Analysis and solution method 1: Assuming that there are X people * * * at the beginning, the total amount of sugar in the two methods remains unchanged, which is 5x+ 10=4× 1.5x-2, and the solution is x= 12, then these sugars * * have12× 5+.
Method 2: After the number of people increases by 1.5 times, each person is divided into 4 pieces, which is equivalent to the original number, and each person is divided into 1.5×4=6 pieces.
With these candies, each person is divided into five pieces, with the extra 10, the less each divided into six pieces and the less two pieces, so the total number of people at the beginning is (10+2)÷(6-5)= 12, so * * has/kloc-.
Two children, A and B, each have a bag of sweets, and each bag of sweets is less than 20. If A gives B a certain amount of sweets, A has twice as many sweets as B. If B gives A the same amount of sugar, A has three times as much sugar as B. So, how many sweets are there in A and B?
According to the meaning of the question, the total amount of sugar should be a multiple of 3 or a multiple of 4, that is, a multiple of 12, because each bag of two bags of sugar does not exceed 20, so the total amount of sugar does not exceed 40. So the total amount of sugar can only be 12, 24 or 36.
If the total amount of sugar is an odd multiple of 12, then the sugar of A is an odd multiple of 12÷(3+ 1)×3=9. Then after giving B the same amount of sugar twice, the sugar of A is 12.
In other words, odd plus even equals even, which is obviously impossible. Therefore, the total amount of sugar cannot be an odd multiple of 12.
Then two children, A and B, can only have even multiples of 12, that is, 24 sweets.
3. There are 42 students in Class A and 48 students in Class B. It is known that in a math exam, the scores are graded according to the percentage system. Results The total math scores of all classes are the same, and the average scores of all classes are integers, and the average scores are higher than 80. How much is the average score of Class A higher than that of Class B?
Solution 1: Because the average score of each class is an integer, the total scores of the two classes are equal, so the total scores are multiples of 42 and 48, so it is a multiple of [42,48] = 336.
Because the average score of Class B is higher than 80, the total score should be higher than 48×80=3840.
Because it is graded according to the percentage system, the average score of Class A will not exceed 100, so the total score should not be higher than 42× 100=4200.
Between 3840 and 4200, only 4032 is a multiple of 336, so the total score of both classes is 4032.
Then the average score of Class A is 4032÷42=96, and that of Class B is 4032÷48=84.
So the average score of Class A is 96-84= 12 higher than that of Class B. 。
Method 2: The average score of Class A is 42 = Class Average × 48, that is, Class A Average × 7 = Class B Average × 8. Because 7 and 8 are reciprocal, the average score of class A is 8 times of a certain number, and the average score of class B is 7 times of a certain number. And because the average scores of the two classes are above 80, they are not higher than 100.
So the average score of Class A is higher than that of Class B 12×(8-7)= 12.
4. A township hydropower station collects electricity charges by households. The specific provisions are: if the monthly electricity consumption does not exceed 24 degrees, it will be charged at 9 cents per degree; If it exceeds 24 degrees, the excess will be charged at 20 points per degree. It is known that family A paid 90.6 points more than family B in a certain month (electricity consumption is calculated by the whole degree). How much did A and B pay respectively?
Analysis and solution If the electricity consumption of both parties exceeds 24 degrees, the difference between their electricity charges should be an integer multiple of 20 cents;
If the electricity consumption of both parties does not exceed 24 degrees, the difference between them should be an integer multiple of 9 cents.
Now 96 points is neither an integer multiple of 20 points nor an integer multiple of 9 points, so the household electricity consumption of A exceeds 24 degrees, and that of B does not exceed 24 degrees.
Assume that household A uses 24+X kWh, household B uses 24-y kWh, 20x+9y=96, x=3 and y = 4.
That is, the electricity consumption of household A is 27 kWh, and that of household B is 20×9= 180 = 1 yuan 8, and that of household A 180+96=276 =2 yuan 7.6.
That is, Party A and Party B each pay 76 cents for electricity in 2 yuan and 65438+80 cents for electricity in 0 yuan.
5. The number of spring outing in primary and secondary schools is an integer multiple of 10. When traveling, the staff of the two schools do not share a car, and each car is as full as possible. It is now known that if both schools rent a coach car with a seat of 14, they need to rent 72 such cars; If both schools rent 19 coaches, the second primary school rents 7 more coaches than the first primary school. How many people will take part in this spring outing?
The analysis and solution assume that the number of spring tourists in the second primary school is m and the number of spring tourists in the first primary school is n, and take19× 6+65438 ≤ m-n ≤19× 8-1,that is, 18, which shows that the second primary school rents seven more vans than the first primary school.
It is also known that two schools need to rent 72 vans with 14 seats, so 70×14+2 ≤ m+n ≤ 72×14, that is, 982 ≤ m+n ≤ 1008.
At the same time, it is known that both m and n are multiples of 10, so there are
The other four groups are because the solutions of m and n are not multiples of 10.
After inspection, only satisfied.
So, 430 people from a primary school and 570 people from a middle school took part in the spring outing.
6. A tourist rowed a boat from the pier of 10 on 15, and he wanted to return to the pier no later than 13. The speed of the river is 1.4 kilometers per hour, and the speed of the ship in still water is 3 kilometers per hour. He rowed for 30 minutes and rested for 15 minutes without changing his direction.
The analytical solution 10 starts from 15 and must be returned no later than 13, so it can be drawn for 2 hours and 45 minutes at most, that is, 165 = 4× 30+3× 15, and four 30 minutes at most.
Downward speed is 3+1.4 = 4.4km/4hr; So the downstream half-hour rowing distance is 4.4×0.5 = 2.2km；;
When the countercurrent velocity is 3- 1.4= 1.6 km /4; Therefore, the upstream half-hour rowing distance is1.6× 0.5 = 0.8km..
After a rest of 15 minutes, the distance of the ship drifting downstream is 1.4×0.25=0.35 km.
In the first case, when starting downstream, it takes at least half an hour to drive 2.2 kilometers, and in the remaining three hours, the inner boat floats downstream by 0.35×3= 1.05 kilometers, so it needs to row 2.2+ 1.05=3.25 kilometers when returning upstream.
3.25÷ 1.6=2.03 125 hours = 12 1.875 minutes. That means at least 30+15× 3+121.875 = 6575. 165 minutes, too late to return on time. Not satisfied.
In the second case, at the beginning of the upstream, the boat will travel 0.8km every half hour, so after three times of upstream, the boat will drift downstream 1.05km when the tourists rest, so it only needs to row 2.4- 1.05 = 1.35km when rowing back. Need 1.35÷4.4≈0.3068 hours ≈ 18.4 1 min. * * * Need 3× 30+3×15+18.41= 65438+. 165 minutes, satisfied.
Therefore, only the second case is satisfied. At this time, the farthest distance is the place where the third countercurrent reaches after two cuts, which is 0.8×3-0.35×2= 1.7 km.
So he can row 1.7 km outside the pier at most.
This machinery factory plans to produce a batch of machine tools. The original plan is to produce 40 units a day, and the task can be completed within the scheduled time. Actually, 48 sets are produced every day, and the task is completed four days ahead of schedule. How many machines are there?
48×[40×4(48-40)]= 960 (Taiwan Province)
8. A printing factory plans to bind a batch of books in 24 days, and bind 12000 copies a day, actually completing the task four days ahead of schedule. How many more books are actually bound every day than originally planned?
12000× 24 ÷ (24-4)-12000 = 2400 (Ben)
9. Of the two brick factories, Factory A originally stored 87,500 bricks, and Factory B stored 4,500 more bricks than Factory A.. One day, A factory sold 25,000 bricks, and B factory sold 3,000 bricks less than A factory. Which factory stores more bricks? How much more?
Brick of Factory A: 87500-25000=62500 (block)
Brick of factory B: (87500+4500)-(25000-3000)=70000 (block)
∴ There are more bricks in Factory B, 70000-62500=7500 (blocks).
10. A basket of apples weighs 45 kilograms. After selling half, the remaining apples weigh 24 kilograms. How many kilograms of apples are there?
(45-24)×2=42 (kg)
1 1. At 8: 00 a.m., Xiaoming rides a bike from place A to place B; At 8: 40am, Xiao Qiang rode his bike from B to A, with a journey of16km. They meet at the midpoint of A and B. What is the distance between A and B?
Solution: This is a problem of walking in opposite directions and meeting for distance. But they didn't start at the same time. If they can start at the same time and find out when they met, they can use mathematical methods.
They met at the midpoint of the distance between the two places, but Xiao Ming walked 40 minutes longer than Xiao Qiang. If they set out at the same time, when they met, Xiao Ming walked less than Xiao Qiang 12÷60×40=8 (km), that is to say, when Xiao Qiang set out, Xiao Ming had already walked 8 km, and they met at 8: 40, because Xiao Ming walked less every hour than Xiao Qiang. Explain that the meeting time of two people is 8÷4=2 (hours), then the distance between A and B is 8+( 12+ 16) × 2 = 64 (kilometers).
A: The distance between A and B is 64 kilometers.
12: Village A and Village B are 3550 meters apart. Xiao Wei walked from village a to village b. Five minutes later, Xiao Qiang rode his bike from Village B to Village A.. 10 minutes later, he met Xiao Wei. Xiao Qiang rides bicycles more than Xiao Wei walks per minute160m. How many meters does Xiao Wei walk every minute?
Solution: If Xiao Qiang walks less 160m per minute and his walking speed is the same as that of Xiao Wei, then Xiao Qiang walks less160× 10 =1600 (m) in10 minutes. The walking distance between Xiao Wei (5+ 10) and Xiao Qiang (10) is 3550- 1600 = 1950 (meters), so the walking distance of Xiao Wei is 1950 \( 5).
Xiao Wei walks 78 meters per minute.
13: The bus starts from Dongcheng and the truck starts from Xicheng at the same time, face to face. Buses travel 44 kilometers per hour and trucks travel 36 kilometers per hour. The bus to the west side is two hours earlier than the truck to the east side. How many hours did the two cars meet on the road after they set off?
Solution: When the bus arrives in the west city, the distance between the truck and the east city is 2×36=72 (km), while the hourly journey of the truck is 44-36 = 8 (km) less than that of the bus, and the time for the bus to travel between the east and west cities is 72÷8=9 (hours), so the distance between the east and west cities is 44×9=396 (km). 396 ÷ (44+36) = 4.95 (hours)
A: The two cars met on the road 4.95 hours after their departure.
14: Both parties set off from Beijing on the same day and rode along the same road to Guangzhou. Party A drives 100 km every day, and Party B drives 70 km on the first day, and then drives 3 km more than the previous day until they catch up with Party A. How many days after Party B leaves?
Solution: Two people start in the same direction at the same time, but at first B is slower than A. When the speed of B increases to the same as A, the distance between them becomes wider and wider. When the speed of B exceeds that of A, the distance between them is getting closer and closer until B catches up with A..
At first, Otsuichi Skywalker was less than A 100-70 = 30 (km). Later, B walked 3 km more every day, and it took 30÷3= 10 (day) to reach the same speed as A, that is, on the first 10 day, the distance between A and B widened day by day, and the 60th.
A: B caught up with A on the 2nd1day after departure.
15: the distance between a and b is 10 km. Both the express train and the local train go from A to B. When the express train leaves, the local train has already traveled 1.5 km. When the express train arrives at B, the distance between the local train and B is still 1 km, so how many kilometers away from B does the express train catch up with the local train?
Solution: The local train traveled 1.5km before the express train left, but when the express train arrived at B, the local train still had 1km, that is, it was 1.5+ 1 = 2.5(km The speed of the express train 1km is 2.5÷ 10=0.25 (km) more than that of the local train.