2. For example, the difference ratio problem: the number A is greater than the number B 12 and A: B = 7: 4, so find two numbers. I have more formulas than you, and multiples are cause and effect. Actual difference of numerator, multiple difference of denominator. Double the quotient and multiply it by their respective multiples to get two numbers. Double the quantity first, 12÷(7-4)=4, so the number a is 4X7=28 and the number b is 4X4= 16.
3, the age problem, the age difference remains unchanged, while adding and subtracting. With the change of age, the multiple is also changing. Grasp these three points and everything will be easy. Example 1: Xiaojun is 8 years old and his father is 34 years old. After a few years, his father is three times older than Xiaojun. Analysis: precession will not change, this year's age is almost 34-8=26, and it will not change in a few years. Knowing the difference and multiple, it becomes the problem of difference ratio. 26÷(3- 1)= 13. In a few years, dad's age is 13X3=39, and Xiaojun's age is 13X 1= 13, so it should be Example 2: Sister 13 years old, brother 9 years old. How old should they be when the sum of their ages is 40? Analysis: precession will not change. The age difference this year is 13-9=4, and it will not change in a few years. After several years, the age sum is 40, and the age difference is 4, which turns into a sum-difference problem. So a few years later, the elder sister's age is (40+4)÷2=22, and the younger brother's age is (40-4)÷2= 18, so the answer is 9 years later.
4. The sum ratio problem is known as a whole. Let's find some examples: the sum of the three numbers A, B and C is 27, and A: B: C =2:3:4. Find the three numbers A, B and C. Formula families want everyone to be together, and separation is principled. Denominator ratio sum, numerator's own. And multiply by the ratio, you deserve it. The denominator ratio sum, that is, the denominator is: 2+3+4 = 9; If the molecule is its own, the ratios of the three numbers A, B and C to the sum are 2÷9, 3÷9 and 4 ÷ 9, respectively; And the multiplication ratio, a is 27X2÷9=6, b is 27X3÷9=9, and c is 27X4÷9= 12.
5. Chickens and rabbits in the same cage: Chickens are not in the same cage, with head 36 and feet 120. Find the number of chickens and rabbits. The formula assumes both chickens and rabbits. How many feet are there? How many feet are missing? Divided by the foot difference, it is the number of chickens and rabbits. When seeking rabbits, it is assumed that all chickens are chickens, so the number of exemptions =( 120-36X2)÷(4-2)=24. When seeking chickens, it is assumed that all rabbits are chickens, so the number of chickens = (4x36- 120) ÷ (4-2) = 65438.
6. Distance problem (1) Example: Party A and Party B walk in opposite directions from two places with a distance of 120km. Party A's speed is 40km/h and Party B's speed is 20 km/h. How long did they meet? The moment the formulas meet, the distance is gone. Divide by the sum of the speeds and you get the time. At the moment of meeting, all the distances have passed, that is, the distance between Party A and Party B is exactly 120km. Divided by the sum of speeds, the time is obtained, that is, the total speed of Party A and Party B is the sum of their speeds of 40+20=60 (km/h), so the meeting time is 120÷60=2 (h) (2) Example of traceability problem: brother and sister go to town from home, the elder sister walks at a speed of 3 km/h, and the slow bird flies first. Divide the distance you go first by the speed difference, and the time is right. Distance to go first: 3X2=6 (km) Speed difference: 6-3=3 (km/h) Catch-up time: 6÷3=2 (h).
7. Concentration problem (1) For example, there is 20 kg of sugar water with a concentration of 15%. How many kilograms of water are added, and the concentration becomes 10%. When adding water to the formula, sugar should be added first, and then sugar water should be added after the sugar is finished. Sugar water minus sugar water is the added water. After adding water, sugar is obtained first. The original sugar content is: 20X 15%=3 kg of sugar. How much sugar water with a concentration of 10% should be used with 3 kg of sugar? 3÷ 10%=30 (kg) sugar water minus sugar water. The amount of sugar water after subtracting the original amount of sugar water is 30-. When adding sugar to the formula, you need water first, and then you need syrup after drinking the water. If you subtract sugar water from sugar water, you can easily solve the problem. Water needs to be added before sugar is added. The original water content is 20x (1-15%) =17 (kg). How much sugar water should there be in 17kg water with a concentration of 20%? 17 ÷ (65438+)
8. Examples of engineering problems: A project will be completed in 4 days by Party A and 6 days by Party B ... After both parties do it at the same time for 2 days, how many days will Party B do it alone? The formula is set to 1, and 1 divided by time is the work efficiency. When you do it alone, your work efficiency is your own. When you do it together, your work efficiency is everyone's efficiency. 1 Subtract what has been done and what has not been done, and divide what has not been done by work efficiency is the result. [1-(1÷ 6+1÷ 4) x2] ÷1÷ 6) =1(days)
9. How many trees should I plant? How do I ask for directions? Subtract 1 directly and the circle is the result. Example 1: Plant trees on a road with a length of 120m, with a spacing of 4m. How many trees have been planted? If the road is straight, planting trees is 120÷4- 1=29 (tree). Example 2: Plant trees at the edge of a circular flower bed with a length of 120m and a spacing of 4m. How many trees have been planted? If the road is round, planting trees is 120÷4=30 (trees).
10, profit and loss problem's formula is full of profit and loss, and the big one subtracts the small one; One profit and one loss, the profit and loss add up. Divided by the difference in distribution, the result is something or people distributed. Example 1: Children divide peaches, each peach 10, and 9 peaches are missing; Eight more than seven per person. How many children and peaches do you want? The formula is: (9+7)÷( 10-8)=8 (person), and the corresponding peach is 8X 10-9=7 1 (person). Example 2: Soldiers carry bullets. 45 rounds, 680 rounds per person; 50 rounds per person is more than 200 rounds. How many soldiers, how many bullets? The total residual problem, the big MINUS the small, that is, the formula is: (680-200)÷(50-45)=96 (person), and the corresponding bullet is 96X50+200=5000 (hair). Example 3: Students distribute books. 10 Each person is missing 90 books; There are eight copies each, and there are still eight left. How many books are suitable for how many students? For the total loss problem, the big one is subtracted from the small one, that is, the formula is: (90-8)÷( 10-8)=4 1 (person), and the corresponding book is 4 1X 10-90=320 (.
1 1. Example of remainder problem: Now the clock shows 18 o'clock. What time is it after the minute hand turns 1990? There are (N- 1) formula residues, the smallest is 1 and the largest is (N- 1). When the cycle changes, don't look at the quotient, just look at the surplus. Analysis: When the minute hand turns one turn, it is 1 hour, and 24 turns are 1 turn of the hour hand, so that the hand returns to its original position immediately. The remainder of 1980÷24 is 22, so it is equivalent to the minute hand rotating 22 times forward, the clockwise hand moving 22 hours forward, the backward 24-22=2 hours, and the clockwise hand pulling back 2 hours. The instantaneous needle is equivalent to 18-2= 16 (point).
12, the formula of cattle grazing problem, assuming that the amount of grass eaten by each cow every day is 1, what is the amount of grass eaten on the first b day of A? How much grass did M eat in the first n days? Subtract the big one from the small one and divide it by the difference of the corresponding days, and the result is the growth rate of grass. The original amount of grass is correspondingly reversed. Formula: a MINUS the amount of grass eaten in b days before b days multiplied by the growth rate of grass. Cattle with unknown grazing amount are divided into two parts: a small part eats new grass first, and the quantity is the proportion of grass; Divide some grass by the number of remaining cows to get the required number of days. The grass grows thick and fast all over the pasture. 27 cows can eat grass for 6 days; 23 cows can eat the grass in 9 days. Q 2 1 How many days will it take to finish the grass? Assume that the daily grazing amount of each cow is 1, the grazing amount of 27 cows for 6 days is 27×6 = 162, and the grazing amount of 23 cows for 9 days is 23×9 = 207. Subtract the small from the large, 207-162 = 45; If the difference between two corresponding days is 9-6=3 (days), the growth rate of grass is 45÷3= 15 (cattle/day); The original amount of grass is derived as follows-formula: the amount of grass eaten in B days before A minus B days multiplied by the growth rate of grass. Grass amount =27X6-6X 15=72 (cattle/day). Cattle with unknown grazing amount are divided into two parts: a small part eats new grass first, the quantity is the proportion of grass, that is to say, 2 1 cow is divided into two parts, and a part 15 cows eat new grass; The remaining 2 1- 15=6 eats the original grass, and the number of days required is: the amount of original grass ÷ distributing the remaining cattle =72÷6= 12 (days).