Primary school sixth grade super S level Olympiad math questions, a primary school six level Olympia

In fact, such a problem, the solution of the column equation is wise, and only one element of the equation is required, and elementary school students can understand it

Solution: Set the scheduled time to x days.

Then the power of A is 1 x and the power of B is 1 (x+8), according to "A and B cooperate for 6 days, and B can do it alone for a few more days, or it can be completed at the scheduled time", the following relationship is listed.

6 (1 x)+x [x (x+8)]=1 may seem complicated, but it's actually easy.

6x+48+x 2=x 2+8x Don't be scared away by x 2, you can actually make an appointment.

2x=48x=24……That's the scheduled time.

In the end, subtract 6 days from the scheduled time is the number of days spent by B-du: 24-6=18 (days) to type more words, in fact, the formula and calculation can be done quickly.

Let A be completed in x days and B in y days, so.

A completes 1 x per day, B completes 1 y per day, and the scheduled time is x days B does it alone, and it takes 8 more days to complete) B completes (x 8) days, so: (x 8) 1 y 1

Obtained by : y x 7

If A and B cooperate for 6 days, B will do it alone in z days):

1／x＋1／y）×6＋1／y×z＝1

Messed up??

Solution: Set the scheduled time to x days.

Then the power of A is 1 x and the power of B is 1 (x+8), according to "A and B cooperate for 6 days, and B can do it alone for a few more days, or it can be completed at the scheduled time", the following relationship is listed.

6×(1/x)+x×[x/(x+8)]=16x+48+x^2=x^2+8x

2x=48x=24……Scheduled time.

Finally, subtract 6 days from the scheduled time to get the number of days spent on Edo: 24-6=18 (days).

The above equation is not used!!

For two days, I didn't think it was right.

The amount of work required for B to complete half of the work 1 2 1 2 2 4 5 1 10 The time required for B to complete half of the time 1 2 1 10 4 5 5 8 After A is completed, B has 1 2 4 5 (1 50 ) 1 5 8) 1 20

60 1 10 1 20 30.

There are still 30 parts in workshop B that are not finished.

1/2 5 = 1/10.

1/2 - 4 1/10 = 1/10.

60 tenths = 600 (pcs).

1/10 4=1/40.

1/40 5=1/8.

1 - 1/2 - 1/8 = 3/8.

Three-eighths 5 = three-tenths.

3 out of 40 4 = 3 out of 10.

1 - 1/2 - 3 out of 10 = 1 in 5.

600 5th = 120 (pcs).

A: There are still 120 parts in workshop B that have not been completed.

Upstairs is not well thought out.

The total is 1, half of the time A is 1 10, the remaining half is 1 15, and B does 2 5 in 1 10, and then 1 2 is 1 40, which means that B is more efficient in the remaining 1 24 time

1 4 was done, and B made 1 4 + 1 2, that is, B did 3 4There are 1 4 left to do.

B is half when A is 2 5, so.

So B still has 150 left undone.

There are still 500 parts left unfinished in workshop B.

A and B two cars at the same time from AB two places, in the opposite direction, 6 hours later in the East Lake Stadium met, if car A late departure hours, car B per hour than the original 3 and 1 2 kilometers less, then the two cars met in the gymnasium, Nuo B car 1 2 hours in advance, car A per hour than the original more kilometers than the original two cars still meet in the gymnasium, ask AB how many meters apart between the two places.

2 (1 12 + 1 15 + 1 20) = 2 (12 60) = 10 (hours) This is the time to unload two shipments

1 1 12 10 1 6 This is the quantity of the goods unloaded by Gang A

1 6) (1 20) 10 3 (hours) The time of the third gang

Assuming that the cargo quantity of both ships A and B is 1, then C helps A and B to complete the task of 2 at the same time.

A's work efficiency = 1 12 = 1 12

B's work efficiency = 1 15 = 1 15

C's work efficiency = 1 20 = 1 20

The work efficiency of A, B and C cooperation = 1 12 + 1 15 + 1 20 = 1 5 The time required for the simultaneous unloading of two ships A and B = 2 (1 5) = 10 hours 10 hours, the workload completed by A = 10 1 12 = 5 6 The workload that C helps A complete = 1-5 6 = 1 6

C helps A unload the goods = 1 6 (1 20) = 20 3 hours = 3 and 1/3 of an hour.

y/12+x/20=1 5y+3x=60y/15+(y-x)/20=1 4y+3(y-x)=60 4y+3y-3x=60 7y-3x=60

5y+7y=120 12y=120 y=1050+3x=60 3x=10

x=10/3

Gang A unloads the goods for 3 hours and 20 minutes.

Set C to help A for time A and help B for time B

1 12 + 1 20)*a+1 12*b=11 15*a+(1 15+1 20)*b=1 Solve the equation.

Set the unloading time for A and the unloading time for B B

1/12（a+b)+1/20a=1

1/15(a+b)+1/20b=1

a=10/3 b=20/3

The three of them have a total of 6 exercise books, 6 texts, 6 large character books, a total of ** = yuan if they are all 1 ** = yuan, all of them are 2 of ** = yuan, all of them are 3 of ** = yuan yuan = 1 text + 2 large character books.

Yuan = 2 exercise books + 1 large character book.

Meta = 1 exercise book + 2 text.

The price of one exercise book, one writing text, and one large print book is (two exercise books and one writing text:

Two texts and one large print:

A workbook and two large print books:

2- Get four exercise books minus one large print book:

*2 Get nine exercise books:

So a workbook:

A book as a text:

A large print book:

Solution: Set up each exercise book x yuan, each text y yuan, and each large character book z yuan. From the title, 3*x+2*y+z= (1).

x+3*y+2*z= (2)

2*x+y+3*z= (3)

Solution, get. x=,y=1,z=

A: ——

Set up an exercise book x yuan, a text y yuan, and a large character book z yuan, which is derived from the title.

3x+2y+z= 1

x+3y+2z= 2

2x+y+3z= 3

1*2, got.

6x+4y+2z= 4

4-2, got.

5x+y=4 51*3,get.

9x+6y+3z= 6

6-3, got.

7x+5y= 7

5*5, got.

25x+5y=20 8

8-7, got.

18x=x=so y=1

Z=A: The unit price of the workbook is RMB, the unit price of the text is 1 RMB, and the unit price of the large-character book is RMB.

I don't know what you think.

Solution: Set up an exercise book, make a text, a large character book, each x, y, z yuan.

3x+2y+z=

x+3y+2z=

2x+y+3z=

The ages of these students are , because 11880=2 2 2 3 3 3 5 11).

Question 1: 108 360 40 = 12 cm.

Question 2: Circumference of a circle = , centimeters.

Question 3: I think it should be a semicircular arch. 40 2 = Last question: Suppose the rabbit can catch up with the turtle after running for x minutes.

35x=1000+10x

25x=1000

x=40

11880=2 2 2 3 3 3 5 11 2 5=10 3 3=9 2 2 3=12 11 Left So their ages are years A: The four students are years old.

40 108 360 = 12cm A: Then the length of the remaining part of the wire is 12 cm.

is the circumference of the circle, and the radius is equal to the circumference of the circle 2, A: The radius of the circle in which the arc is located is 27 cm.

c 2 = 1 2 d 1 2 Answer: The arc length of the arch is centimeters.

1000 (35-10) = 40 (minutes) Answer: A rabbit can catch up with a turtle by running for 40 minutes.

This doesn't seem to be an Olympiad problem... Have you ever seen what the real sixth grade Olympiad looks like?

Hello! 11880 =2×2×2×3×3×3×5×11 = 9×10×11×12

So the four students are all years old.

(35-10) 40 minutes.

Decompose 11880 into prime factors to get 11880=2*2*2*3*3*3*5, and these four students are 9, 10, 11, and 12

1 First calculate how many percent of 108 degrees is 360 degrees.

That is, 108 360 = 30 and then (1-30) * 40 can be 2 120 ° 360 ° = 1 3

Then 2 is sufficient.

The span of the 3 arches is actually the diameter of the circle where the bridge is made.

That is, the 40* turtle road.

Speed difference = 35-10 = 25

Let's come up with a more difficult to think about: as we all know, straight lines and circles are composed of countless points, may I ask, who has more points in straight lines and circles? (There are answers, no kidding).

A and B and repair completed 1/3 of the whole project in 5 days, and the sum of the efficiency of A and B is 1 3 5 = 1 15

B C and repair 2 days to complete the remaining 1 4, this 1 4 is the whole project: 2 3 * 1 4 = 1 6So the sum of the efficiency of Ethyl-Propylene is 1 6 2 = 1 12

A and C have and repaired for 5 days, and these 5 days are completed: 1-1 3-1 6 = 1 2, so the efficiency of A and C is 1 2 5 = 1 10

The efficiency of each person can be found: e.g. the efficiency of A = (1 15 + 1 12 + 1 10) 2-1 12 = 1 24

B is: 1 40

C is: 7 120

Knowing efficiency, we know how much each person is doing.

B worked 5 + 2 = 7 days, so B completed 7 40 of the whole project.

So B's remuneration is 600x7 40=105 yuan.

And then it took 5 days for A-C and repair to start?