Party A and Party B arrive from two places 3.2 kilometers apart. If they go in the opposite direction, they will meet in two hours; If you go in the same direction, meet in eight hours, and find the speed of Party A and Party B. ..
The original speed of the cart was 30 km/h, and now it is accelerating at a constant speed, increasing by 20 km/h; The original speed of a car was 90 km/h, but now it starts to slow down at a constant speed, which is10 km/h. How long will it take for the two cars to have the same speed? What is the speed at this time?
Use the equation.
Two students, A and B, play math games. The rules are as follows: First, Party A quotes a number that is not zero, then Party A quotes a number that is 1 less than Party B, and then Party B quotes a number that is twice that of Party A for the second time, and so on. The winner who reported 0 first, prophet A reported 0 for the fourth time, and Party A reported 0 for the first time ().
Please list equations or algebraic expressions and formulas.
According to the regulations of civil aviation, a passenger in different cabins can carry up to 20 kilograms of luggage free of charge, and the baggage ticket is purchased at the rate of 65438+ 0.5% of the air fare per kilogram for the country-seeking part. A passenger brought 35 kilograms of luggage and paid 1323 yuan together with the luggage, so he asked for the passenger's ticket money.
Kangaroos and rabbits have a jumping competition. Kangaroos jump 412 meters at a time, and rabbits jump 2 3/4 meters at a time. They jump once a second. During the race, from the starting point, there is a balloon every 12 3/8 meters. When one of them steps on the balloon, the game is over, and the first person to step on it wins. Asked when the other jumped a few meters.
Kangaroos can hit the fourth balloon after jumping 1 1 time. At this time, he jumped 49.5 meters and the rabbit jumped 30.25 meters.
After jumping nine times, the rabbit can hit the second balloon. At this time, the rabbit jumped 24.75 meters and the kangaroo jumped 40.5 meters.
Therefore, the rabbit won. At the end of the race, the kangaroo jumped 40.5 meters and the race took 9 seconds.
I haven't done primary school questions for a long time, and I feel quite fun.
In order to prevent "bird flu", Aunt Wang, a professional chicken farmer, is preparing disinfectant. Instructions on the disinfection powder bag: preparation requirements: the concentration of poison is greater than or equal to 2%.
Note: This product must be completely melted.
Shelf life: 2 years at room temperature. Aunt Wang uses 8 kilograms of disinfectant every day, so how many kilograms of disinfectant can she prepare at most a day?
If the concentration should be equal to 2%, because the topic requires the most.
8 ÷ 2% = 400 (kg) ... This is a disinfectant.
400-8 = 392kg ... This is clear water.
Because the question requires the kilograms of disinfectant, it is only necessary to answer the kilograms of disinfectant.
A: She can prepare up to 400 kilograms of disinfectant a day.
Boss Li sells these two kinds of biscuits with different sizes and the same thickness. After a few days of buying, more and more people buy big biscuits, and fewer and fewer people buy small biscuits. Why?
Attached:
Big sesame seed cake: 4 cents each, 16 cm.
Small sesame seed cake: 3 cents each, 12 cm.
An adult needs about 3 kilograms of water every day. According to the information below, please calculate how many kilograms of water an adult needs to drink every day. (Keep one decimal place)
Attached:
14% body manufacturing
39% comes from solid matter.
47% depends on drinking water.
4. Xiaohua blows soap bubbles once every minute, just blowing 100 bubbles each time. After the soap bubbles were blown out, half of them burst after one minute, but 1/20 did not burst after two minutes, and all the soap bubbles burst after two and a half minutes. When Xiaohua blew out 100 new soap bubbles for the second time, how many soap bubbles did not burst?
The area of the pie is π× (16/2) 2 = 64π.
The cake area is π× (12/2) 2 = 36 π.
64π/4 & gt； 36π/3 (that is, a dime can buy a bigger cake)
So buying pies is cost-effective.
3.3× 47% =1.41=1.4 (kg)
4.100× (1+1/2+1/20) =155 (pieces)
Use the symbols and brackets of four operations to make the following formula hold.
2 2 2 2 2 2 2 2 2 2=2008
impossible
Because the biggest case of 10 2-4 operation is multiplication.
The result is 1024.
So the maximum value is 1024.
It can't be 2008
With the cooperation of Party A, Party B and Party C, a job was completed in 15 days.
Party A and Party B cooperate 10 day, and Party C will do the rest, which will be completed in 30 days.
Party A and Party C will cooperate for 20 days, and Party B will do the rest, which will take another 8 days.
Party A, Party B and Party C are required to do it alone. How many days will it take?
15 *( 1/x+ 1/y+ 1/z)= 1( 1)
10 *( 1/x+ 1/y)+20 * 1/z = 1(2)
20 *( 1/x+ 1/z)+8 * 1/y = 1(3)
(2)*3-( 1)*2 can get z=30, (2)*2-(3) substitute z=30 to get y=36, and then substitute yz (1) to get x= 180.
Therefore, Party A needs 180 days, Party B needs 36 days and Party C needs 30 days. ..
Attached, I just read the first two answers, which are different from my answer to the second condition.
Judging from the title, condition 2: Party A and Party B cooperate 10 day, Party C does it, and it will be completed in 30 days. The 30 days here should include 10 days that Party A and Party B did first, that is, Party C only did 20 days. If it takes 30 days to refer to C alone, then the second condition should be described as the third condition, that is, C "needs" 30 days.
It is said that Li Sangong opened a shop, and many guests came to the store. There are more than 7 guests in each room, and 9 guests in each room are empty. How many rooms are there?
If the room is set to X and the guest is set to Y, then
Y=7X+7
Y=9(X- 1)
X=8，Y=63。
There are 65,438 RMB +50 RMB from 0 yuan, 2 yuan and 5 yuan with face value 1 16 yuan. It is known that 1 yuan has two more tickets than 2 yuan. How many tickets are there for each of the three kinds of RMB?
A, chicken cage and rabbit cage method
Since 1 yuan is two more than that of 2 yuan, please save these two first, and the title will become 48, *** 1 14 yuan.
At this time, because 1 yuan is as much as 2 yuan, we changed them into 1.5 yuan denomination.
The title has changed again, becoming 48, 1 14 yuan, 5 yuan 65438 and 0.5 yuan 65438.
If it's all from 5 yuan, 48 ***48*5=240 yuan.
For every increase of 1.5 yuan, the amount is 5- 1.5=3.5 yuan.
* * * Less than 240-114 =126 yuan.
1.The number of sheets in 5 yuan is 126/3.5=36.
It is equivalent to 18 in 0 yuan and 18 in 2 yuan. If I return two 1 yuan to him, then 1 yuan is 20.
5 yuan's is 50-20- 18= 12.
Second, with the idea of ordinary equation:
Suppose there is X in 2 yuan, then there is X+2 in 1 yuan and 50-X-(X+2) in 5 yuan.
The total amount is:
X * 2+(X+2)* 1+[50-X-(X+2)]* 5 = 1 16 .
X= 18，
Then 65438+18+2 = 20 in 0 yuan and 50-18-20 in 5 yuan =12.
There are two ports on a river, which are 45 kilometers apart. Every day, two passenger ships A and B with the same speed leave two ports at the same time and travel in opposite directions. On this day, when Ship A set out from the port, it dropped something, floated on the water and went downstream. Four minutes later, how many hours after the departure of Ship A 1 km and Ship B, are you expected to meet this thing?
The best answer is "this thing floats on the water and floats downstream." After 4 minutes, the distance is A 1km, "A is in front, and the floating objects are behind. This is a" catch-up problem ". The speed difference between a and floating objects is: a speed+water speed-floating object speed (water speed) = A speed, and a speed is 4 minutes 1km.
The distance between the floating object and B is 45km, and B is upstream, which is a typical encounter problem. The sum of velocities is: b velocity-water velocity+floating object velocity (water velocity) = B velocity, and b velocity is also1km for 4 minutes, 45× 4 = 180min = 3 hours.
3. 15 years ago, the father's age was seven times that of his son. 10 years later, the father's age was twice that of his son. How old are father and son this year?
15 years ago, the father's age was seven times that of his son, that is, the age difference was seven times that of his son.
After 10 years, the age of the father is twice that of the son, that is, the age difference is 2- 1= 1 times that of the son, that is, the age of the son after 10 years is exactly equal to the age difference, that is, the age of the son after 10 years is equivalent to 15 years.
15+20=25 years My son's age is equivalent to 6- 1=5 times that of my son five years ago.
25/5=5 years old (15 years ago)
5+ 15=20 years old (now)
1. The distance from mountain to mountain is 1200m. The average speed of floret is 60m per minute when it goes up the mountain and 120m when it goes down the mountain. What is the average speed of small flowers going back and forth?
It's time to go up the mountain.
1200/60 = 20 o'clock
Downhill time is
1200/120 =10 point
The average speed of floret going back and forth is per minute.
1200× 2 ÷ (20+ 10) = 80m。
2. A truck can transport ore 20 times a day in sunny days and only 12 times a day in rainy days. It is continuously transported 1 12 times, with an average of 14 times a day. How many sunny days are there?
* * * Shipped.
1 12 ÷ 14 = 8 days.
Suppose it rains for 8 days, it should be shipped.
12× 8 = 96 times
Less than now.
1 12-96 = 16 times
There is more luck in sunny days than in rainy days.
20- 12 = 8 times
sunny day
16 ÷ 8 = 2 days
3. Party A and Party B shall report from 1 to 20/in turn, and each party may report to 1 or 2 once. It is stipulated that whoever gets to 20 first will win. If you want to win, should you report first or later? How many times? How to report it in the future?
Two people can apply for one round.
1+2 = 3 numbers
20÷3=6……2
So you can win by reporting first.
If you bid 2 first, you can win as long as the sum of two people is 3 (the other party bids 1, you bid 2, the other party bids 2, and you bid 1).
1, clever addition and subtraction
100+99-98-97+96+95-94-93+……+8+7-6-5+4+3-2- 1
Analysis: This is a comprehensive problem of adding and subtracting several numbers. There are 100 addition and subtraction items. If you want to simplify the calculation, you can combine two arrays into one group through the exchange of sequence and sequence, and * * * can be combined into 50 groups, each with a value of 2.
Solution: Primitive = (100-98)+(99-97)+(96-94)+...+(4-2)+(3-1)
=2×50= 100
Note: you can also combine four numbers into a group in order to get.
100+99-98-97=96+95-94-93=……
=4+3-2- 1=4
The original formula can be combined into 25 groups, and the value of each group is 4, and the result is equal to
4×25= 100。
2. Clever multiplication and division
9999×2222+3333×3334
Analysis: The problem of decomposing 9999 into 3333×3 has the same factor as that of 3333×3334, and the operation can be reversibly simplified by multiplying distribution rate.
Solution: 9999×2222+3333×3334
= 3333×3×2222+3333×3334
= 3333×(6666+3334)
= 3333× 10000
= 33330000。
3. Solve application problems with hypothetical methods.
There are 52 students in Class Four (2). They went boating in the park and rented a boat of 1 1. Each big boat takes six people, and each small boat takes four people, which is just full. How many big boats and small boats are there for rent?
Analysis: Suppose all the big ships are rented, because each big ship takes six people, then 1 1 ship * * takes 66 people, which is more than the original class 14 people. The reason for the change is that each boat takes four people. Now suppose that it takes six people, and each boat takes two more people. Obviously, the number of boats has increased.
Solution: What is the number of ships?
(6× 1 1-52)÷(6-2)=7 (pieces)
The number of large ships is
1 1-7=4 (pieces)
There are four big boats and seven small boats.
4, column equation to solve application problems
Two people, Party A and Party B, produce parts. Party A produces 8 hours and Party B produces 6 hours. Party A produces 88..a A more than Party B, and the output per hour is always 2 less than that of Party B.. B how much does it produce per hour?
Analysis: The output of 88 parts is only 8 hours and 6 hours. According to this quantitative relationship equation, the key is to know how many parts A and B produce per hour. From the topic conditions, we can know that "A produces 2 less per hour than B". If B produces X every hour, A produces (X-2) every hour. In this way, we can list the equations and find out the working efficiency of B.
Solution: Suppose B produces X units per hour, then A produces (X-2) units per hour.
(X-2)×8-6X=88
8X- 16-6X=88
2X=88+ 16
X=52
A: B produces 52 units per hour.
5. Profit and loss issues
The kindergarten teacher distributed candy to the children. If each child gives five sweets, there will be 22 more sweets. If each child is divided into seven pieces of sugar, there will be less 18 pieces of sugar. How many children and some sweets are there?
Analysis: The total number of sweets is the same as the number of children. If everyone has five, the remaining 22; If everyone scores more (7-5), less 18. Here, because the second time scored 2 points more than the first time, the result of each score was different (22+ 18). In this way, (22+ 18)÷(7-5) is the number of children.
Solution: The number of children is
(22+ 18)÷(7-5)=20 (piece)
The total number of sweets is
5×20+22= 122 (piece)
Or 7×20- 18= 122 (pieces)
A: 20 children, candy 122.