Today, Shi Ming Little Book Boy talks to you about the "pit" in mathematical operation-a common mistake-prone point.
First, the trip problem-"car crossing the bridge" and midway rest
Example 1. It takes 33 seconds for a train to pass through a tunnel with a length of 1200m, and 3 seconds for another train with a length of 150m and a speed of 50 m/s to pass through a bridge with a speed of 50m/s. Then, how long does it take for the train to pass through a bridge with a length of 600m after it has slowed down by half?
18 second
B 20 seconds
C 30 seconds
D 36 seconds
fundamental analysis
According to the question, two trains passed by by mistake, which is equivalent to meeting each other, so the distance passed by is the sum of the lengths of the two trains and the speed is the sum of the speeds of the two trains. Let the train speed be Vm/s and the car body length be L. Because it takes three seconds for two trains to pass by by mistake, (L+ 150)/3=(V+50) and L=3V. It takes 33 seconds for the train to pass through the 1200m tunnel, and the total distance is the sum of the train length and 1200. So L+ 1200=33V, that is, L=3V, that is, 1200=30V, V = 40m/s ... If the train speed is halved, the speed is 20m/s, and the total distance after passing the bridge is the length of the car body and the length of the bridge S = 600+3× 40 = 720m. So, the answer to this question is D.
Error/trap analysis
The length of the wrong train is the sum of the body lengths of the two trains, and the body length can not be ignored when crossing the tunnel.
Ex. 2: Lao Li goes out for a walk and exercises at 9 o'clock every morning on time. He walked 6 kilometers at a speed of 3 kilometers per hour and took a 5-minute break every 20 minutes. Then Lao Li kept exercising until () he went home.
One piece 1 1: 20
B 1 1: 25。
C 1 1: 30
D 1 1: 45
fundamental analysis
According to the meaning of the question, Lao Li's speed is 3 kilometers per hour and the distance is 6 kilometers, that is, if he doesn't have a rest, it will take Lao Li 6 hours to walk 6 kilometers. Take a break every 20 minutes, that is, 60÷20=3 times per hour, because the last walk does not involve rest time, that is, 120-minute trip requires 5 breaks, and the total rest time =5×5=25 minutes. So the time for Lao Li to practice Taoism is 9: 00 +2 hours and 25 minutes = 1 1: 25. So the answer to this question is B.
Error/trap analysis
Take a five-minute break every 20 minutes. From going out to exercise to going home, the last walk home does not involve rest. The rest time during the journey can't be ignored. Note whether it is necessary to calculate the total travel time.
Second, the probability problem-reverse calculation
Example 3. In the graduation season, there are four boys, A, B, C and D, in dormitory 306. They want to take a photo with the head teacher, Mr. Song. If Mr. Song is asked to sit in the middle, and students A and B cannot sit next to each other, how many sitting methods are there?
A 8 species
B 12 species
C 16 species
D 24 species
fundamental analysis
Option 1: According to the questions, there are five teachers taking a group photo, and the teacher wants to be in the middle, so there are two people on the left and right of the teacher. A and B can't sit next to each other, that is, AB is not adjacent, so A and B are on the left and right sides of the teacher, and C and D are on the left and right sides of the teacher. Let A choose 1 from the remaining four positions, and use C(4, 1)=4 sitting methods, then B can only choose one from the two positions that are never next to A, use C(2, 1)=2 sitting methods, and the remaining two CD players choose from the remaining two positions, and they are arranged in A.
Solution 2: Reverse consideration. Teacher Song is sitting in the middle. The total sitting method of four students in ABCD is A (4,4) = 24. If two students in AB sit next to each other, they can choose two positions on the left or right of Teacher Song after binding, C(2, 1)=2, the internal position of the two students is A (2,2) = 2, and the other side of Teacher Song is CD II. There are 24-8= 16 situations where people do not sit next to each other. So the answer to this question is C.
Error/trap analysis
The probability problem can be solved from the front or from the back (usually when the front is difficult to solve), but it should be noted that the latter solution should be considered comprehensively or not too much calculation.
Third, the issue of economic profit-profit rate.
Example 4: The purchase price of a commodity is 5% lower than last month, but the supermarket still sells at last month's price, and the profit rate has increased by 6 percentage points. Then the profit rate of the supermarket selling the goods last month is:
12%
B 13%
C 14%
D 15%
Basic problem solving
Let the purchase price of goods last month be 100, and the price of goods be x, then the purchase price of goods this month is 95, and (x-95)/95-(x-100)/100 = 0.06, that is, x/95-x/656. So the answer to this question is C.
Error/trap analysis
Generally speaking, the profit rate in mathematical operation = profit/cost, and the profit rate in data analysis = profit/income.
Fourth, the cycle date problem-interval cycle, date calculation
Example 5. A, B and C all go to the gym to exercise every once in a while. A goes every two days, b every four days and c every seven days. In April 10, the three met. When is the next meeting?
May 28(th)
June 5(th)
July 24(th)
July 25(th)
fundamental analysis
According to the stem, if A goes to the gym every two days, then the cycle of A going to the gym is 2+ 1=3 days, if B goes to the gym every four days, then the cycle of B going to the gym is 4+ 1=5 days, if C goes to the gym every seven days, then the cycle of C going to the gym is 7 days, then A, B and C go to the gym at the same time. On April 10, the three met, and the next meeting in the gym was after 105. At this time, there are 20 days left in April, 3 1 day in May, 30 days in June and 24 days left in July. So the answer to this question is C.
Error/trap analysis
Go once every n days, with a period of n+ 1 day; Go once every n days and cycle for n days.
The Problem of verb (Verb's Abbreviation) Concentration —— The Application of Cross Method
Example 6. Mix a few grams of existing brine with the concentration of 12% and 24%, and add 50 grams of water to make 600 grams of brine with the concentration of 18%. Then the mass ratio of the original brine with the concentration of 12% and 24% is:
6: 05
B 1
C 5:6
The fourth chapter section 7
fundamental analysis
Solution 1: Suppose that there is x grams of salt water with a concentration of 12% and y grams of salt water with a concentration of 24%. After mixing, 50 grams of water was added to obtain 600 grams of saline water with the concentration of 18%, so the total amount of solution was 600=x+y+50, and the salt content before and after mixing was 12% x+24%. So, the answer to this question is D.
Analysis 2: According to the stem of the question, 600g of brine with the concentration of 18% is obtained by mixing, and the concentration of brine with the concentration of 12% and 24% before adding water is 600×18%/(600-50) =108. So, the answer to this question is D.
Error/trap analysis
The mass of solute remains unchanged before and after solution mixing, but the mass and concentration of solution change. In this problem, 50 grams of water is added after mixing the two solutions, and the brine with the concentration of 18% is obtained, which cannot be directly calculated when using the cross method.
The above are common mistakes in mathematical operations, and some basic formulas in mathematics are attached. I wish you all the best.
1, engineering problems:
Workload = work efficiency × working hours, and total workload = sum of various workloads.
2. Travel problems:
Distance = speed × time
The formula of encounter pursuit problem is: encounter distance =(V 1+V2)× encounter time, pursuit distance =(V 1-V2)× pursuit time.
The formula of running water problem: using water flow = ship speed+water speed, using water flow = ship speed-water speed.
The formula of round-trip encounter problem is: two encounters of cross-strait type S=3S 1-S2, and two encounters of single-shore type S=(3S 1+S2)/2(S 1 is the first encounter distance and S2 is the second encounter distance).
3. Economic profit:
Profit = selling price-cost, profit rate = profit/cost, total profit = single profit × sales volume = total selling price-total cost.
4, permutation and combination:
Calculation formula of permutation and combination: a (n, m) = n * (n-1) * (n-2) * ... * (n-m+1); C(n，m)= n *(n- 1)*(n-2)*……*(n-m+ 1)/m！ ; C(n，m)=C(n，n-m).
The difference and application between classification principle and step-by-step principle: addition for classification and multiplication for step.
Common permutation and combination methods: binding method, insertion method, partition method, dislocation arrangement method and round table arrangement method.
5, the principle of tolerance:
Two exclusive formulas A+B-A∩B= all-none.
Three-set exclusion formula A+B+C-(A∩B+B∩C+A∩C)+A∩B∩C= None.
Three sets of deformation formula A+B+C- (which only belongs to the sum of the values of two sets at the same time) -2×A∩B∩C= all-none.
6, concentration problem:
Solution = solute+solvent, concentration = solute/solution
7, cattle grazing problem:
Y=(N-x)×t, where y represents the number of grasslands, n represents the number of cattle, x represents the growth rate of grass, and t represents the time required for cattle to eat this grass.
8. Basic mathematical formulas
Sum of (1) frequent test sequences
Natural sequence:1+2+3+...+n = n * (n+1)/2. [In natural sequence, the number of numbers = (large number-decimal number)+1]
Arithmetic progression with tolerance d: a [n] = a [1]+(n-1) d; s[n]=(a[ 1]+a[n])/2×n； s[n]= na[ 1]+n(n- 1)/2×d .
(2) Numeric characteristics of multiples of 2, 3 and 5
Multiply of 2 = number divisible by 2: the last digit of the number is even;
Multiply of 5 = number can be divisible by 5: the last digit of the number is 0 or 5;
Multiply of 3(9) = The number is divisible by 3(9): the sum of the digits of each bit of the number is a multiple of 3(9).
(3) congruence theorem
Difference and subtraction are the same (choose the least common multiple of divisor and then "subtract")
Sum plus sum (choose the least common multiple of divisor and then "add sum")
The remainder is the same as the remainder (choose the least common multiple of the divisor and then "add the remainder")
(4) Diversity characteristics
If a:b=m:n(m and n are coprime), then a is a multiple of m, b is a multiple of n, and a+b(a-b) is a multiple of m+n(m-n).