First, the problem of chickens and rabbits in the same cage:
Basic question type: There are 30 chickens and rabbits in the cage, one with *** 100 legs. Q: How many chickens and rabbits are there?
The solution to this problem is: suppose all 30 rabbits are chickens, then * * * has 2x30=60 legs, and 100-60=40 legs are missing. Because each rabbit has 4-2=2 legs more than chickens, rabbits * * * have 40/2=20, and chickens * * *.
Of course, it can also be reversed. Assuming that all 30 rabbits are rabbits, there are 120 legs, which is 20 more. Because chickens have two legs less than rabbits, there are 10 chickens.
There are many similar problems, but they are all changed from the basic questions, as follows:
The club has 30 sets of chess, only for 100 children. Chess is played by every two people and checkers is played by every six people. How many chess and checkers are there?
Second, the engineering problems:
Basic question:
When both parties complete a project, it takes 3 days for Party A to do it alone and 6 days for Party B to do it alone. How many days does it take for Party A and Party B to complete it together?
Solution:
The daily workload of Party A is 65,438+0.3 of the total engineering quantity, and the daily workload of Party B is 65,438+0.6 of the total engineering quantity. The daily workload of two people working together =1/3+1/6 =1/2, so it takes two days for both parties to complete the project together.
There will be many changes in this question, such as how many days does A work first, and then B starts to work; Or Party A and Party B work together for one day, and Party B works alone, and so on. But the idea of solving the problem is the same. Set the total workload to 1, and then calculate.
Third, the problems encountered:
Basic question type: The distance between Party A and Party B is 20km, the speed of Party A is 6km/h, and the speed of Party B is 4km/h, both of which start in the same direction at the same time. How long will it take to meet?
Solution: This is relatively simple, 20/(6+4)=2.
There are many changes in this kind of questions. Usually a starts for a while before b starts. Or how far is the meeting place from A?
Fourth, the pursuit of problems:
Basic problem: A's speed is 10 km/h, B's speed is 15 km/h, and A starts 2 hours first. How long will it take B to catch up with A?
Solution: A departs for 2 hours, the traveling distance is 10x2 = 20km, and the speed of B is 15- 10 = 5km/h, so the catching-up time is 20/5=4 hours.
There are many changes in this topic, such as the famous water discharge problem. When the water injection pipe is turned on 10 minute, you can fill a bath and clean it in 20 minutes. Ask how many minutes can two pipes be opened at the same time. This topic can be done according to the following ideas: water injection speed 110, drainage speed1.
Verb (abbreviation of verb) water flow problem:
Basic question type: The distance between Party A and Party B is 300 kilometers, the ship speed is 20 kilometers per hour, and the current speed is 5 kilometers per hour. How long does it take to go back and forth?
Solution: Downstream, speed 20+5 = 25km/h, time 300/25 = 12h. If you come back against the current, the speed is 20-5 = 15km/h and the time is 300/ 15 = 200.
Basic concept: Travel problem studies the movement of objects, and it studies the relationship between the speed, time and travel of objects.
Basic formula: distance = speed × time; Distance ÷ time = speed; Distance/speed = time
Key question: determine the position in the journey.
Meeting problem: speed sum × meeting time = meeting distance (please write other formulas)
Pursuit problem: Pursuit time = distance difference ÷ speed difference (write other formulas)
Running water problem: downstream travel = (ship speed+water speed) × downstream time = (ship speed-water speed )× downstream time.
Downstream speed = ship speed+current speed = ship speed-current speed.
Still water velocity = (downstream velocity+upstream velocity) ÷2 Water velocity = (downstream velocity-upstream velocity) ÷2
Running water problem: the key is to determine the speed of the object, refer to the above formula.
Bridge crossing problem: the key is to determine the moving distance of the object, refer to the above formula.
For reference only:
Sum and difference problem formula
(sum+difference) ÷2= larger number;
(sum and difference) ÷2= smaller number.
Sum-multiple problem formula
And present (multiple+1)= a multiple;
Multiple x multiple = another number,
Or sum-a multiple = another number.
Formula of differential multiple problems
Difference ÷ (multiple-1)= smaller number;
Smaller number x multiple = larger number,
Or decimal+difference = large number.
Average problem formula
Total quantity/total number of copies = average value.
General travel problem formula
Average speed × time = distance;
Distance/time = average speed;
Distance-average speed = time.
The formula of reverse travel problem can be divided into "meeting problem" (two people start from two places and walk in opposite directions) and "parting problem" (two people walk with their backs to each other). Both of these problems can be solved by the following formula.
(speed sum) × meeting (leaving) time = meeting (leaving) distance;
Meet (leave) distance ÷ (speed sum) = meet (leave) time;
Meet (leave) distance ÷ meet (leave) time = speed and.
Formula of the problem of traveling in the same direction
Catch-up (pull-out) distance ÷ (speed difference) = catch-up (pull-out) time;
Catch up (pull away) the distance; Catch-up (pull-away) time = speed difference;
(speed difference) × catching (pulling) time = catching (pulling) distance.
Formula of train crossing bridge problem
(bridge length+conductor) ÷ speed = crossing time;
(Bridge length+conductor) ÷ Crossing time = speed;
Speed × crossing time = sum of bridge and vehicle length.
Navigation problem formula
(1) general formula:
Still water speed (ship speed)+current speed (water speed) = downstream speed;
Ship speed-water speed = water flow speed;
(downstream speed+upstream speed) ÷2= ship speed;
(downstream speed-upstream speed) ÷2= water flow speed.
(2) Formula for two ships sailing in opposite directions:
Downstream speed of ship A+downstream speed of ship B = still water speed of ship A+still water speed of ship B.
(3) Formula for two ships sailing in the same direction:
Hydrostatic speed of rear (front) ship-Hydrostatic speed of front (rear) ship = the speed at which the distance between two ships is reduced (expanded).
After calculating the speed of narrowing or expanding the distance between the two ships, solve it according to the above formula.
Engineering problem formula
(1) general formula:
Efficiency × working hours = total workload;
Total workload ÷ working time = working efficiency;
Total amount of work ÷ efficiency = working hours.
(2) Assuming that the total workload is "1", the formula for solving engineering problems is:
1÷ working time = the fraction of the total amount of work completed in unit time;
1What is the score that can be completed per unit time = working time.
(Note: When solving engineering problems with the hypothetical method, you can arbitrarily assume that the total workload is 2, 3, 4, 5 ... Especially when the total workload is assumed to be the least common multiple of several working hours, the fractional engineering problem can be transformed into a relatively simple integer engineering problem, and the calculation will become simpler. )
Formula of profit and loss problem
(1) A surplus (surplus) and a deficit (deficit), the formula can be used:
(profit+loss) ÷ (the difference between two distributions per person) = number of people.
For example, "children divide peaches, and each person 10 is less than 9 and more than 8." Q: How many children and peaches are there? "
Solution (7+9)÷( 10-8)= 16÷2
=8 (a) ........................................................................................................................................................................
10×8-9=80-9=7 1 (pieces)
Or 8×8+7=64+7=7 1 (pieces) (omitted)
(2) Both times are surplus (surplus), and the formula can be used:
(large surplus-small surplus) ÷ (the difference between two distributions per person) = number of people.
For example, "soldiers carry bullets for marching training, each carrying 45 rounds and more than 680 rounds; If each person brings 50 rounds, then 200 rounds more. Q: How many soldiers are there? How many bullets are there? "
Solution (680-200)÷(50-45)=480÷5
=96 (person)
45×96+680=5000 (hair)
Or 50×96+200=5000 (hair) (omitted)
(3) If twice is not enough (loss), the formula can be used:
(big loss-small loss) ÷ (the difference between two distributions per person) = number of people.
For example, "send a batch of books to students, each with 10 copies, with a difference of 90 copies;" If each person sends 8 copies, there are still 8 copies left. How many students and how many copies are there? "
Solution (90-8)÷( 10-8)=82÷2.
=4 1 (person)
10×4 1-90=320 (this) (omitted)
(4) If one time is not enough (deficit) and the other time is just used up, you can use the formula:
Deficit (the difference between two distributions per capita) = number of people.
(Example omitted)
(5) One time there is surplus, and the other time it is just used up. This formula can be used to:
Surplus (the difference between two distributions per person) = number of people.
(Example omitted)
Formula of chicken and rabbit problem
(1) Given the total number of heads and feet, find the number of chickens and rabbits:
(total number of feet-number of feet per chicken × total number of heads) ÷ (number of feet per rabbit-number of feet per chicken) = number of rabbits;
Total number of rabbits = number of chickens.
Or (number of feet per rabbit × total head-total feet) ÷ (number of feet per rabbit-number of feet per chicken) = number of chickens;
Total number of chickens = rabbits.
For example, "Thirty-six chickens and rabbits, enough 100. How many chickens and rabbits are there? "
Solution1(100-2× 36) ÷ (4-2) =14 (only)
36- 14=22 (............................. chicken only).
Solution 2 (4×36- 100)÷(4-2)=22 (only) ............................................................................................................................
36-22= 14 (................................ rabbit only).
(short answer)
(2) Given the difference between the total number of chickens and rabbits, when the total number of chickens is greater than that of rabbits, the formula can be used.
(number of feet per chicken × total head-foot difference) ÷ (number of feet per chicken+number of feet per rabbit) = number of rabbits;
Total number of rabbits = number of chickens
Or (the number of feet per rabbit × the total number of heads+the difference between the number of feet of chickens and rabbits) ÷ (the number of feet per chicken+the number of feet exempted from each chicken) = the number of chickens;
Total number of chickens = rabbits. (Example omitted)
(3) Given the difference between the total number of feet of chickens and rabbits, when the total number of feet of rabbits is greater than that of chickens, the formula can be used.
(the number of feet per chicken × the total number of heads+the difference between the number of feet of chickens and rabbits) ÷ (the number of feet per chicken+the number of feet per rabbit) = the number of rabbits;
Total number of rabbits = number of chickens.
Or (the number of feet per rabbit × the total number of heads-the difference between the number of feet of chickens and rabbits) ÷ (the number of feet per chicken+the number of feet per rabbit) = the number of chickens;
Total number of chickens = rabbits. (Example omitted)
(4) The following formula can be used to solve the gain and loss problem (the generalization of the chicken-rabbit problem):
(65438 points +0 number of qualified products × total number of products-total score achieved) ÷ (score for each qualified product+deduction for each unqualified product) = number of unqualified products. Or the total number of products-(points deducted for each unqualified product × total number of products+total realized score) ÷ (points deducted for each qualified product+points deducted for each unqualified product) = unqualified.
For example, "workers who produce light bulbs in a light bulb factory are paid according to their scores." Each qualified product will be deducted 4 points, and each unqualified product will be deducted 15 points. A worker who produces 65,438+0,000 light bulbs gets 3,525 points. How many are unqualified? "
Solution1(4×1000-3525) ÷ (4+15)
=475÷ 19=25 (pieces)
Solution 21000-(15×1000+3525) ÷ (4+15)
= 1000- 18525÷ 19
= 1000-975=25 (pieces) (omitted)
(The "gain and loss problem" is also called the "glassware transportation problem", and the transportation fee for each intact piece is ×× yuan. The damaged piece not only does not pay the transportation fee, but also needs to pay the cost of ×× yuan ... The above formula can obviously be applied to its solution. )
(5) The problem of chicken-rabbit exchange (the problem of finding the number of chickens and rabbits after knowing the total number of feet and the total number of feet after chicken-rabbit exchange) can be solved by the following formula:
[(sum of total feet twice) ÷ (sum of feet of each chicken and rabbit)+(difference of total feet twice) ÷ (difference of feet of each chicken and rabbit) ÷ 2 = number of chickens;
(sum of total feet twice) ÷ (sum of feet of each chicken and rabbit)-(difference of total feet twice) ÷ (difference of feet of each chicken and rabbit) ÷ 2 = number of rabbits.
For example, "There are some chickens and rabbits, and * * * has 44 feet. If you change the number of chickens into the number of rabbits, * * * has 52 feet. How many chickens and rabbits are there? "
Solution [(52+44) ÷ (4+2)+(52-44) ÷ (4-2)] ÷ 2
=20÷2= 10 (only applicable)
〔(52+44)÷(4+2)-(52-44)÷(4-2)〕÷2
= 12÷2=6 (only applicable)
Tree planting problem formula
(1) The problem of planting trees on the unclosed line;
Interval number+1= number of trees; (planting trees at both ends)
Road length-interval length+1= number of trees.
Or interval number-1= number of trees; (No planting at both ends)
Road length ÷ section length-1= number of trees;
Road length ÷ number of sections = length of each section;
Length of each section × number of sections = road length.
(2) The problem of planting trees on closed lines:
Road length/interval = number of trees;
Road length/number of intervals = road length/number of trees
= length of each interval;
Length of each section × number of sections = length of each section × number of trees = road length.
(3) Planar tree planting:
Total area/area per tree = number of trees
Formulas for Solving Fractions and Percentages
Comparison number ÷ standard number = score (percentage) rate corresponding to comparison number;
Number of growth ÷ standard number = growth rate;
Reduction number ÷ standard number = reduction rate.
perhaps
The difference between two numbers ÷ the smaller number = a few more (one percent) (increase);
The difference between two numbers ÷ the larger number = a few (hundredths) (minus).
Reciprocal formula for increasing or decreasing percentage (percentage)
Growth rate ÷( 1+ growth rate) = reduction rate;
Reduction rate ÷( 1- reduction rate) = growth rate.
How much smaller than the area of Jiaqiu? "
This is an application problem to find the reduction rate according to the growth rate. According to the formula, the answer can be as follows
What percentage? "
This is an application problem of finding the growth rate from the reduction rate. According to the formula, the answer can be as follows
Solution to the application problem of comparison number
Standard number × percentage rate = comparison number corresponding to percentage rate;
Standard number × growth rate = growth number;
Standard number × reduction rate = reduction number;
Standard number × (sum of dichotomy) = sum of two numbers;
Standard number × (difference of dichotomy) = difference of two numbers.
Formula for solving the application problem of standard number
Contrast number ÷ Score (percentage) corresponding to contrast number = standard number;
Growth number ÷ growth rate = standard number;
Reduction number ÷ reduction rate = standard number;
Sum of two numbers and sum of two rates = standard number;
The difference between two numbers ÷ the difference between two rates = standard number;
Formula of square matrix problem
(1) solid square: (number of people on each side of the outer layer) 2= total number of people.
(2) Hollow square:
(number of people on each side of the outermost layer) 2- (number of people on each side of the outermost layer -2× number of layers) 2= number of hollow squares.
perhaps
(Number of people on each side of the outermost layer-number of layers) × number of layers× 4 = number of hollow squares.
Total number of people ÷4÷ layers+layers = number of people on each side of the outer layer.
For example, there is a three-story hollow square with 10 people on the outermost layer. How many people are there in the whole square?
Scheme 1 is regarded as a solid square first, so the total number of people is
10× 10= 100 (person)
Then calculate the square of the hollow part. From the outside to the inside, every time you enter a floor, the number of people on both sides is less than 2, and then you enter the fourth floor. The number of people on each side is
10-2×3=4 (person)
Therefore, the number of squares in the hollow part is as follows.
4×4= 16 (person)
So the number of people in this hollow phalanx is
100- 16=84 (person)
Solution 2 directly applies the formula. According to the formula of the total number of people in the hollow square.
(10-3)×3×4=84 (person)
There are many kinds of interest rate problems. The calculation formulas of common simple interest and compound interest problems are introduced as follows.
(1) simple interest problem:
Principal × interest rate× term = interest;
Principal ×( 1+ interest rate× term) = principal and interest;
Principal and interest and cash (1+ interest rate × term) = principal.
Annual interest rate ÷ 12= monthly interest rate;
Monthly interest rate × 12= annual interest rate.
(2) compound interest:
Principal ×( 1+ interest rate) Number of deposit periods = principal and interest.
For example, "someone deposits 2400 yuan with a term of 3 years, and the monthly interest rate is 10.2 ‰ (that is, monthly interest 1.02). After three years, how much is the principal and interest? "
Solution (1) is calculated at the monthly interest rate.
3 years = 65438+2 months ×3=36 months
2400×( 1+ 10.2%×36)
=2400× 1.3672
= 328 1.28 (yuan)
(2) Use the annual interest rate.
Change the monthly interest rate to the annual interest rate first:
10.2‰× 12= 12.24%
Seek principal and interest again:
2400×( 1+ 12.24%×3)
=2400× 1.3672
= 328 1.28 yuan (omitted)