18+1= 27-8, so it is not difficult to give the above four answers.

5. Answer: (1) Divide three of the nine squares into two parts:

Commentary: This problem of dividing sugar is very interesting. Let's do math first. Nine pieces of candy were divided equally among the four children.

Because there is a sentence in the title that limits the method of segmentation, that is, "each candy can only be cut into two parts at most."

Pay attention to this "restriction".