Grade 6 Olympiad Problems Maximum and Minimum Questions 7 .

1.A number is rounded to the percentile, and the result is, what is the maximum number of five decimal places? What is the minimum?

Maximum: Minimum:

2.If the sum of three numbers is 144, what is the maximum product of the three numbers that meet such a rule?

3.The sum of the length, width, and height of a cuboid is equal to 36 decimeters, when the length, width, and height of this cuboid are the largest?

Length = 12 dm Width = 12 dm Height = 12 dm.

4.There is a three-digit number that can form six different three-digit numbers, and the sum of them is 2886, what is the smallest of them?

The smallest of them is: 139

5.There are 34 tons of goods, transported from city A to city B, the load capacity of large trucks is 5 tons, the load capacity of small trucks is 3 tons, the fuel consumption of the two trucks is 10 liters and liters respectively, how many large trucks and small trucks are used to transport the least fuel consumption?

7 big trucks!

1. Maximum: , Minimum:

2. Knowing the sum of three numbers, finding the maximum value of the product of three numbers can be regarded as the sum of the three sides of the known hexahedron, and finding the maximum area of the regular hexahedron, that is, the maximum area of the cube is 144 3=48,48=110592 (I don't know what you mean by asking to list **).

3. Same as above, the length, width and height are all 12

4. Let the three numbers be a, b, c, then the six three-digit numbers of 6 are: abc, acb, bac, bca, cab, cba, it is not difficult to see that a, b, c three numbers are in the single digit, ten, hundred each appear 2 times, then the sum of the six three-digit numbers is (a+b+c) 200+(a+b+c) 20+(a+b+c) 2=2886, easy to get a+b+c=13, then the smallest three-digit number that meets this condition is 139

5. Set up a large car and B small car. 5a+3b=34, easy to obtain, only two solutions, a=5, b=3; a=2,b=8.Fuel consumption:

10a+, the two sets of solutions into the fuel consumption formula to obtain the minimum fuel consumption solution is: a=5, b=3(If you don't take into account the implicit conditions of exactly the end of the shipment, 7 large trucks are the most fuel-efficient).

Fourth, I don't understand. The title of the fourth question is not clear! Whether the fuel consumption is only transported once and back or once should be clearly stated.

There is a formula for the pursuit problem.

Pursuit distance Speed difference = pursuit time.

(Unit: km).

2.Set to catch up with time t, got.

20t=15t+10

t=2, then the distance already traveled when catching up = 20 2=40, so the distance from place b = 60-40 = 20

1. Solution: Let the first 2 minutes be (x-2) If Xiao Ming's speed is known to be 15 kilometers per hour, that is, 250 meters per minute, then Xiaohua's speed is 250 meters 3=750 meters.

The column equation is 250 x + 20000 = 750 x and the solution is x = 40

x-2)=38

Therefore, Xiao Hua is 2 minutes before Xiao Ming, and the distance between the two is 250 38 + 20000-750 38 = 1000 (meters).

2. Solution: If the time of Xiaohu is x hours when catching up, then 10+15x=20x solution gets x=2

Therefore, when Xiaohu catches up with Xiaoqiang, the distance from B is 60-20 2=20 (km).

5 times we only consider the worst-case scenario:

1. Divide into 3 groups first, use the balance to weigh the two groups of 33 each, if balanced, then in the third group, if unbalanced, then in the heavier group.

2. The worst-case scenario is that the heavy ball is in the third group. It is divided into , 3, measured by the same method for the second time, and then divided the group containing the heavy ball into , measured again by the same method, and then divided into

5. The worst case is that the heavy ball is in the third group of 2, and then it will be weighed once.

So weigh at least 5 times, and you will be able to find out the overweight ball, and the key to the problem is to divide it into three groups, and weigh two groups with the same number of balls, which is the best way to weigh it.

First time, first layer: efgh; 2nd layer: i j k l

abcd mnop

The second time, the first layer is the left half of the second layer of the previous one: j i, the second layer: fe, the third layer: gh, and the fourth layer: kl

nm ba cd op

The third time, the first layer is the previous fourth layer lower part OP, the second layer CD, the third layer BA, and the fourth layer NM

5th layer JI, 6th layer FE, 7th layer GH, 8th layer KL

Fourthly, the first layer is the right half of the eighth layer L, the second layer H, the third layer E, the fourth layer I, the fifth layer M, the sixth layer A, the seventh layer D, the eighth layer P, the ninth layer O, the tenth layer C, the eleventh layer B, the twelfth layer N, the thirteenth layer J, the fourteenth layer F, the fifteenth layer G, and the sixteenth layer K.

Actually, it's a matter of order, I hope it helps you

You should fold it yourself and try it! I've folded it over and it's c!

There are 9 + 2 90 + 3 900 = 2889 numbers from 1 to 999. Remove the number of numbers that are not 0 and get it.

Then make up these numbers into three digits, that is, 000 to 999, after the number is not 0, it will not increase, but the number of 0 changes, we calculate the number that is not 0, and then subtract from 2889, there is only the number of 0.

This gives 3 1000 = 3000 numbers, where each number appears 3000 10 = 300 times.

Then the number that is not 0 is 300 9 = 2700, and the number of 0 is 2889-2700 = 189.

Add the 3 zeros in 1000, so from 1 to 1000, the number 0 will appear 189 + 3 = 192.

Two digits: 9

Three-digit number: (10+9)*9=171 four-digit number: 1

10 to 90 appears 9 times, 110 to 190 9 times, and 210 to 290 9 times. 101..109 appears 9 times, 201....209 appears 9 times, and the analogy is 9*9=81 times.

Appears 18 times. 1000 appears 3 times, in total.

9*10+18+3+81=192 times.

1-100 appears 11 times.

101-200 appears 20 times.

901-1000。。。21 times.

A total of 192 times.

Question 1: There is a pool, the first time to release 2 5 of the total water, the second time to release 40 cubic meters, the third time to release the remaining 2 5, there is 66 cubic meters of water left in the pool, how many cubic meters of water can be stored in the whole pool?

Solution: If the whole pool can store x cubic meters, then:

x（1-2/5)-40]*(1-2/5)=66[3x/5-40]*3/5=66

9x/5-120=330

x=1050/9=

That is, the whole pool can store cubic meters of water.

Question 2: The kindergarten will distribute a basket of apples to the children, if the children in the kindergarten are divided into 5, there will be 6 fewer for each child; If it is assigned to the children in the nursery class, each person is divided into 4 more children, and it is known that the large class has 2 fewer children than the small class, so how many apples are in this basket?

Solution: If there are x people in the small class, then there are (4x+4) apples in this basket, then:

5（x-2) -6=4x+4

5x-16=4x+4

x=20, that is, there are (4x+4)=4*20+4=84 apples in this basket.

A total of 62 people from Class A and B participated in the science and technology group activities. The 1 5 number of participants in Class A was 2 less than the 1 4 participants in Class B. How many people from Class A and Class B participate in the Science and Technology Group activities?

Solution: If there are x people in class A and (62-x) people in class B, then:

x/5=(62-x)/4-2

4x-5(62-x)=-40

4x-310+5x=-40

x=30 means that there are 30 people in class A and (62-x)=62-30=32 people in class B.

1.How many cubic meters are left for the second time: 66 (1-2 5) = 110 cubic metersHow many cubic meters are left for the first time: 110 + 40 = 150 cubic meters, all 150 (1-2 5) = 250 cubic meters.

2。Because the large class has 2 fewer people than the small class, after the small class is distributed according to the same number of apples as the large class, there are 4 2 4 left, according to the profit and loss problem algorithm of one profit and one loss, (profit + loss) (large number - decimal) = number of people, large class (4 + 6 + 4 2) (5-4) = 18, 18 apples 5-6 = 84.

3。From the fact that 1 5 of the number of participants in class A is 2 less than 1 4 of the number of participants in class B, it can be seen that 1 5 plus 2 people of A is the same as 1 4 of the number of people in class B. And after 62 + 5 2s, there are 5 + 4 such 1 copies, so:

B's 1 4 = (62 + 5 2) (5 + 4) = 8 people, B = 8 1 4 = 32 people.

A = 62-32 = 30 people.

You don't need an equation.

Solution: [66 (1-2 5)+40] (1-2 5)=150 (3 5)=250

Solution: If the number of people in the large class is the same as that of the small class, that is, if no one is sent to the small class, there will be 18 less, and if no one is sent to the small class and no one is 4, there will be 4 more, so the number of people in the small class is: (18 + 4) (6-4) = 11 So the apple has: 4 * 11 + 4 = 48

Solution: Number of people in class B: =72 (9 4)=32

Design: The whole pool can store x cubic meters of water.

Set up: Kindergarten class A, small class B.

5a-6=4b+4

a+2=ba=18 b=20

Count the rest.

Set up: Person A A, Person B B.

a+b=62

1. Solution: The whole pool can store x cubic meters of water.

1-2/5) x - 40 ] 1-2/5 ) 66

3/5 x-40) *3/5 ＝ 66

9/25 x＝90

x 2502, solution: set up a large class of x people, then the small class is x + 2 people.

5x+6=（x+2）*4-4

x=18 apples have a total of 18*5-6=84 (pcs).

3. Solution: If there are x people in class A, there are 62-x people in class B.

62-x）* 1/4 — 1/5 x=29/20 x=

x 30 then class B has 62-30 = 32 (people).