6th grade math problem help Urgent

32, up to 64 + 160 + 96 = 320 students that each person will get the same number of prizes, each person will get the same number of prizes, each person will get the same number of prizes, each prize is 64 32 2 electric toys, 1603 2 5 bags of model aircraft information, 96 32 3 copies of popular science books.

2. The original one tree every 3 meters, now it should be changed to one tree every 4 meters, and the least common multiple of 3 and 4 is 12, that is, every 12 meters do not need to move, then there are (33-1) * 3 12 + 1 9 trees do not need to move. 24 newly dug tree pits 33-9.

3. The accounts are: 72 barrels, a total of ( ) yuan.

4. The least common multiple obtained is 180, and the first group of monkeys has at least 180 12 15 (only) It is known that the first group of monkeys has more than 40, then 180 needs to be expanded by 3 times to meet the requirements, and the peanuts have a total of 180 3 540 (grains).

Then there are 540 12 + 540 15 + 540 18 45 + 36 + 30 113 monkeys.

The greatest common factor of 160 96 (remember to use short division) is awarded to 32 students, each with a score of 2+5+3=10

2.(33-1)*3=96 (m)3*4=12 96 12+1=9 (trees)33-9=24 (pieces).

4.First find out the least common multiple of 12 15 18 is 180, 180 12 = 15 In the forty tens there are only 3 * 15 = 45, so there are 45 monkeys, 45 * 12 = 540 peanuts 540 18 = 30 540 15 = 36 45 + 30 + 36 = 113

Set A to be x years old and B y years old.

y-(x-y)=4

x+(x-y)=61

x=42 y=23

The youngest one is 23 years old.

The younger ones are x years old, and the older ones are y years old.

y-(x-4)=x

x+(61-y)=y

x=23y=42

The youngest one is 23 years old.

I'll just do it step by step.

1.3 days, 5 days, and 6 days look at Xiao Ming and Xiao Qiang like this, and what is sought is the least common multiple of the number of days on duty.

5*6=30 days can be on duty at the same time.

Therefore, there are 28 days in February of 06, and there are 13 days in February from the 16th, so the same day on duty again should be 30-13=17 days in March.

So the column formula is: 5*6-28+16-1=17

2.Let the x edges be long in meters, then the column formula (1 dm = 1 m).

x=100, so 100.

Length of a row = meters.

3.then the volume of each glass bead is v

4v=(280-200)

v = 20 (cubic centimeters) (1 cubic centimeter = 1 ml).

4.From the meaning of the title, since 200 meters and resting for 1 minute, the speed of A is (200 100) + 1 = 3 minutes, and the speed of 200 meters per minute is 200 3 meters.

B's velocity (200 80) + 1 = minute, and 200 meters per minute = 200 meters per minute.

The speed difference between A and B = 200 3-400 7 = 200 21

So let's say X minutes A catches up with B, and when A runs one more lap than B, it just catches up with B.

200x/21=800

x=84, so catch up with B after 84 minutes Because 84*80 200=not divisible, you can catch up in 84 minutes.

The least common multiple is 30

So 30 days after being on duty at the same time, it will be on duty again at the same time.

That is, March 18.

2) 1 meter = 10 decimeters.

10*10*10=1000 (blocks).

m)3)(280-200) 4=20 (cubic centimeters)4) The average speed of A is 200 3, and the average speed of B is 200 After setting xmin, A catches up with B.

200x/3-200x/

x=84（min）

84 3 = 28

84 or more are divisible, so at 84 minutes, A and B are resting, that is, A catches up with B at 83 minutes.

1.5*6=30 (least common multiple of 3,5,6) 15+30=45

45-28=17 (28 days in February 2006) Answer: March 17.

2.1 cubic meter 1000 cubic decimeters.

1000 1 1000 pcs.

arranged in a row, the base area is square meters).

Length=1 m).

1 ml = 1 cubic centimeter.

4.Speed and distance meeting time.

800 4 200 (m).

Speed difference, chase distance, catch up time.

800 20 40 (m).

Large number (and difference) 2

A: (200 40) 2 120 (m).

B: 200 120 80 (m).

800 (120-80) = 20 minutes.

A catches up with B after 20 minutes.

You don't even give a bounty.

And the process continues.

1.Xiao Ming: Go to work on February 15, go to work on February 18, go to work on February 21, go to work on February 24, go to work on February 27, go to work on March 2 (2006 is not a leap month, so February is the 28th), go to work on March 5, go to work on March 8, go to work on March 11, go to work on March 14, go to work on March 17, go to work on March 20, ......(and so on).

Xiao Gang: Go to work on February 15, go to work on February 20, go to work on February 25, go to work on March 2, go to work on March 7, go to work on March 12, go to work on March 17, ...... on March 17(and so on).

Xiaoqiang: Go to work on February 15, go to work on February 21, go to work on February 27, go to work on March 5, go to work on March 11, go to work on March 17, go to work on March 23, ...... on March 23(and so on).

We will find that they are at work at the same time on March 17, so they go to work at the same time on March 17.

1dm 1dm = 1 square dm, 10dm 10dm = 100 square dm, 100 1 = 100 (pcs).

100x1=100dm=1m

A: 100 of these square blocks are required; Lined up in a row and 1 meter long.

80 4 = 20ml = 20 cubic cm

A: The volume of each glass sphere is 20 cubic centimeters.

m) 800 20 40 (m).

A: (200 40) 2 120 (m).

B: 200 120 80 (m).

Speed and distance meeting time.

800 4 200 (m).

Speed difference, chase distance, catch up time.

800 20 40 (m).

Large number (and difference) 2

A: (200 40) 2 120 (m).

B: 200 120 80 (m) 800 (120-80) = 20 minutes.

Answer: A catches up with B after 20 minutes.

Question 1 The least common multiple is 30, which should be a month later, but February 2006 has 28 days, so the answer is March 17.

In the second question, the volume of a block is cubic meters, so 1 small cube is arranged in a row with a total length of 1000 meters.

The third question is that the volume of water rising is the total volume of the glass beads, then (280-200) 4=20 cubic centimeters.

The fourth question is known from the question: The average speed of A is 200 3, and the average speed of B is 200, and after setting xmin, A catches up with B.

200x/3-200x/

The solution is x=84

84 3 = 28 84

The above are all divisible, indicating that A and B are resting at 84 minutes, that is, A catches up with B at 83 minutes.

In the first question, the equivalent of finding their least common multiple is =3*5*2=30 days.

Because 2006 is not a leap year, there are 28 days in February, so the next shift is at 30-(28-15)=17, so on March 17.

The second question, because 1 meter = 10 decimeters.

Therefore, a total of 10 * 10 * 10 = 1000 small squares are arranged in a row, and the length is 1000 * 1 = 1000 decimeters = 100 meters The third question, (280-200) 4 = 20 ml = 20 cubic centimeters The fourth question, A runs 100 meters per minute and rests for 1 minute every 200 meters, so A needs 3 minutes to run every 200 meters.

B runs 80 meters per minute and rests for 1 minute every 200 meters, so B needs 200 80+1 = minutes for every 200 meters run.

Set xmin after A catches up with B.

200x/3-200x/

x=84（min）

84 3 = 28

84 or more is divisible, so A and B are resting at 84min, so A catches up with B after 84-1=83 minutes.

One percent, because of 3 5 boys, then the proportion of boys who are excellent in sports in the whole school is 5% 3 5 = 3%, the standard is excellent in sports, 3 4 is boys, and the outstanding students in the whole school have 3% (3 4) = 4%, subtract 3% of boys, and get 1%.

Capacity: 500 people.

Male: 500x3 5=300 (person).

5% of the boys are excellent in sports: 300x5% = 15 (people) 3 4 of the students who have "attained" the "standard" in the whole school.

Outstanding students in the whole school: 15 3 4 = 20 (people) Girls who met the standards "Students who achieved the standards"Students who achieved the outstanding results: 20-15 = 5 (people) The number of students who "met the standards" and won the outstanding students in the whole school accounted for the total number of students in the school

5÷500x100%=1%

Write only the largest one. The length of the rectangle is the diameter, and the width is the radius. The product of the diameter multiplied by the radius is 20, so the square of the radius is 10. The area of the semicircle is 10* square centimeters.

4 Scenarios!! Don't write dead.

Master and apprentice do 3 15 = 1 5 together for 3 days, then the apprentice completes 7 30-1 5 = 1 30 in 2 days, and 1 60 in 1 day

If this batch of parts is made by the apprentice alone, it will take 1 (1 60) = 60 days.

Hope it helps!

First the master did it for 3 days, then the apprentice did it for 5 days, which is equivalent to 3 days of master-apprentice cooperation, and then the apprentice did it for another 5-3 = 2 days.

Therefore, the work efficiency of the apprentice is:

7/30-1/15x3）÷2=1/60

It takes 1 1 60 = 60 days for the apprentice to complete alone.

Total workload = 30 units = > 2 units in 1 day = 3 days for > master and 6 units in 3 days.

However, 3 days for the teacher + 5 days for the apprentice, a total of 7 units = > 2 days for the apprentice to do 1 unit = > apprentice for 1 day to do the unit = > if the apprentice does it alone, it will take 30 days.