Solution: Answer with profit and loss problem's thought.
The quotient is (96-80) (120-19) =16, so the dividend is 120 16+80 = 2000.
2. There are four different natural numbers, in which the sum of any two numbers is a multiple of 2 and the sum of any three numbers is a multiple of 3. Find the smallest four natural numbers that satisfy the conditions.
Solution: The sum of any two numbers is a multiple of 2, which means that these numbers are all even or all odd.
The sum of any three numbers is a multiple of 3, which means that the remainder of these numbers divided by 3 is the same.
The minimum natural number that satisfies the condition, because 0 is a natural number. So I think the result is 0,6, 12, 18.
3. On the circular runway, point A starts from point A and point B reverses at the same time. Six minutes later, they met, another four minutes later, A arrived at point B, and another eight minutes later, they met again. How many minutes does it take for A and B to make a round trip?
Solution: It takes 8+4 = 12 minutes for Party A and Party B to make a circle. It's six minutes' journey for B and only four minutes for A. ..
Therefore, it takes 12 minutes for line B and 1264 = 8 minutes for line A, so it takes 8+ 12 = 20 minutes for line A. It takes 2046 = 30 minutes for line B to make a turn.
Party A and Party B walk in opposite directions on the same highway. The speed of Party A is 1.5 times that of Party B. It is known that Party A passes through the post office at 8 am and Party B passes through the post office at 10 in the morning. When did Party A and Party B meet halfway?
Solution: We regard Bank B's 1 hour trip as 1 copy.
Then at 8 am, the distance between Party A and Party B is 10-8 = 2.
So when we met, Party B made 2 (1+ 1.5) = 0.8 copies, and 0.860 = 48 minutes.
So we met at 8: 48.
5. Both Party A and Party B start to climb the mountain from the foot of the mountain at the same time, and go down immediately after reaching the top of the mountain. They go downhill twice as fast as they go uphill. When Party A reached the top of the mountain, Party B was still 400 meters away from the top of the mountain, and when Party A returned to the bottom of the mountain, Party B was just halfway up the mountain. Find the distance from the top of the mountain to the bottom of the mountain.
Solution: Assuming that both parties can continue to improve, the speed ratio of both parties is (1+12): (1+1/22) = 6: 5.
So when A travels to the top of the mountain, B travels 5/6, so the distance from the top of the mountain to the bottom of the mountain is 400 (1-5/6) = 2400 meters.
6. A bus carries some passengers from the starting point. The number of passengers getting off at the first stop is 1/7 (including a driver and two ticket sellers), the number of passengers getting off at the second stop is 1/6, and the number of passengers getting off at the sixth stop is 1/2.
Solution: Finally, there are 1+ 1+2 = 4 people. So the total number of people in the car is
4 ( 1- 1/2) ( 1- 1/3) ( 1- 1/6) ( 1- 1/7) = 28.
Then, there are 28-3 = 25 passengers in the starting bus.
7. There are three grasslands with an area of 4 mu, 8 mu and 10 mu respectively. The grass on the grassland is as thick and grows as fast. The first grassland can feed 24 cows for 6 weeks, and the second grassland can feed 36 cows 12 weeks. How many weeks can the third grassland feed 50 cows?
Scheme 1: Let each cow eat 1 serving of grass every week.
The first piece of grassland is 4 mu, which can feed 24 cows for 6 weeks.
It shows that 244 = 6 cows per mu can eat for 6 weeks.
The second grassland covers an area of 8 mu and can be eaten by 36 cows 12 weeks.
It shows that each acre of grassland can feed 368 = 9/2 cows 12 weeks.
Therefore, every acre of grassland should be (9/212-66) (12-6) = 3 copies per week.
So 66-63 = 18 original grass per mu.
Therefore, the grass of the third grassland is 18 10 = 180, and the perimeter is 3 10 = 30.
So the third grassland can feed 50 cows 180 (50-30) = 9 weeks.
Scheme 2: Let each cow eat 1 serving of grass every week. Let's change the subject.
There is a piece of grassland 1 mu, which can feed 244 = 6 cows for 6 weeks and 368 = 9/2 cows 12 weeks. How many weeks can it feed 50 10 = 5 cows?
Therefore, the weekly grass length (9/212-66) (12-6) = 3 copies.
Original grass (6-3) 6 = 18,
Then 5 cows eat 18 (5-3) = 9 weeks is enough.
8.B is between A and C, A is from B to A, after departure 1 hour, B to C, after departure 1 hour. C suddenly remembered to inform A and B of something important, so he chased A and B by bike from B. It was known that A and B had the same speed, and C was three times as fast as A and B.
My thoughts are as follows:
If you chase B first and then return, the time is 1 (3- 1) 2 = 1 hour.
Return after chasing armor again for 3 (3- 1) 2 = 3 hours.
* * * It takes 3+ 1 = 4 hours.
If you chase the armor first and then return, the time is 2 (3- 1) 2 = 2 hours.
Chase B, chase 3 (3- 1) 2 = 3 hours and then return.
* * * It takes 2+3 = 5 hours.
So it takes the least time to chase B first. So, we started after we caught up.
9. A knife, 3 yuan. If Xiaoming buys this knife, the ratio of money between Xiaoming and Xiao Qiang is 2: 5; If Xiao Qiang bought this knife, the ratio of their money was 8: 13. How much money does Xiaoming have?
Solution 1:
Xiao Ming buys it, and the money left by Xiao Ming is 2 (2+5) = 2/7 of the money left by two people.
If Xiao Qiang bought it, Xiao Ming's money would be 8 (8+ 13) = 8/2 1.
So Xiaoming's remaining money accounts for 2/78/2 1 = 3/4 of his original money.
So Xiaoming's original money was 3 (1-3/4) = 12 yuan.
Solution 2:
If Xiao Ming buys it,
The remaining (8+ 13) (2+5) 2 = 6 copies,
Used 8-6 = 2 copies.
So Xiao Ming has 328 = 12 yuan.
10. The circumference of the circular runway is 500m. Both a and b start clockwise from the starting point. A runs every minute 120m, and B runs every minute 100m. Both stop every 200m 1min, so how many minutes does it take for A to catch up with B for the first time?
Solution: I have two understandings of this topic.
1. Party A and Party B live in the same place for the first time after their departure.
Then the first line will have a departure time of 200 120+ 1 > 2 minutes after 200 meters.
At this time, B needs 2 minutes, and it also needs 1002 = 200 meters.
In other words, B walked for 2 minutes, and then stopped at 200 meters with A who was resting.
Second, A ran 500 meters more than B and caught up.
Because after the trip, A walked 500 meters more than B,
Then it means taking 500200 = 2 100 more, that is, twice.
That is, the distance between a and b is 500+ 1002 = 700 meters.
To chase 700 meters, A needs to walk 700 (120- 100) = 35 points.
Line A needs to rest for 35 minutes, 35 120200- 1 = 20 minutes.
So * * * needs 35+20 = 55 points.