2. Three groups 180 people. The sum of the number of people in the first and second groups is 20 more than that in the third group, and the first group is 2 less than that in the second group. Find the number of the first group.

Analysis: key points: First, consider the first and second groups as a whole! Taking the third group as a whole, we call this method "splitting into two", that is, three questions are transformed into two questions, and the numbers of the first group and the second group are obtained first, and then the numbers of the first group are obtained. This is also a question of sum and difference.

Solution: (180+20) ÷ 2 = 100 (people)-the number of people in the first and second groups.

(100-2) ÷ 2 = 49 (people)-Number of people in the first group.

Comprehensive: [( 180+20) ÷ 2-2] ÷ 2 = 49 (persons)-Number of people in the first group.

A: The first group consists of 49 people.

4. In a subtraction formula, the sum of the minuend, the subtraction and the difference is equal to 120, and the subtraction is three times the difference, so what is the difference?

Analysis: This is a problem of multiplication. Subtraction is triple difference, so the minuend is quadruple difference, so the sum of the minuend, the minuend and the difference is eight times difference, which should be equal to 120, so the difference = 120 ÷ 8 = 15.

Solution:120 ÷ (1+3+1+2) = 15 A: The difference is equal to15.

6. Fifty students attended the party. The first girl who attended the meeting shook hands with all the boys, the second girl missed only one boy, the third girl missed only two boys, and so on. The last girl who attended the meeting shook hands with seven boys. How many boys are there among these students?

Analysis: This is a sum-difference problem. We can think of it this way: If there are six more girls in this class, the last girl should only shake hands with 1 boy. There are as many boys as girls at this time, so there are more boys than girls (7- 1)! The number of boys is:

Solution: (50+6) ÷ 2 = 28 (person). The number of boys is 2 8.

Note: There is another solution, 7+6+5+4+3+2+ 1 = 28 (person).

My analytical method is not clear yet. Please correct me.

8. Extra-curricular books A, B and C * * 100. The number of a is divided by the number of b, the number of c is divided by the number of a, the quotient is 5, and the remainder is also 1. So how many books does B have?

Analysis: This is a double problem. After reading the question, you will understand, "The number of A, B and C is 100, A is more than five times that of B, and C is more than five times that of A, 1. What are the numbers of A, B and C? " . Namely: b is 1 time; A is more than five times that of B.1; C is (5×5) times that of B (1× 5+ 1) 6. Then the difference between 100 and (1+6) is (1+5+5× 5) times, so we can find out what b is.

Solution: [100-1-(1× 5+1)] ⊙ (1+1× 5) =

10. There are 108 pieces of goods, which are stored in the warehouse in four piles. Twice the number of the first pile is equal to half the number of the second pile, 2 less than the number of the third pile and 2 more than the number of the fourth pile. How many pieces are stored in each pile?

Analysis: If we regard the first pile as 1 time, then we can calculate that the second pile is (2×2)4 times, the third pile is 2 times more and the fourth pile is 2 times less, so a * * * is exactly 1+4+2+2 = 9 times (the third pile and the fourth pile

Solution: (108+2-2) ÷ (1+2× 2+2) =108 ÷ 9 =12 (pieces)-the first pile.

12× 2× 2 = 48 (pieces)-the second pile; 12× 2+2 = 26 (pieces)-the third pile; 12× 2-2 = 22 (pieces)-the fourth pile;

A: Each pile 12 pieces, 48 pieces, 26 pieces and 22 pieces.

12. The car, horse and cannon in China chess represent different natural numbers respectively. If: car-horse = 2, gun-car = 4, gun-horse = 56, what is "car+horse+gun"?

Analysis: This is a differential problem. According to the topic, the horse is 1 time, the car is 2 times, and the gun is 4 times, so the multiple difference between the gun and the horse is (2×4- 1)7 times, and the difference between the number of the gun and the horse is 56. According to the formula of difference problem, the values of car, horse and gun can be obtained respectively.

Solution: 56 ÷ (8- 1) = 8 horses;

8×2 = 16- car

16× 4 = 64 guns

8+ 16+64 = 88- car+horse+gun answer: the sum of car, horse and gun is 88.

14. Two students, A and B, originally planned to study by themselves at the same time every day. If A increases the self-study time by half an hour and B decreases the self-study time by half an hour, then the self-study time of six days in B is only equal to that of one day in A.. Q: How many minutes do A and B plan to teach themselves every day?

Analysis: Problems of different times. In the past, time was the same. Now A is half an hour, and B is half an hour short. Now the difference between the two numbers is (30+30) 60 minutes, and now the difference is (6- 1) 5 times. In this way, you can find B's daily self-study time, plus 30 minutes, you can get the planned time of self-study every day.

Solution: (30+30) ÷ (6-1)+30 =12+30 = 42 (minutes) A: I originally planned to study by myself for 42 minutes every day.

It involves four or more objects, the quantitative relationship is known, it is not convenient to use directly, and it is a complex differential problem related to other knowledge.

Typical problem

1. There are four classes in Grade Four. Except Class A, the total number of students in the other three classes is 13 1. The total number of the other three categories except category D is134; The total number of students in Class B and Class C is less than that in Class A and Class D 1. How many students are there in these four classes?

Answer: Use 1 31+134 = 265, which is the sum of 1 A and d and 2 B and c. Because the total number of classes B and C is less than that of classes A and D/kloc, we use 265-65438.

2. There are four numbers, and the sum of every three numbers is 45, 46, 49 and 52 respectively. What is the smallest of these four numbers?

Answer: think about it. What if I add up four numbers? In fact, every number has been added three times! Everyone must remember this idea! (45+46+49+52)÷3=64 is the sum of these four numbers, and the topic requires the smallest number, so I subtract 52 (the largest sum of some three numbers) from 64 to be the smallest number, which is equal to 12.

Insert a number between two digits and it becomes three digits. For example, insert the number 6 in the middle of 72, and it becomes 762. Some two-digit numbers are inserted in the middle, and the three-digit number is nine times that of the original two-digit number. Find all these two-digit numbers.

Answer: for this question, we must first judge what the unit is. This number multiplied by 9 is equal to the original unit, which means that the unit can only be 0 or 5! Look at 0 first, and you will soon find that it doesn't work, because 20×9= 180, 30×9=270, 40×9=360 and so on. No matter how many times 9 is multiplied, the result is that the hundred digits are always less than ten digits, so you can only be 5. After a simple calculation, it is not difficult to find that 15, 25, 35, 45 are numbers that meet the requirements.

4. A class bought several exercise books with the unit price of 0.5 yuan. If these exercise books are only for girls, each person will get an average of15; If these exercise books are only given to boys, everyone will get 10 on average. Then, divide these exercise books evenly in the class. How much does each person have to pay?

Answer: This kind of question is too simple for a student who has studied engineering problems, but engineering problems are the content of grade six. What should fourth-grade students do? We can think like this: I suppose there are two girls in the class (use your head, why not set up 1 girl? ), then a * * * has 30 exercise books, and then it is concluded that there are 3 boys, and 30÷(2+3)=6 means that everyone should have 6 exercise books, so everyone has to pay 3 yuan money.

The zookeeper distributed peanuts to three groups of monkeys. If only the first group is assigned, each monkey can get 12. If only the second group is given, each monkey can get 15 capsules; If you only give it to the third group, each monkey will get 20 pills. Then if you give it to three groups of monkeys on average, how many pills will each monkey get?

Answer: Like the previous question, I want to find the number 1, which is a multiple of 12, a multiple of 15 and a multiple of 20. Can you find it? The smallest one that can be found is 60, so I assume that * * * has 60 peanuts, and then I can work out that the first group has 5 monkeys, the second group has 4 monkeys and the third group has 3 monkeys, so that * * * has 5+4+3= 12 monkeys and 60÷ 12=5.

6. An integer, minus 4 times the remainder after dividing by 5, is 154, so what is the original integer?

Answer: First of all, if the dividend is divided by the divisor, the remainder must be less than the divisor. Therefore, in this problem, the remainder is definitely not greater than 4, which determines that the original integer can only be: 154+4×0, 154+4× 1, 154+4×2,154+4.

7. Many parents (father or mother, not teachers) and teachers accompanied some primary school students to participate in a math competition. There are 22 known parents and teachers, with more parents than teachers, more mothers than fathers, 2 more female teachers than mothers, and at least/kloc-0 per male teacher. How many fathers are there among these 22 people?

A: There are more parents than teachers, so the number of teachers is less than 22÷2= 1 1, that is, it does not exceed 10, and the number of parents is not less than 12. In at least 12 parents, mothers are more than fathers, so mothers are more than 12÷2=6, that is, not less than 7. Because there are two more female teachers than mothers, there are not less than 9 female teachers, but at most 10 teachers and at least 1 0 male teachers, so there must be 10 teachers (9 female teachers, 1 male teachers) and 7 mothers in 12 parents.

8. There are 20 questions in a math exam. It is stipulated that 2 points will be deducted for a correct answer, 1 point for a wrong answer, and no points will be scored for unanswered questions. After the exam, Xiao Ming got 23 points. He wants to know how many questions he made wrong, but only remembers that the number of unanswered questions is even. Please help Xiao Ming calculate. He answered several questions wrong.

Answer: 20 questions, if done correctly, you can get 20×2=40 points. If you don't answer 1 question, 2 points will be deducted, and if you answer a wrong question, 3 points will be deducted. Xiao Ming got 23 points, 40-23 points less than the total score =17 points. Because the unanswered questions are even, we can first think that if there are 0 unanswered questions, 17 is wrong, but 17÷3=5…2, which is impossible! Think again, two questions were not done, 4 points were not done, 17-4= 13 points were lost because of mistakes, 13 = 4 … 1 is impossible! If you don't do four questions, subtract eight points. 17-8=9 points is because you made fewer mistakes, and 9÷3=3, so three questions are wrong.

9. The price of a commodity is: 1 cent for each piece, 4 points for every five pieces and 7 points for every nine pieces. Xiao Zhao's money can buy 50 at most, and Xiao Li's money can buy 500 at most. How much is Xiao Li's money more than Xiao Zhao's?

Answer: Do the math in your head first. Is nine sevens the most cost-effective? Look at Xiao Zhao first: 50÷9=5…5, so he has 5×7+4=39 cents; Look at Xiao Li: 500÷9=55…5, so he has 55×7+4=389 points, so Xiao Li has 389-39=350 points more than Xiao Zhao. Never think that (500-50)÷9×7=350 is enough. For example, if I change 500 to 400, the method is wrong!

10. A kindergarten has the smallest class size, with 27 middle class and 6 large class. There are 25 boxes of oranges in the Spring Festival, with no more than 60 oranges and no less than 50 oranges in each box. The single digit of the total number of oranges is 7. If everyone divides it into 19, the number of oranges is not enough. Now everyone in the big class has one more point than everyone in the middle class, and everyone in the middle class has one more point than everyone in the small class, which is just enough. How many oranges did everyone in the big class get at this time? How many people are in the small class? This question is the most difficult one in this lecture! ! ! )

Answer: First of all, the number of oranges is between 1250(=25×50) and 1500(=25×60). Please help me to see the difference between the following two ways to divide oranges. (1) each big class a+ 1, each middle class a, and each small class a-1; (2) One for each class, big, middle and small. The first method, I asked the children in the big class to take out 1 to make up for the children in the small class, and each person made up 1. Because there are six more people in the big class than in the small class, there are six more oranges in the end.

If I take out six oranges from all the oranges, I can make the first division in the original problem become my second division. Because the total number of oranges is 7, and the unit after subtracting 6 is 1, so many oranges can be distributed to all children and everyone has the same number, so the total number of people and the number of oranges distributed to everyone are odd! !

But obviously 19 per person is not enough, so it can only be 17, 15, 13 and so on. Of course, 15 is impossible (because after any number is multiplied by 15, everyone is either 5 or 0). Let's see if it is 65438. 1250 ÷ 13 = 96 ...2, then there are at least 96 people, so the total number of large classes and small classes is at least 96-27=69. However, the number of small classes will at least not exceed 27 in the middle class, so the total number of large classes and small classes should not exceed 27+(27+6)=60, which is contradictory to my previous results! So everyone can't be less than 13, which means everyone should be 17 apple.

Now the total number of apples is 7-6= 1, and everyone has 17 apples, so the total number of apples should be 3! ! Look again: 1250÷ 17=73…9, 1500÷ 17=88…4, then we can find that the total number must be 83. Because if it's 73, the oranges haven't been eaten yet. Therefore, both the large class and the small class are 83-27=56 people. Using the formula of sum-difference problem, we can quickly get that the number of small classes is: (56-6)÷2=25.

1 1. Put a cube on the table with a number on each side, and the sum of the two numbers on the opposite side is equal to 13. Xiao Zhangcan sees the top and both sides, and the sum of the three numbers he sees is18; Li Can Jr. saw the top surface and the other two surfaces. The sum of the three numbers he saw is 24. What is the number of this side stuck on the table?

Answer: think about it first. If I add 24 to 18, what faces will I get? Is the sum of four sides and two top surfaces! The sum of the four sides should be: 13+ 13=26, then the number of top surfaces can be calculated as: (18+24-26)÷2=8, so the number of bottom surfaces is: 13-8=5.

12. The map on the left is the road map. There are a large group of children in A. They are going east or north. At every intersection from A, half of it goes north and the other half goes east. If 60 children have been to intersection B, how many children have been to intersection C?

Answer: We try to assume that A has 1 child. What happened to the two children? We find that there will be half a child behind, which is not acceptable. So we assume that there are four, eight, 16 children, and we still find that there will be half a child behind. Then we assume that A has 32 children! Use your head: Why are the numbers 1, 2,4,8,16,32? Are there any rules in these figures? Finally, after calculation, it can be found that C has 8 children passing by and B has 10 children passing by, but in fact B has 60 children passing by, so the result should be 6 children and A has 32 children! So there are 8×6=48 children passing through point C.

13. The football used in the game is black and white, in which the black leather is a regular Pentagon, the white leather is a regular hexagon, and the sides of the black Pentagon and the white hexagon are equal. The sewing method is as follows: sewing the five edges of each black leather with the edges of five white leather respectively; Of the six sides of each white leather, three are sewn together with the sides of black leather, and the other three are sewn together with the sides of other white leather. If a football has 12 pieces of black regular pentagonal leather on its surface, how many pieces of white regular hexagonal leather should this football have?

Answer: 1. How many sides does black leather have: 12×5=60? These 60 edges are stitched together with white leather. For white leather, three of the six sides of each white leather are sewn together with the side of the black leather, and the other three sides are sewn together with the sides of other white leather, so half of all the sides of the white leather are sewn together with the black leather, so the white leather should have 60× 2 = 65438+.

14.5 empty bottles can be exchanged 1 bottle of soda. A classmate in class drank 16 1 bottle of soda, and some of them were replaced with the rest of the empty bottles. How many bottles of soda should they buy at least?

Answer: Generally speaking, you can think like this: first buy 16 1 bottle of soda, and then use this 16 1 empty bottle for 32 bottles of soda (161÷ 5 = 32 …/kl After drinking 125 empty bottles, use 25 empty bottles for 5 bottles of soda, then use 5 empty bottles for 1 bottle of soda, and finally use this empty bottle and the remaining 4 empty bottles for another bottle of soda. This is always the case.

15. There are three piles of apples, of which the number of apples in the first pile is more than that in the second pile, and the number of apples in the second pile is more than that in the third pile. If one apple is taken from each pile, the number of the first pile is three times that of the second pile among the remaining apples. If you take out the same number of apples from each pile, there are 34 apples left in the first pile, and the number of apples left in the second pile is twice as much as that in the third pile. What is the maximum sum of three piles of apples?

A: This kind of question is the same as Question 10. It's easier said than done. The key is how to make the fourth grade students understand!

Start with the first condition: take one from each pile of apples. Among the remaining apples, the number of the first pile is three times that of the second pile. At this time, it is assumed that the second pile is 1 apple, and the first pile is 3 apples, with a difference of 2 apples. Let's look at the second condition: take out the same number of apples from each pile, so that there are 34 apples left in the first pile, and the number of apples left in the second pile is twice that of the third pile. Because the same number of apples are taken from each pile, the second pile is still 2 apples less than the first pile, so these 2 copies should be less than 34 copies (why can't you think about it yourself? ) So one copy is at most 16, so in the second condition, the second pile has 34- 16×2=2, and the third pile has 2÷2= 1, so when we return to the first condition, the second pile should be 1 copy. So at the beginning, there were 49, 17 and 16 respectively, and the total * * * was 49+ 17+ 16=82.

Example 1: After the autumn harvest, Hongxing Farm stored 56,000 Jin of grain in two warehouses respectively. As we all know, the first warehouse stores three times as much food as the second warehouse. How many kilograms of grain do you want to store in each warehouse?

Analysis and solution: We can regard the grain quantity stored in the second warehouse with smaller capacity as 1, then the grain quantity stored in the first warehouse is 3, and the total grain quantity in the two warehouses is 56,000 Jin, which is equivalent to the grain quantity stored in the second warehouse, so we can get the grain quantity stored in the second warehouse.

(1) Grain quantity in the second warehouse: 56000 ÷ (3+1) =14000 (kg).

(2) Grain quantity in the first warehouse: 4000×3=42000 (kg)

Answer: The first warehouse stores 42,000 Jin of grain, and the second warehouse stores14,000 Jin of grain.

Example 2: There are 526 pear trees, peach trees and walnut trees in the orchard, with 24 pear trees, twice as many as peach trees, and fewer walnut trees 18. How many pear, peach and walnut trees are there?

Analysis and decomposition:

As can be seen from the known situation, there are 24 pear trees, twice as many as peach trees, and there are 18 walnut trees, which are less than peach trees. It can be seen that if we regard the number of peach trees as 1, we can know the share of other trees. Adding 18 walnut trees is equal to peach trees, that is, walnut trees also account for 1 share, and then subtracting 24 from pear trees is equal to twice that of peach trees, and peach trees also account for 2 shares. If you do this, the total number of trees will become (526+ 18-24)=520, which is exactly 4, equivalent to 4 peach trees.

(526+ 18-24)÷(2+ 1+ 1)

=520÷4

= 130 (tree)

Peach trees only account for one share, so there are 130 peach trees.

Pear tree: 130×2+24=284 (plant)

The walnut tree is:130-18 =112 (tree).

Answer: There are 284 pear trees, 0/30 peach trees, 0/2 walnut trees1/kloc-0.

Example 3: Divided by the divisor, the quotient is 4 and the remainder is 3. The sum of dividend, divisor, quotient and remainder is 155. What are the dividend and divisor?

Analysis: First subtract the quotient and the remainder with 155, and the remainder is the sum of the dividend and the divisor: 155-4-3= 148.

The dividend is four times that of the divisor, which is 3 more. If the divisor is regarded as 1, then the dividend is 4 more than 3, as shown in the following figure.

The divisor accounts for 1, so the divisor is (148-3)÷(4+ 1)=29.

The dividend for 4 shares is greater than 3, so the dividend is 29×4+3= 1 19.

A: The dividend is 1 19 and the divisor is 29.

Example 4: The number of original books in Class 4 (1) is the same as that in Class 4 (2). Later, Class 4 (1) bought 1 18 new books, and Class 4 (2) took out 70 books from the original books of this class and gave them to the first-grade students. At this time, Class 4 (65438+)

(1) Later, there were more books in Class Four (1) than in Class Four (2):118+70 =188.

(2) The extra 188 only accounts for 2 volumes, so each book is: 188÷(3- 1)=94 volumes.

The number of original books is: 94+70= 164 (there are as many original books in two classes).

A: The original books of both classes are 164.

Example 5: Father is 39 years old. A few years ago, my father was four times older than my son.

Analysis: The age difference between father and son is 39- 12=27. Because the age is always the same, the age of being a father is a son.

Four times the age of the son, the son's age is:

27(4- 1)= 9 (age)

12-9=3 (year)

Three years ago, my father was four times as old as my son.

Example 6: Two cold storages, A and B, store 6250 boxes of eggs. After 65,438+065,438+000 boxes were shipped from A, B had 350 boxes of eggs, twice as many as the remaining eggs in A. How many boxes did A and B originally store?

(i) How many cases are there in each copy?

(6250-1100-350) ÷ (1+2) =1600 (box)

(ii) Eggs originally stored in vault A:

1600+1100 = 2700 (box)

(iii) Eggs originally stored in vault B:

1600×2+350=3550 (box)

A: Warehouse A originally stored 2,700 boxes of eggs, while Warehouse B originally stored 3,550 boxes of eggs.

In the final exam, a pupil averaged 9 1.5 in Chinese and mathematics, and he knew the numbers.

Academic performance is 5 points more than Chinese performance. How many points do you want for both courses?

Solution 1: Chinese and math * * * have 9 1.5×2= 183 (points).

Chinese score: (183-5)÷2=89 points.

Math score: (183+5)÷2=94.

Option 2: Mathematics is 5 points more than Chinese, so the math score is 2.5 points higher than the average score, and the Chinese score is 2.5 points lower than the average score.

So: Chinese score: 9 1.5-2.5=89 (points).

Math score: 9 1.5+2.5=94.

A: 89 points for Chinese and 94 points for mathematics.

Fourth, the practice part.

1.There are 5,200 cubic meters of water in Pool A and 2,400 cubic meters of water in Pool B. If the water in Pool A flows into Pool B at a speed of 44 cubic meters per minute, after a few minutes, the water in Pool B is three times that of Pool A.. ..

2. Divide 1296 into four numbers: A, B, C and D. If A adds 2, B subtracts 2, C multiplies 2, and D divides 2, the four numbers are equal. What are these four numbers?

3. Liushumo Village has 5 10 mu of paddy fields and 230 mu of dry fields. It is planned to change some dry fields into paddy fields this winter and next spring, so that the mu of paddy fields in the village will reach three times that of dry fields. How many acres of dry land should be changed into paddy fields?

4. The distance between City A and City B is 135km. Xiao Zhang rides a bike from city A to city B at 7 am, and Xiao Li rides a motorcycle from city B to city A at 8 am. Zhang and Li met on the road at 6:5438+00 in the morning. If the speed of a motorcycle is three times that of a bicycle, what are the speeds of a motorcycle and a bicycle per hour?

5. The sum of numbers A and B is 80, and the sum of five times the number A and three times the number B is 3 14. What are the numbers a and b?

6. Warehouse A and warehouse B * * * store 84,500 kilograms of soybeans, and take out 6,500 kilograms from warehouse A. After taking out 4,000 kilograms from warehouse B, the remaining soybeans in the two warehouses are exactly equal. How many kilograms of soybeans were originally stored in warehouse A and warehouse B?

7. A batch of oil needs 20 vehicles if it is transported by A-class tanker, and 25 vehicles if it is transported by B-class tanker. It is known that Type A oil tanker is 2 tons more than Type B oil tanker. How many tons of oil do you want?

8. Divide 16 1 by two numbers so that the sum of the two numbers is seven times the difference between the two numbers. What are these two numbers?

9. Both brothers are a little silly, thinking that only when they are one year older, others will not grow up. One day, my brother said to his younger brother, "I will be twice your age in three years." The younger brother said, "No, I will be as old as you in three years." How old were they then?

10, the number of watermelons shipped by fruit shops is four times that of cantaloupes. If you sell 130 watermelons and 36 cantaloupes every day, there will be 70 watermelons left after the cantaloupes are sold out. Q: How many watermelons and honeydew melons did the fruit shop send?

An analysis of the answers and ideas of verbs (the abbreviation of verb)

1. Solution: When the water volume of pool B is three times that of pool A, the total water volume of the two pools is still 5200+2400=7600 cubic meters. If pool A is regarded as 1, there should be three portions of water in pool B at this time, and the volume of water in each portion of 1 should be:

7600( 1+3)= 1900 (m3)

Therefore, there is 1900 cubic meters of water in pool a. ..

The water from pond A is 5200- 1900=3300 (m3).

So the time is 3300÷44=75 (minutes).

A: After A:75 minutes, the water in Pool B is three times that in Pool A. ..

2. Solution: When four numbers are equal, each number can be regarded as "1", then

As can be seen from the figure, the number of a was originally 1, which was 2 less;

The number b was originally 1, 2 more;

The number c was originally 0.5;

Ding's number turned out to be two copies.

Therefore, it can be concluded that:

( 1296+2-2)÷( 1+ 1+0.5+2)

= 1296÷4.5

=288

Therefore, the number A is 286, the number B is 290, the number C is 144, and the number D is 576.

3. The total number of fields is 565,438+00+230 = 740. After transformation, the upland field is regarded as "1", so paddy field accounts for 3, so each field should be 740÷(3+ 1)= 185 (mu).

Therefore, the reformed dry land is: 230- 185=45 (mu).

4. If Xiao Zhang's distance per hour by bike is regarded as "1", Xiao Li has to walk three times per hour by motorcycle.

* * * 9 copies, so the distance of each copy is: 135÷9= 15 (km).

Therefore, the bicycle speed is15km; 15× 3 = 45km, and the speed of motorcycle is 45km.

5. Solution: The sum of the two numbers A and B is 80, and the sum of the three times of the two numbers is 80×3=240.

And the sum of five times the number a and three times the number b is 3 14.

Therefore, the two multiples of a are: 3 14-240=74.

∴ A number is: 74÷2=37.

The number b is 90-37=43.

6. Rough solution: 84500-6500-4000=74000 (kg)

74000÷2=37000 (kg)

37000+65000=43500 (kg)

37000+4000=4 1000 (kg)

Answer: Warehouse A and Warehouse B originally stored 43,500 kilograms of soybeans and 4 1 1,000 kilograms of soybeans respectively.

7. Rough solution: 20×4=40 (ton)

25-20=5 (vehicles)

40÷5=8 (ton)

8×25= 100 (ton)

A: This batch of oil is 65,438+000 tons.

9. Briefly explain: From what my brother said, we can know that my brother is three years younger than my brother.

Judging from what my brother said, when he was 3 years older, he was twice as old as my brother is now.

That is, when the elder brother is 6 years older than the younger brother, the elder brother is twice as old as the younger brother, so the younger brother is 6 years old this year and the elder brother is 6+3=9 years old this year.

10. Suppose 36 cantaloupes are sold every day, and the number of watermelons sold is 36×4= 144. Then cantaloupe and watermelon must be sold out at the same time.

In fact,144-130 =14 pieces are sold less every day.

When the cantaloupes are sold out, there are more than 70 cantaloupes, so:

70÷ 14=5 (days) sold melons for 5 days.

36×5= 180 pieces 180×4=720 (pieces)

So the fruit shop has 720 watermelons, 180 cantaloupes.

720+ 180=900 (pieces)

A: There are 900 watermelons and honeydew melons sent by the fruit shop.