Two 6th grade application questions. You ll get someone to help you

Divide a bunch of oranges among several children, if each person divides 3, then there are 8 more oranges, if each person divides 5, then the last person gets no less than 3 oranges, ask how many children, how many oranges?

Solution: If there are A children, then there are 3A+8 oranges.

According to the title. 5≥3a+8-5（a-1）≥3

5≥3a+8-5a+5≥3

8≥-2a≥-10

4 a 5a = 4 or 5

So there are 4 or 5 children.

Then there are 20 or 23 oranges.

If you haven't learned about inequalities, you can do so.

The last person gets no less than 3 oranges, that is, 3, 4, 5 oranges.

When you get 3.

3a+8-5（a-1）=3

2a+13=3

a=5 when 4 are obtained.

3a+8-5（a-1）=4

2a+13=4

a=9 2 (off-topic.)

When 5 are obtained.

3a+8-5（a-1）=5

2a+13=5

a=4 or so, let the last child get b.

3a+8-5（a-1）=b

b=13-2a

a=（13-b）/2

b = 3, 4, 5 are substituted separately, and only when b = 3 and 5 a has a positive integer solution.

There are x children. Not less than three, but not more than five, so, if each person divides five, then the last person gets three or four oranges.

Equations can be listed according to the title:

3x+8=5(x-1)+4 (if there are four) solves: x=

But there can't be half a person, so if each person is divided into five, then the last person can only get three oranges.

Equations can be listed according to the title:

3x+8=5（x-1)+3

Solution: x=4

Then, oranges have 3x+8=5 (pcs).

Elementary school students' application methods are used to solve application problems, and students are not yet able to list equations to solve application problems.

From the original 3 per person to 5 per person, each person has increased by 2, and the original 8 more can be divided into 8 2 = 4 people, so the child should have 4 + 1 = 5 or 4 people, and the orange should have 3 * 5 + 8 = 23 or 3 * 4 + 8 = 20.

There are x children with y oranges.

y-3x=8 y-5（x-1）>=3

Simultaneous equations are fine, and there may be a couple of possibilities, which I didn't calculate.

Suppose the younger brother is x years old this year, and the older brother is 55-x years old.

When the elder brother is x years old, it is (55-x-x=55-2x) years ago, when the younger brother is the age.

x-（55-2x）=3x-55

x=2(3x-55) x=22 55-22=33Answer: My brother is 33 years old this year, and my brother is 22 years old.

Let the younger brother's previous age be x, then the younger brother's current age is 2x, and the brother's current age is 55-2x, according to the age difference, it can be obtained: 55-2x-2x=2x-x, x=11, the younger brother is now 22 years old, and the elder brother is now 33 years old.

Understand no, if there is a problem, then QQ

There was a year when the elder brother was the younger brother's age this year, and the older brother was exactly twice the age of the younger brother. It shows that the younger brother is twice as old as the younger brother this year, twice as old as the younger brother at that time, and now it is 3 times that of the younger brother at that time, and the two brothers are 5 times that of the younger brother who used to have a year, that is, 55 5 = 11 There was a year when the younger brother was 11 years old, the elder brother is now 33 years old, and the younger brother is now 22 years old.

Equation: Let the younger brother be x years old now, then the elder brother is 55-x years old, and there was a year when the elder brother was x years old, then the difference in the middle is 55-x-x=55-2x years, and the younger brother should be x-(55-x-x)=3x-55, from the age of the elder brother at that time is exactly twice the age of the younger brother, we can know: 2(3x-55)=x, the solution:

5x=110；x = 22 years. Then the elder brother is 55-22 = 33 years old.

The elder brother is x, the younger brother y, after n years, x=2y

x=y+nx+n+y+n=55

Solve: x=22, y=11, n=11

1.Solution: Suppose you have to go x in total.

1/3x+(3/10*2/3)x+(1/3*3/4)x+13=x1/3x+1/5x+1/4x+13=x

47/60x+13=x

13/60x=13

x=60A: A total of 60 to go.

2.Solution: Set to contain x grams of gold.

1/19x+（720-x）/10=50

9/190x+72=50

9/190x=-33

x=696/2/3

820-696 2 3 = 123 1 3 (grams) Answer: 696 2 3 grams of gold, 123 1 3 grams of silver.

60+80=140(km)

A: 140 kilometers apart.

Hope that helps.

3.80 4 = 20 20 * 3 = 60 km.

The rest of the questions really won't.

1、（1-1/3）*3/10=1/5

13 (1-1 3-1 4-1 5) = 6060-13 = 47 km.

2. If the alloy contains x kilograms of gold, then it contains 770-x kilograms of silver.

x/19+(770-x)/10=50

x = 570 kg.

It contains 570 kilograms of gold and 200 kilograms of silver.

(4 3) = 60 km.

80 + 60 = 140 km.

The two buses were 140 meters apart when they departed.

In the first question, let Uncle Wang travel a total of x kilometers, x (one-third x + two-thirds times three-tenths x + one-third times three-quarters x) 13, and solve the equation to get x 60

In the second problem, let this alloy gold account for x grams, then silver accounts for 770 x grams, and the equation is 1/19 x + (770 x) times 1/10 50, and the equation is solved, x 570, 770 x 200

The third question can be known from the question, the distance of the motorcycle line is three-quarters of the car, so the motorcycle line is 80 * three-quarters, 60 kilometers, so the distance between the two cars is 60 + 80 140 (kilometers), here it should be noted that the unit should be unified, so it must be turned into 14000 meters, in order to write the answer.

1.A class of students went on an outing of 18 kilometers to Beishan. There is only one car, and it needs to be divided into two groups, group A takes the car first, and group B walks.

The car goes to place A, group A gets off and walks, the car returns to Xiaochai to pick up group B, and finally the two groups arrive at Beishan at the same time. The speed of the car is known to be 60 km/h, and the walking speed is 4 km/h. Find the distance from station A to Beishan.

Set: The distance from station A to the north stool Tongshan is 1.

Speed ratio: car: walking = 60:4 = 15:1

Whole process: 1 + (15-1) 2 + 1 = 9 copies.

Per serving: 18 9 = 2 (km).

The distance from Station A to Beishan is 2 km.

2.A and B respectively set off from A and B at the same time, and A and B met for the first time at the place where A exceeded the midpoint 50 meters, and A and B immediately turned back and walked back after arriving at A and B, and as a result, A and B met for the second time at 100 meters away from A, seeking the distance between A and B.

Let half of the distance between A and B be 1, and the distance between A and B will be 2.

First encounter: A walk: 1 part + 50 meters.

The second time you need: A walk: 3 parts +150 meters.

The second encounter A walks 2 parts -100 meters.

3 parts +150 m = 2 parts - 100 m.

1 portion = 250 meters.

2 parts = 500 meters.

The distance between A and B is 500 meters.

The eldest picks are 2 times the second, then the total number of the eldest and the second is a multiple of 3, and the third pick is the least, but there are more than 5, then the number of the third child is removed from the 46 and the rest should be a multiple of 3. Because the third child picked the least, but also more than 5, so the third picked at least 7, 46-7 = 39, at this time the second picked 39 (2 + 1) = 13, the eldest picked 13 * 2 = 26.

The third child may also pick 10 spring ties, 46-10 = 36, at this time the second child picked 36 (2 + 1) = 12, and the eldest picked 12 * 2 = 24.

The third child may also pick 13, 46-13 = 33, and the second child picks 33 (2+1) = 11, which is not the least picked by the third child.

So there are two scenarios.

The third picked at least 7, 46-7=39, then the second picked 39 (2+1)=13, and the eldest picked 13*2=26.

The third child may also pick 10, 46-10 = 36, at this time the second child picked 36 (2+1)=12, and the eldest picked 12*2=24.

In the 6th grade of primary school, I learned a one-dimensional equation, and this problem can be set to the total number of processing as x, then.

2 5*x+75=13 20*x gets x=300

The remaining is 300 * (1-13 20) = 105

13 20-2 5=5 20 75 5 20=300 (pieces) 300 7 20=105 (pieces).

It's a little more complicated,