2 application questions will be fast

1.A-B-C-D.

A fetches water for one minute, B and C wait for one minute, for a total of 3 minutes. B hit the water for 2 minutes, C Ding waited for 2 minutes, a total of 4 minutes, C hit the water for 3 minutes, D waited for 3 minutes, and D hit the water for 4 minutes.

Shared fetching time: 1 + 2 + 3 + 4 = 10 minutes.

So: the sum of the time required to queue and fetch water is at least one minute.

The minimum sum of the time spent queuing and fetching water is 1+2=3 minutes.

The minimum sum of queuing and fetching time is 1+2+3+=6 minutes.

The minimum sum of the time spent queuing and fetching water is 1+2+3+4=10 minutes.

The shortest sum of the total queuing and water fetching time is 1+3+6+10=20 minutes.

2.It takes 8 minutes for two people to sweep the floor + 2 minutes for 1 person to wipe the blackboard + 6 minutes for 1 person to carry water = 8 minutes for 2 people to mop the floor + 10 minutes for 2 people to wipe tables and chairs = 10 minutes for 1 person to get to the garbage = 5 minutes.

So it is: 8 + 10 + 5 = 23 minutes.

Solution 2: 2 people to sweep the floor 8 minutes + 1 person to carry water 6 minutes + 1 person to garbage 5 minutes = 11 minutes.

10 minutes for 2 people to mop the floor + 10 minutes for 2 people to wipe tables and chairs = 10 minutes.

It takes 2 minutes for 1 person to wipe the blackboard.

So it's 11 + 10 + 2 = 23 minutes.

1 4 = 4 (minutes) 2 3 = 6 (minutes) 3 2 = 6 (minutes) 4 1 = 4 (minutes).

4 + 6 + 6 + 4 = 20 (minutes).

Answer: In the order of A, B, C, and D, the sum of the large water time is the smallest, and the shortest time is 20 minutes.

The second question is whether to take out the garbage must be done after sweeping the floor.

1) Order: A, B, C, D.

Time: 1 4 2 3 3 2 4 1 20 (min) 2) 1) Wipe the blackboard 2 + lift water 6 + sweep 8 + take out the garbage 5 (2) Team wipe tables and chairs 10 + mop 10

21 minutes in total.

Question 1: 20 minutes.

In the order of A, B, C, and D.

1 4 = 4 (minutes) 2 3 = 6 (minutes) 3 2 = 6 (minutes) 4 1 = 4 (minutes).

4 + 6 + 6 + 4 = 20 (minutes).

The second question is 18 minutes.

It takes 8 minutes for 2 people to sweep the floor, 10 minutes for 2 people to mop the floor, 2 minutes for 1 person to wipe the blackboard, 6 minutes for 1 person to carry water, 5 minutes for 1 person to reach the garbage, and 10 minutes for 2 people to wipe tables and chairs. Could you please arrange it and see how it can save time and be reasonable?

Number four people in this group: ABCD

a Sweeping (8) Mopping (10) 18 min.

b Same as a.

c-Erase the blackboard(2) Take out the trash(5) Wipe the table and chairs(10) 17 minutes.

d-Lift water(6) Wipe tables and chairs(10) 16 minutes.

In ABCD, 18 minutes for A is the longest, so the answer is 18 minutes.

Private car 192km = 80km h = m s Jeep 21m s Private car ( m s ) Jeep ( 21m s ) Private car Average speed is faster 2012-03-22 21:55:42 Added:

Back to 002 : sol 192 km h = (192 km hour) * (1 hour 3600 seconds) * (1000 m 1 km) = 192 * 1000 3600 (m s) "There seems to be a leak here to remove the slag seepage = m second >21

The average speed of a private car = 192 The average speed of a jeep = 21m s because of the 21m s = the average speed of a private car is faster.

001 (aka dydx ( Elementary 5 ) mins= seconds has been excepted. resol the car run=192 =192*1000 grip spine ( =

cor to 3 sig fig jeep ran 21m/s because jeep's speed < the car's speed so the car 's rate is faster

It takes hours to cover 192 kilometers by private car; Jeeps travel an average of 21 meters per second, which car has the fastest average speed? (Application Questions) sol 192 km/h = (192 km/h) * (1 hour 3600 seconds) * (1000 m 1 km) = 192 * 1000 3600 (m/s) = m/s >21 Private cars are faster.

1] In the autumn harvest season, the farm plans to recruit a group of people to harvest 12,000 tons of grain, the farm originally had 10 people, and plans to increase the number of people by 50 people, each person can harvest 10 tons of wheat per day.

n=10+50=60 people.

d = 12000 60 10 = 20 days.

2]] In the autumn harvest season, the farm plans to recruit a group of people to work, the farm originally had 10 people, and plans to increase the number of people to 50 people.

n = 50 people.

d = 12000 50 10 = 24 days.

Simple four operations, which should be understood in the 4th grade, I hope it will be useful or hidden for you.

1: This is a problem for finding the least common multiple.

If there are 3 fewer students participating in the competition, the least common multiple that can be divisible at the same time is 120

Therefore, at least 123 students participated in the rope skipping competition.

2:3 p.m.

One hour = 60 minutes, 60 is not divisible by 9, so 120 minutes have passed since 2 p.m. without lights, not divisible by 9, and 180 minutes after 3 p.m. without lights, divisible by 9, lights on!

1: First of all, if there are no extra people in each group of 6 people, 8 people in each group, and 10 people in each group, then the least common multiple of 6, 8, and 10 60 must be met, so the multiples of 60 are all eligible, that is, 60n (n is a positive integer), because according to each group of 6 people, each group of 8 people, and each group of 10 people, there are 3 people left, so there are 60n+3 students participating in the god jumping competition.

2: It is also a problem of finding the least common multiple. One is to ring every 60 minutes, the other is to measure every 9 minutes, then the least common multiple of 9 and 60 is 180, 180 9 = 20, that is, the 20th light after 12 o'clock noon, 180 60 = 3, that is, the third ring after 12 o'clock noon, so the next time both the bell and the light are on is 180 minutes after 12 o'clock at noon, that is, at 3 o'clock in the afternoon.

A common multiple of 6,8,10 + 3,123

After the least common multiple of 9 and 60, 15:00

Students participating in the God Jumping Competition are divided into groups of 6 people, 8 people in each group, and 10 people in each group, and there are 3 people left, and how many students participate in the God Jumping Competition.

Solution If the least common multiple is 120, then the students participating in the god jumping competition are: 120 + 3 = 123 people.

An electronic clock, every hour to the hour, every 9 minutes to go on the light, at 12 noon, it both bell and light, when is the next time both bell and light?

It is best to solve this kind of problem with a system of equations, and the solution process is as follows: set up equipment A x pieces, B equipment y pieces, then there are Piqing production a: 5x+6y=50 pieces, and Tsai Min produces Nian Zhi b:

10x+20y=140 pieces, two types of consortia, find x=4 y=5 rent z=200x+300y, substitute x, y into z=2300, problem solving....

When x minutes, the price is the same for both methods.

x=70 When he plays ** for less than 70 minutes per month, use the first way, and use the second way when he plays more than 70 minutes.

For exactly 70 minutes, you can do both.

If your phone bill is less than 42 yuan, you can do the first one, and if it is more than 42 yuan, you can do the second one.

If he has more than 70 minutes of talk time per month, then use option 2, and if it is less than 70 minutes, use option 1, formula: 21 (

Get x=70

When x<70 selects (1) there is no monthly fee, ** fee per minute when x=70 selects (1) or (2).

When x>70 chooses (2) the monthly rent is 21 yuan, ** fee per minute, which way does he take to save money?

2.The original area: 36 24 = 864 (square meters), the current area: 40 28 = 1120 (square meters), increase: 1120-864 = 256 (square meters).

3.10000 5 4 = 500 (m) 500 500 = square kilometers).

5.256 pages).

6.Yuan) choose mine.

, that is, 62 of them.

50 * 50 = 2500 square meters = square kilometers kilogram page.

1.4cm=, combining two together is a square, on the long side: 5).

On the wide side: 1)...Total:

12 2 2 = 48 (pcs) 2Original area: 36 24 = 864 (square meters), current area:

40 28 = 1120 (square meters), increase: 1120-864 = 256 (square meters).

3.10000 5 4 = 500 (m) 500 500 = square kilometers) 4 50×

5.256 pages).

6.Meta).

Solution: Set: There are x copies every day, and Lele has 3 4x copies.

3/4x-14 =5/16x

3/4x -5/16x=14

7/16x=14

x=323/4x=24

There are 32 books every day, and there are 24 books for Lele.

Your adoption is the driving force for me to move forward!

If you don't understand, you can ask until you understand the question!

Solution: Set: There are x copies every day, and Lele has 3 4x copies.

3/4x-14=5/16x

12/16x-5/16x=14

7/16x=14

x=14*16/7

x=32 Answer: There are 32 books every day, and there are 24 books for Lele.

Solution: If there are x copies every day, then Lele has 3 4x copies.

3 4x-14=5 16x, find x=32.

A: There are 32 books every day, and there are 24 books in Lele.