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The fewer math problems and answers in grade five, the better.
Elementary school fifth grade olympiad-quick calculation and clever calculation

In daily life and solving mathematical problems, we often have to calculate. In math class, we learned some simple calculation methods, but if we are good at observation and diligent in thinking, we can find more ingenious calculation methods in calculation, which will not only make you calculate well and quickly, but also make you smart and alert.

Example 1: Calculation: 9.996+29.98+169.9+3999.5.

Solution: It seems that the addition in the formula cannot be calculated by the simple calculation method learned in math class. However, as long as these numbers increase a little, they will become ten, a hundred or a thousand. After rounding these figures off, it is easy to calculate. Of course, remember that you have to subtract as much as you add when rounding.

9.996+29.98+ 169.9+3999.5

= 10+30+ 170+4000-(0.004+0.02+0. 1+0.5)

=42 10-0.624

=4209.376

Example 2: Calculation:1+0.99-0.98-0.97+0.96+0.95-0.94-0.93+…+0.04+0.03-0.02-0.01.

Solution: The number in the formula starts with 1 and decreases by 0.0 1 in turn until the last number is 0.0 1. Therefore, there are 100 numbers in the formula * *, and the operations in the formula are to add two numbers, then subtract two numbers, add two numbers, and then subtract two numbers ... in this order.

Because the arrangement and operation of numbers are very regular, according to the law, we can consider adding a bracket to every four numbers. Is there a certain regularity in the operation results of each group of numbers? It can be seen that if the number 1 in each group is subtracted from the number 3 and the number 2 is subtracted from the number 4, each of them gets 0.02, making a total of 0.04. Then the result of each group number (that is, each bracket) operation is 0.04, and the whole formula 100 is divided into 25 groups, and its result is the sum of 25 and 0.04.

1+0.99-0.98-0.97+0.96+0.95-0.94-0.93+…+0.04+0.03-0.02-0.0 1

=( 1+0.99-0.98-0.97)+(0.96+0.95-0.94-0.93)+…+(0.04+0.03-0.02-0.0 1)

=0.04×25

= 1

If you can flexibly use the law of commutation, you can also add parentheses in the calculation according to the following methods:

1+0.99-0.98-0.97+0.96+0.95-0.94-0.93+…+0.04+0.03-0.02-0.0 1

= 1+(0.99-0.98-0.97+0.96)+(0.95-0.94-0.93+0.92)+…+(0.03-0.02-0.0 1)

= 1

Example 3: Calculation: 0.1+0.2+0.3+…+0.8+0.9+0.10+0.1+0./kloc-0+2+…+0.

Solution: The numbers in this formula are arranged like a arithmetic progression, but if you look closely, it is actually composed of two arithmetic progression. 0. 1+0.2+0.3+…+0.8+0.9 is the first arithmetic progression, and each number behind it is 0. 1 more than the previous one. And 0.10+0.1+0.12+…+0.19+0.20 is the second arithmetic progression, and each number behind it is 0.0/kloc-0 more than the previous one.

0. 1+0.2+0.3+…+0.8+0.9+0. 10+0. 1 1+0. 12+…+0. 19+0.20

=(0. 1+0.9)×9÷2+(0. 10+0.20)× 1 1÷2

=4.5+ 1.65

=6. 15

Example 4: Calculation: 9.9× 9.9+ 1.99

Solution: Of the two factors of 9.9×9.9 in the formula, one factor is enlarged by 10, and the other factor is reduced by 10, and the product remains unchanged, that is, the product can become 99× 0.99; 1.99 can be divided into the sum of 0.99+ 1 After this change, the calculation is simpler.

9.9×9.9+ 1.99

=99×0.99+0.99+ 1

=(99+ 1)×0.99+ 1

= 100

Example 5: Calculation: 2.437× 36.54+243.7× 0.6346

Solution: Although the factors of the two multiplications in the formula are different, the numbers of 2.437 in the previous multiplication and 243.7 in the next multiplication are the same, but the positions of decimal points are different. If the decimal points of two factors in one multiplication are moved in opposite directions by the same number, so that the two numbers become the same, simple calculation can be made by multiplication and division.

2.437×36.54+243.7×0.6346

=2.437×36.54+2.437×63.46

=2.437×(36.54+63.46)

=243.7

* Example 6: Calculation:1.1×1.2×1.3×1.4×1.5

Solution: Although several numbers in the formula are arithmetic progression, the formula is not summation, and the result of this formula cannot be calculated by arithmetic progression summation.

Students who usually pay attention to accumulating calculation experience may notice that the product of multiplying 7, 1 1 and 13 is 100 1, and a three-digit multiplied by 100 1 is their product.

1. 1× 1.2× 1.3× 1.4× 1.5

= 1. 1× 1.3×0.7×2× 1.2× 1.5

= 1.00 1×3.6

=3.6036

Calculate the following questions and write a simple calculation process:

1.5.467+3.8 14+7.533+4. 186

2.6.25× 1.25×6.4

3.3.997+ 19.96+ 1.9998+ 199.7

4.0. 1+0.3+…+0.9+0. 1 1+0. 13+0. 15+…+0.97+0.99

5. 199.9× 19.98- 199.8× 19.97

6.23.75×3.987+6.0 13×92.07+6.832×39.87

*7.20042005×20052004-20042004×20052005

*8.( 1+0. 12+0.23)×(0. 12+0.23+0.34)-( 1+0. 12+0.23+0.34)×(0. 12+0.23)

Calculate the following questions and write a simple calculation process:

1.6.734- 1.536+3.266-4.464

2.0.8÷0. 125

3.89. 1+90.3+88.6+92. 1+88.9+90.8

4.4.83×0.59+0.4 1× 1.59-0.324×5.9

5.37.5×2 1.5×0. 1 12+35.5× 12.5×0. 1 12

Fifth grade, the second volume of mathematical Olympic examination questions

Name class score

Calculate the following questions in a simple way.

20.36-7.98-5.02-4.36 1 17.8÷2.3-4.88÷023

9.56×4. 18-7.34×4. 18-0.26×4. 18

1. There are 123 children. Divide them into a group of 12 or a group of 7, just finished eating, there is nothing left. It is also known that the total population is about 15. So, how many groups of 12 people are there? How many groups are there in a group of seven?

2. The average score of Zhang Ni's five exams is 88.5, and the full score of each exam is 100. How many times does Zhang Ni have to get full marks in order to get the average score above 92 as soon as possible?

The sum of the ages of the father and his three sons is 108 years old. If we wait another six years, the age of the father is exactly equal to the sum of the ages of the three sons. How old is my father?

4. Processing a batch of parts, originally planned to process 80 parts a day, just to finish the task on schedule. Due to the improvement of production technology, 100 pieces are actually processed every day, which not only completes the processing task four days ahead of schedule, but also processes more 100 pieces. How many parts did they actually process?

5. A pool can hold 8 tons of water, and the pool is equipped with a water inlet pipe and a water outlet pipe, which are two-pronged, and put a pool of water in 20 minutes. It is known that the water inlet pipe injects 0.8 tons of water into the pool every minute. How many tons of water does the water pipe discharge per minute?

6. Cut the wire into 15 segments. One part is 8 meters long and the other part is 5 meters long. The total length of 8 meters is 3 meters more than that of 5 meters. How long is this wire?

7. Divide a big fish into three parts: head, body and tail. The fish tail weighs 4 kilograms. The weight of fish head is equal to the weight of fish tail plus half the weight of fish body, and the weight of fish body is equal to the weight of fish head plus the weight of fish tail. How much does this big fish weigh?

8. The gym needs to pay 287 yuan to buy 5 soccer balls and 4 basketballs. 154 yuan buys 2 soccer balls and 3 basketballs. So how much do you spend on a football and a basketball?

9.5 yuan has RMB *** 14 and RMB *** 100. How many 5-yuan coins and 10 yuan?

10, someone climbed from village a to the top of village b, and it took him 7 hours to walk 30.5 kilometers. He went up the mountain at a speed of 4 kilometers per hour and down the mountain at a speed of 5 kilometers per hour. If the speed of going up and down the mountain remains the same, how long will it take to return from village B to village A along the original road?

1 1, Party A and Party B walk in opposite directions at the same time, with Party A riding at a speed of16km and Party B riding a motorcycle at a speed of 65km. A meets B at a distance of 62.4 kilometers from the starting point. How many kilometers is AB?

12, tortoise and rabbit race, the rabbit runs 35km per minute, and the tortoise climbs10m per minute. On the way, the rabbit slept and woke up to find the tortoise 50 meters in front of him. How long will it take the rabbit to catch up with the tortoise?

13. On the 600-meter-long circular runway, brother and sister run clockwise at the same starting point at the same time and meet every 12 minutes. If the speed of two people is the same, or starting from the original starting point at the same time, my brother runs counterclockwise, then they will meet every 4 minutes. How many minutes does it take for two people to run a lap?

14. In still water, the speed of ships A and B is 20km and 16km/h respectively. Two ships set off from a port one after another, and B set off two hours earlier than A. If the water speed is 4km per hour, how many hours after A sets off, will it catch up with B?

15. It takes 40 seconds for a train to cross a 440m bridge and 30 seconds to cross a 3 10/0m tunnel at the same speed. What is the speed and length of this train?

16, a bookshelf is divided into upper and lower floors, and the number of books on the upper floor is four times that of the lower floor. After taking five books from the lower level and putting them on the upper level, the number of books in the upper level is exactly five times that of the lower level. How many books are there on the ground floor?

17, there is 1800 kg cargo, which is divided into three cars: A, B and C. It is known that the number of kilograms loaded by A car is exactly twice that of B car, and B car carries 200 kg more than C car. Party A, Party B and Party C each have one copy.

Inclusion and exclusion

1. There are 40 students in a class, of which 15 is in the math group, 18 is in the model airplane group, and 10 is in both groups. So how many people don't participate in both groups?

Solution: There are (15+18)-10 = 23 (people) in the two groups.

40-23= 17 (person) did not attend.

A: There are 17 people, and neither group will participate.

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There are forty-five students in a class who took the final exam. After the results were announced, 10 students got full marks in mathematics, 3 students got full marks in mathematics and Chinese, and 29 students got no full marks in both subjects. So how many people got full marks in Chinese?

Solution: 45-29- 10+3=9 (person)

A: Nine people got full marks in Chinese.

3.50 students stand in a row facing the teacher. The teacher asked everyone to press 1, 2,3, ..., 49,50 from left to right. Let the students who are calculated as multiples of 4 back off, and then let the students who are calculated as multiples of 6 back off. Q: How many students are facing the teacher now?

Solution: multiples of 4 have 50/4 quotients 12, multiples of 6 have 8 50/6 quotients, and multiples of 4 and 6 have 4 50/ 12 quotients.

Number of people turning back in multiples of 4 = 12, number of people turning back in multiples of 6 ***8, including 4 people turning back and 4 people turning back from behind.

Number of teachers =50- 12=38 (person)

A: There are still 38 students facing the teacher.

4. At the entertainment party, 100 students won lottery tickets with labels of 1 to 100 respectively. The rules for awarding prizes according to the tag number of lottery tickets are as follows: (1) If the tag number is a multiple of 2, issue 2 pencils; (2) If the tag number is a multiple of 3, 3 pencils will be awarded; (3) The tag number is not only a multiple of 2, but also a multiple of 3 to receive the prize repeatedly; (4) All other labels are awarded to 1 pencil. So how many prize pencils will the Recreation Club prepare for this activity?

Solution: 2+000/2 has 50 quotients, 3+ 100/3 has 33 quotients, and 2 and 3 people have 100/6 quotients.

* * * Preparation for receiving two branches (50- 16) * 2 = 68, * * Preparation for receiving three branches (33- 16) * 3 = 5 1, * * Preparation for repeating branches (2+).

* * * Need 68+5 1+80+33=232 (branch)

A: The club has prepared 232 prize pencils for this activity.

5. There is a rope with a length of 180 cm. Make a mark every 3 cm and 4 cm from one end, and then cut it at the marked place. How many ropes were cut?

Solution: 3 cm marker: 180/3=60, the last marker does not cross, 60- 1=59.

4cm marker: 180/4=45, 45- 1=44, repeated marker: 180/ 12= 15,15-/kloc-.

Cut it 89 times and it becomes 89+ 1=90 segments.

A: The rope was cut into 90 pieces.

6. There are many paintings on display in Donghe Primary School Art Exhibition, among which 16' s paintings are not in the sixth grade, and 15' s paintings are not in the fifth grade. Now we know that there are 25 paintings in Grade 5 and Grade 6, so how many paintings are there in other grades?

Solution: 1, 2,3,4,5 * * has 16, 1, 2,3,4,6 * * has15,5,6 * * has 25.

So * * has (16+ 15+25)/2=28 (frame), 1, 2,3,4 * * has 28-25=3 (frame).

A: There are three paintings in other grades.

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7. There are several cards, each with a number written on it, which is a multiple of 3 or 4. Among them, cards marked with multiples of 3 account for 2/3, cards marked with multiples of 4 account for 3/4 and cards marked with multiples of 12 account for 15. So, how many cards are there?

Solution: The multiple of 12 is 2/3+3/4- 1=5/ 12, 15/(5/ 12)=36 (sheets).

There are 36 cards of this kind.

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8. How many natural numbers from 1 to 1000 are divisible by neither 5 nor 7?

Solution: multiples of 5 have 200 quotients 1000/5, multiples of 7 have quotients 1000/7 142, and multiples of 5 and 7 have 28 quotients 1000/35. The multiple of 5 and 7 * * * has 200+ 142-28=3 14.

1000-3 14=686

A: There are 686 numbers that are neither divisible by 5 nor divisible by 7.

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9. Students in Class 3, Grade 5 participate in extracurricular interest groups, and each student participates in at least one item. Among them, 25 people participated in the nature interest group, 35 people participated in the art interest group, 27 people participated in the language interest group, 12 people participated in the language interest group, 8 people participated in the nature interest group, 9 people participated in the nature interest group, and 4 people participated in the language, art and nature interest groups. Ask how many students there are in this class.

Solution: 25+35+27-(8+ 12+9)+4=62 (person)

The number of students in this class is 62.

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10, as shown in Figure 8- 1, it is known that the areas of three circles A, B and C are all 30, the areas of overlapping parts of A and B, B and C, and A and C are 6, 8 and 5 respectively, and the total area covered by the three circles is 73. Find the area of the shaded part.

Solution: The overlapping area of A, B and C =73+(6+8+5)-3*30=2.

Shadow area =73-(6+8+5)+2*2=58.

A: The shaded part is 58.