Sixth grade Olympiad math application questions, please write down the arithmetic

1. There are a total of x people participating in this competition. Then the number of people who pass the second game is 6x 7, and the number of people who fail is x 7.

The number of people who pass the first game is 6x 7-2 and the number of people who fail is x 7+2. The equation (6x 7-2)-4(x 7+2)=2 is obtained.

Solve the equation: 6x 7-2-4x 7-8=2

2x/7=12

x=422, let the cost of this product be x yuan. The original profit was 60 (100-x); Now the price is reduced by 100x4% = 4 yuan, then 3x4 = 12 pieces are sold. The profit is (60+12)(100-4-x), and the equation 60 (100-x) = (60+12) (100-4-x) is obtained

Solve the equation: 60 (100-x) = 72 (96-x).

500-5x=576-6x

x=763, let the 2-cent coin be x, the 1-cent coin be y, and the 5-cent coin be z, and the equation x+y+z=26 is obtained

2 cents are exchanged for 2x 5 5 cents, and 1 cent is exchanged for y 5 5 cents, giving the equation 2x 5 + y 5 + z = 11

2x+y+5z=55 x+4z=29 Since 1 cent and 2 cents are all exchanged for 5 cents, they should both be multiples of 5, and neither is zero. Under these conditions, there is only one answer: 1 point 15 pieces, 2 points 5 pieces, 5 points 6 pieces.

4 meters = 64 decimeters. The length, width and height of the iron box are 5, 5, 6 decimeters, respectively. The area is 2x (5x5 + 5x6 + 5x6) = 170 square decimeters.

1. If the number of failures is x, then (4x+2)+x=7(x-2) solution x=8, a total of 7*(8-2)=42 people participated in this competition.

2. Set the cost to x, then after the price reduction, the selling price is 100 * (1-4%) = 96 yuan, and the order is 60 + 100 * 4% * 3 = 72 pieces.

60 (100-x) = 72 (96-x), and the solution is x = 76

3. Let 1 cent x piece, 2 cent y piece, 5 cent z piece, then x+y+z=26, x+2y+5z=55, z<11, the solution gets x=15, y=5, z=6, so, the original 5 cent coin 6.

4. According to the principle of the maximum volume of the cube of the same area, the iron bar needs to be cut as equal as possible, and the side length of the cuboid is ,,, the required iron sheet area is: square meters.

If it can be truncated into 12 equal parts, then the cube side length is.

His surface area is: (sq.m.)

So, at least square meters of iron sheet are needed.

1.Let him do it for x days.

24x+24=40x-40

16x=64

x=4(24*4+24) 4=30 (pcs).

2.The cost of the first product is $x.

1+20%)x*90%+(1+15%)(2200-x)*90%=2200+131

x=1200

3.The armor was done for x days first.

x/16+(14-x)/12=1

3x+56-4x=48

x=84.Suppose the speed at which students walk is x kilometers per hour x = 48 * 12 60

The least common multiple of x = and 12 is 60, 60 10 = 6 times 15 and 12 The least common multiple is 60, .,60 15 = 4 times 6-4 = 2 times.

2*10=20 cm.

1. Solution: If the original rice is x kg, then the existing rice is (1 + 20%) x kg 1 3x + 260 = (1 + 20%) x

The solution is x=300

Then the existing rice is: 300 (1+20%)=360 kgA: 360 kg of existing rice.

2. Solution: There are 2 1s in 1-10

There are 10 1's in 11-20

There are 8 1s in 21-91

There is 1 1 in 100

So in total: 21 1s.

A: A total of 21 1's were written.

3. Solution: If class A has x books, then class B has (296-x) books.

5 13) x + (1 4) (296-x) = 95 to get x = 156

A: There are 156 books in Class A.

Solution: 1:260 (2 3 1 5) 300 kg 2: 10 10 1 21.

3: 95 4 296 84 cops.

84 (5 13 4 1) 156 .........The number of books in Class A is 296,156,140 .........Number of books in Class B.

1. Set the original rice a kilogram, there is a title:

1-2 3)*a+260=(1+1 5)*aA solution: a=300 kg.

Piece. 3. If class A has book A, then class B has book 296-a, book 5 13*a+1 4(296-a)=95

Solution: a=156

That is, there are 156 books in class A and 140 books in class B.

Solution: 5 (1) class boys: 30 * (1-20%) = 24 (people) x people in each class.

x-24) 7=(x-30) 5 -- the ratio of the number of boys in class 52 to the number of girls in class 15 is 5:7).

5(x-24) =7(x-30)

5x-120=7x-210

x=452x=2*45=90 (person).

This kind of Olympiad problem should be combined with arithmetic and equations, so that it is easier to understand!

Hope it helps!

There are more girls on May 2 than on May 1: 30 20% = 6 people.

Both classes have the same number of people.

Then there are 6 more girls on May Day than there are boys on May 2.

52 boys: 6 (7-5) 5=15.

These two classes, each class has: 30 + 15 = 45 people.

Male students in a class: 30 (1-20%)=24 (people) more: 30-24=6 (people).

Per serving: 6 (7-5) = 3 (persons).

Second class boys: 3 5 = 15 (people).

Class 1 girls: 3 7 = 21 (person).

Total: (24+21) 2=90 (person).

The number of boys in the May 1st class is 30*80%=24, which is 6 fewer than the number of girls in the 52nd class.

Due to the equal number of students in the two classes, there were 6 fewer boys in Class 52 than girls in Class 15.

The number of girls in class 51 is 6 (2 7) = 21.

The total number of students in the May 1st class is 24 + 21 = 45

The total number of students in the two classes is 45+45=90

1. Suppose: If B completes the project every day, then A completes x+1 304(x+x+1 30)+5x=1 every day

x=1 15 then x+1 30=1 10

A1 1 10 = 10 (days).

B1 1 15 = 15 (days).

2. Set: The distance between the two cities of A and B is 1 (it can also be set to x, and it will be about in the end) 1998 speed 1

Now the speed is 10 198 (1+30%) 1+25%) 1+20%)1 [10 198 (1+30%) 1+25%) 1+20%)]]hours).

3. Reverse 100 (30+20)=2 (hours) and 100 (30-20)=10 (hours) in the same direction

The minimum is 2 hours in the opposite direction, and the maximum is 10 hours in the same direction.

4. If the original large pile has x tons, the small pile is 24-x tons.

x(1-1/4)=24-x+4

x=16 then 24-x=8

Answer: The large pile was 16 tons and the small pile was 8 tons.

Hope I can solve your problem.

(4-3)x(3+4)=24x7=1682, apricot+peach+pear=1500

3 apricots 5 = peach - 30 = pear + 15 peach = 3 apricot 5 + 30 pear = 3 apricot 5-15 substitution.

Apricot + 3 apricot 5-15 + 3 apricot 5 + 30 = 1500 11 apricot 5 = 1485 apricot = 675 3 apricot 5 = 405

Peach = 435 pear = 390

3. Li Mingjiao (Chinese) and (Politics) classes.

Chen Cong teaches (Mathematics) and (Art) classes.

Sun Liang teaches (geography) and (**) classes.

The first instruction is equivalent to adding 125+ to Mickey Mouse's distance, and the second instruction is equivalent to adding another 125+

The nth instruction is equivalent to an additional 125+

A total of 125n+ is added to the n instructions

If Donald wants to win the race, the increased distance must be greater than 2500 meters. Got the inequality.

125n+>2500

125n+>2500 multiply 2

250n+>5000 with division.

20n+n^2+n>400

n^2+21n>400

n^2+21n+(21/2)^2>400+441/4n+21/2)^2>

n+21 2> Remove the negative roots.

n > should be at least 13 times.

One: 1/3 1 4 = 1/12 2: 24 1/12 = 288 (pcs).

Yes A Yes B Poor efficacy to the corresponding total amount.

Efficacy, efficacy, response rate.

One. A and B are completed together in four days.

4x12/1=3/1

24 3 1 = 72.

1.Permutation, combination of combinations: Pick two out of eight people who are not in the order c(n=2,m=8)=(8*7) (2*1)=28

2. Male players: (5*4) (2*1)=10 female players (4*3) (2*1)=6 (2 female players are selected from four female players), a total of 10*6=60 schemes.

3.From. 1+3+5 weighs as much as 2+4+8.

It is seen that 1 3 5 and 2 4 8 each have a bottle of salt water or no salt water, but according to the first and second sentences there is no salt water, it is impossible to say that 6 7 is not salt water

The first sentence 5+6 is heavier than 7+8 5 6 may be salt water, but the last sentence shows that 6 7 is not salt water, so 5 is salt water, then there is one bottle left in 1 2 Choose one of them, look at the last sentence 1 3 5 There can only be one bottle of salt water, and if you are sure that 5 is salt water, then the remaining bottle is No. 2.