Think about solving problems:
According to the known conditions, a table is 288 yuan more than a chair, which is exactly (10- 1) times the price of a chair, so the price of a chair can be obtained. According to the price of chairs, we can get the price of a table.
Answer:
Solution: the price of the chair:
288( 10- 1)= 32 (yuan)
The price of a table:
32× 10=320 (yuan)
A table 320 yuan, a chair 32 yuan.
Three boxes of apples weigh 45 kilograms. A box of pears is 5 kilograms heavier than a box of apples. How much do three boxes of pears weigh?
Think about solving problems:
You can first find out that the weight of 3 boxes of pears is more than that of 3 boxes of apples, plus the weight of 3 boxes of apples, which is the weight of 3 boxes of pears.
Answer:
Solution: 45+5×3=45+ 15=60 (kg)
Three boxes of pears weigh 60 kilograms.
3. Party A and Party B walked across two places at the same time. Four hours later, they met four kilometers from the midpoint. A is faster than B. How many kilometers is A faster than B per hour?
Think about solving problems:
According to the meeting at a distance of 4 kilometers from the midpoint, and the speed of A is faster than that of B, it is known that A walks 4×2 kilometers more than B, and it takes 4 hours to meet. You can work out how many kilometers A is faster than B per hour.
Answer:
Solution: 4×2÷4=8÷4=2 (km)
A: A is 2 kilometers faster than B per hour. 20 17 xiaoshengchu mathematics classic application questions and answers 20 17 xiaoshengchu mathematics classic application questions and answers.
Li Junhe Zhang Qiang spent the same money to buy the same pencil. Li Jun asked for 13 pencils, Zhang Qiang asked for 7 pencils, and Li Jun gave Zhang Qiang 0.6 yuan money. How much is each pencil?
Think about solving problems:
According to the fact that two people spent the same money to buy the same kind of pencil, Li Jun asked for 13 and Zhang Qiang asked for 7, indicating that everyone should get (13+7)÷2, while Li Jun asked for 13, which was 3 more than he deserved, so he gave Zhang Qiang and 0.6 yuan money.
Answer:
Solution: 0.6 ÷ [13-(13+7) ÷ 2] = 0.6 ÷ [13-20 ÷ 2] = 0.6 ÷ 3 = 0.2 (.
A: Every pencil is 0.2 yuan.
At 8 o'clock in the morning, two buses, A and B, set off from two stations at the same time and walked in opposite directions. After a while, two buses reached both sides of a river at the same time. Because the bridge over the river is being repaired, vehicles are forbidden to pass. The two cars need to exchange passengers and then return to their respective departure stations by the same route. When they arrived at the station, it was already 2 pm. Car A travels 40 kilometers per hour and car B travels 45 kilometers per hour. How many kilometers are there between these two places? (The exchange time is omitted)
Think about solving problems:
According to the fact that two cars leave two stations at 8: 00 am and return to the original station at 2: 00 pm, the travel time of the two cars can be calculated. According to the speed and driving time of the two cars, the total distance traveled by the two cars can be calculated.
Answer:
Solution: 2: 00 pm is 14: 00 20 17 xiaoshengchu mathematics classic application question and answer article 20 17 xiaoshengchu mathematics classic application question and answer from/article/wk-7850000136016. .
Round trip time: 14-8=6 (hours)
Distance between the two places: (40+45)×6÷2=85×6÷2=255 (km)
Attendant: The distance between the two places is 255 kilometers.
6. The school organized two extracurricular interest groups to go to the suburbs. The first group walked 4.5 kilometers per hour, and the second group walked 3.5 kilometers per hour. 1 hour later, the first group stopped to visit an orchard, which took 1 hour, and then chased the second group. How long will it take to catch up with the second group?
Think about solving problems:
When the first group stopped to visit the orchard, the second group traveled [3.5-(4.5-3.5)] kilometers, which is the distance that the first group must catch up with. It is also known that the first group is faster (4.5-3.5) kilometers per hour than the second group, from which the catch-up time can be calculated.
Answer:
Solution: The distance between the first group and the second group:
3.5-(4.5-3.5)=3.5- 1=2.5 (km)
Time taken for the first group to catch up with the second group:
2.5÷(4.5-3.5)=2.5÷ 1=2.5 (hours)
A: The first group can catch up with the second group in 2.5 hours.
7. There are two warehouses, A and B, each of which stores 32.5 tons of grain on average. The tonnage of grain stored in warehouse A is 5 tons less than that in warehouse B. How many tons of grain are stored in warehouse A and warehouse B respectively?
Think about solving problems:
According to the fact that the tonnage of grain stored in warehouse A is 5 tons less than that in warehouse B, we can know that if the tonnage of grain stored in warehouse A increases by 5 tons, the tonnage of grain stored in warehouse B is 4 times that of warehouse B, and the total grain storage will also increase by 5 tons. If the tonnage of stored grain in warehouse B is regarded as 1 times, the total tonnage of stored grain is (4+ 1) times, from which the tonnage of stored grain in warehouse A and warehouse B can be calculated.
Answer:
Solution: store the grain in warehouse b:
(32.5× 2+5) ÷ (4+1) = (65+5) ÷ 5 = 70 ÷ 5 =14 (ton)
The warehouse stores grain;
14×4-5=56-5=5 1 (ton)
Warehouse A stores 5 1 ton of grain, and warehouse B stores 14 ton of grain.
8. Team A and Team B built a 400-meter-long road together. Team A completed four days from east to west and team B completed five days from west to east. Team A takes more lessons than team B every day10m, 20 17 junior high school math classic application problems and answers. How many meters do Team A and Team B repair every day?
Think about solving problems:
According to the fact that Team A repairs10m more than Team B every day, it can be considered that if Team A repairs for 4 days and Team B repairs for 4 days, the total length will be reduced by 410m, which is equivalent to Team B (4+5).
From this, we can get the number of meters repaired by team B every day, and then get the number of meters repaired by the two teams every day.
Answer:
Solution: b the number of meters repaired every day:
(400-10× 4) ÷ (4+5) = (400-40) ÷ 9 = 360 ÷ 9 = 40 (meters)
Number of meters repaired by Team A and Team B every day:
40×2+ 10=80+ 10=90 (m)
A: Two teams repair 90 meters every day.