Mathematics 6th Grade Equation Problems 2 6th Grade Mathematics Equation Problems

Solution:1Let's say he made x dollars.

Solution: x=43625

Answer: (omitted) 2(1) Set: It takes x hours to arrive.

45x=15+15+15

Solution: x=1 1*60=60 minutes > 42 minutes!

Answer: (omitted) 2) Set: It takes x hours to arrive.

45x=15+(15-5) (15+45)*45*2 solution: x=2 3 2 3*60=40 minutes Less than 42 minutes!

A: (omitted).

2.The second question de the first scheme does not imply arrival.

The second one is okay.

Just calculate the time and compare it with 42.

It is necessary to convert 45 kilometres into kilometres.

**The money required is: 40 10,000 400,000;

**A handling fee is required: (40 10,000) ,000;

The proceeds from the sale are: 45 10000 450000;

The commission required for selling is: (45 10,000) ,375;

Total earn: 450,000 400,00 3,000 3,375 43625;

2 A: (2) The plan allows all fans to arrive at the stadium on time.

Process: All based on the time taken by the second group to arrive (starting with walking).

It took 15 60 1 3 (hours) to get to the pitch from 15 km to the pitch with a car that was not broken, and 15 1 3 5 (km) on foot

The car sends the first fans back as soon as they arrive, and on the way to meet the fans on foot can be listed

Let x be the time when the car returns to the car and the fans on foot meet, then there is:

45x＋15x＝10

60x＝10

x＝1/6；

That is, the first group of fans arrived and immediately returned on the way to meet the fans on foot, and the fans walked for 1 6 hours. The distance covered at this time was 15 1 6 km).

That is, the foot fans transfer to the car when there are still kilometers from the stadium.

The time from the transfer to the time it takes to arrive at the stadium is hours).

In summary, the total multi-use time of scheme 2 is 1 3 1 6 1 6 2 3 (hours) 40 (minutes) < 42 minutes.

So all the fans can hint at arriving at the stadium.

Yuan. Solution: (45-40)*10000-40*10000*2The second option is to get 8 fans to get to the stadium within the specified time.

Solution: Option 1:

At a distance of 15 kilometers from the stadium, if the car drops off the first group of people, and the second group waits for the car to return to pick up, then it must take 45 (15*3) = 1 hour, because there are only 42 minutes left in the game, that is, it is impossible to get to the stadium within the allotted time.

Scheme 2: while the car sends away the first batch of people, the second group of people walks in the direction of the field at an average speed of 15 kilometers, then when the car turns back to pick up the second group of people, the second group of people has walked (15 45) * 15 = 5 kilometers at this time, then the car will pick up the second group of people at a distance of 10-[10 (15 + 45)]*15 = kilometers. In this way, the total time it takes to send 8 fans by car is 15 45+ hours, or 40 minutes.

Solution: Let's say that after X years, Dad's age is exactly right.

Xiao Ming. [then after x years, Dad's age is [40+x] years old, and Xiao Ming is [12+x] years old].

According to the title, got.

40+x=3[12+x]

40+x=36+3x

4=2xx=2

A: After 2 years, my father is exactly 3 times older than Xiao Ming.

If you still don't understand, add QQ and ask me.

Solution: After X years, my father's age is exactly three times that of Xiao Ming.

12+x)*3=40+x

36+3x=40+x

3x=4+x

2x=4x=2

A: After 2 years, my father is exactly 3 times older than Xiao Ming.

If there are x people who process frames, there are (30 x) people who process lenses. According to the equation of the question: 72x=96 (30 x).

72x=2880-96x

168x=2880

Solution: x number of people is an integer, so x=17

Solution: There are x classes.

9x+8=2x5x

Solve it yourself, and the number of classes will come out.

There are x volleyball, 2x basketball, and 2x class y

9y=2x-8

5y=x gives y=8 x=40

There are x classes, Y basketball, and Z volleyball.

Equation 1: 2*z=y

Equation 2: 2*5*x=9*x+8

Equation 3: y+z=9*x+5*x+8

There are 8 classes, 80 basketball, 40 volleyball.

Solution: Remember that there are x volleyballs, y basketballs, and a class of a.

From the title: y=2x

x=5a9a+8=y

Solution: x=40, y=80, a=8

Solution: Let the first box of sugar originally have x grains of sugar, and the second box of sugar originally had (x-11-11) grains of sugar.

3(x-11-11-2)=x+2

3(x-24)=x+2

3x-72=x+2

3x-x=2+72

2x=74x=37

The second box = x-11-11 = 37-11-11 = 15 capsulesAnswer: The first box of sugar originally had 37 pieces of sugar, and the second box of sugar originally had 15 pieces of sugar.

The first box has x seeds, and the second box has y seeds.

x-11=y+11 x+2=3(y-2) gives x= 37 y=15

Answer:..

Solution: If the first box has x grains, then the second box has (x+2) 3+2 capsules.

x-11=(x+2)/3+2+11

x=x/3+24+2/3

2x/3=74/3

x=37x+2)/3+2=15

A: The first box of candy is 37 capsules, and the second box is 15 capsules.

Solution: Let the first box originally have x box, and the second box originally had y box.

From the title, x-11 = y-11

x+2=3(y-2）

solution, x=37

y=15A: The first box originally had 37 grains, and the second box originally had 15 capsules.

The first box is set to x, the second box is set to y, yes.

x-11=y+11---1

x+2=3(y-2)--2

Simultaneous two-style, solution.

y=15,x=37

Let the sugar in the first box be x, then the second box is [x-11].

It can be seen from the title.

x-11-2]*3=x+2

The answer is your own.

The first box of x grains, the second box of y grains.

x-11=y+11

x+2=3(y-2)

x-y=22

3y-x=8

3y-y=22+8

2y=30y=15

x=22+y=22+15=37

The first box is x, and the second box is y

x-11=y+11

x+2=3(y-2）

The solution is x=37 and y=15

The first box of sugar was 37 capsules, and the second box of sugar was 15 capsules.

Suppose the first box has sugar A grains and the second box has sugar B grains, then there is the following equation:

a-11=b+11

a+2=3（b-2）

The solution yields: a=37, b=15

Let the first box have x grains and the second box have Y grains.

According to the conditions. x-11=y+11

So y=x-22

Then the second box has x-22 capsules.

Again, according to the conditions.

3*(x-22-2)=x+2

x=37 so y=x-22=15A:.

Solution 1: Set the total number of people to x. Then the female classmates have (people.

x=120 female classmates: (person.

2 solution: set a fountain pen x only. Then B fountain pen 100-x only.

6x=4（100-x）

x=40100-40=60.

Therefore, there are 40 fountain pens for A and 60 fountain pens for B.

The male students who participated in the group gymnastics performance accounted for the total number of people in the Spring Festival, and then how did the male students add 15 people, at this time the male students accounted for 4 9 Please count, how many female students participated in the group gymnastics performance?

Solution: If there are x people participating in the group gymnastics performance, then there are (1-37%) x female students participating in the group gymnastics performance

5x/8=5x/9+75/9

5/8-5/9)x=75/9

5/72*x=75/9

x=75/9*72/5=120

That is, the female students who participated in the group gymnastics performance were (1-37%) x = people.

2.A stationery store has a total of 100 fountain pens of A and B, and it is known that each fountain pen is 6 yuan, and each has 4 yuan, and the money used in the two pens of A and B is the same.

Solution: A fountain pen purchases x counts, then B fountain pen purchases (100-x) counts, then:

6x=4(100-x)

6x=400-4x

x = 40 that is, a fountain pen purchase of 40 pieces, B pen purchase (100-x) = 100-40 = 60 pieces.

The first problem is not solved by the equation, the difference method is used, so let's figure it out for yourself, first make it into a fraction, which is three-eighths, use four-ninths minus three-eighths, calculate that it is five-sevenths, and then divide 15 by five-sevenths, and calculate that it is 216, and then multiply it by 216 to calculate that it is 135, and this is the first problem, and the key to the equation is to grasp the equivalence relation.

The second question is very simple, if A pen x is 100-x then B is 100-x The price is the same then it is 6x=4 (100-x) This is very easy to grasp the relationship is The unit price of A pen A sells = the unit price of B's pen The quantity of B sold Result x=40 Answer: 40 A pens 60 B pens.

1.Let's take the female student who participates in the performance as X

x/(x=75

2.Let x be the number of strokes.

6x=4(100-x)

x = 40 (40 fountain pens are restocked).

100-40=60 (B fountain pen purchase 60 pieces).

The grain in warehouse A is twice that of warehouse B, warehouse A ships 350 tons per day, warehouse B transports 250 tons per day, and after a few days, the grain in warehouse B is just finished, and warehouse A still has 900 tons left, how many tons of grain are there in each of the two warehouses?

Solution: After X, the grain in warehouse B is just out of transportation, and there are still 900 tons left in warehouse A.

900+350x=2×250x

900+350x=500x

900=150x

x=6, so warehouse A used to have grain: 350 6 + 900 = 3000 tons, and warehouse B and A used to have grain: 3000 2 = 1500 tons.

Set a number of days to be x days.

The grain in warehouse A is twice as large as that in warehouse B.

2 x 250x = 900 + 350xx = 6 days.

A 3000 B 1500

Let B have x tons, then A has 2x tons, and let y day B be transported, and there is a system of equations x = 250y

2x=350y+900

If the solution is x=1500 and y=6, then the original grain of A is 2x=3000 tons, and the original grain of B is x=1500 tons.

x+(x+15)=95

2x=80x=40

x+15=55

Tomato plot 40 square meters.

Eggplant 55 square meters.

Set 10x tons of elephants and x tons of cattle.

10x-x=

9x=x=tons.

10x = 5 tons.

5 tons for elephants, tons for cattle.

1 by the title x+x+15=95

x=40 so 40 square meters for tomato and 55 square meters for eggplant.

2 If the ox weighs x tons, the elephant is 10 x tons.

By the title 10x-x=

x = so the elephant weighs 5 tons and the ox weighs a ton.

1：x+x+15=95,x=40,x+15=55;2: Let the ox be x, the elephant is 10x, so, 10x x=; x=,10x=5

The original grain station had 500 (1-3 8) = 800 kg of rice.