Math problems, six up, six on math problems

These two problems are similar to the chickens and rabbits in the same cage: you can use equations, you can also use comprehensive calculations, the equations are easy to understand, use the equations first:

1) Solution: If there are x bicycles, then there are 18-x tricycles: (1 bicycle has 2 wheels, 1 tricycle has 3 wheels).

2x=3（18-x）+1

The solution is x=11

18-x=7

So there are 11 bicycles and 7 tricycles.

There are x tables for singles and 12-x for doubles.

1 singles table with 2 people, 1 doubles table with 4 people) 2x = 4 (12-x) + 12

The solution is x=10

12-x=2

So there are 10 tables for singles and 2 for doubles.

Comprehensive equation: (1) (2 18 1) (2 3) = 7ps: The equation is easy to understand, but the equation learned in the sixth grade of primary school is still a simple equation, they can't solve the equation with parentheses, and there are equations with unknown quantities on both sides of the equation, so the equation is still a bit difficult for primary school students, and the comprehensive equation is not easy to understand, but the student can accept this formula.

If you want to understand this formula, you can read the problem of chickens and rabbits in the same cage

Number of bicycles: Number of tricycles.

Both bicycles and tricycles must be whole numbers.

So bicycles and tricycles can only be separated.

Do the math, if it's 10,8

Then the bicycle wheel 20 and the tricycle wheel 24 do not fit the topic.

If it's 11,7

22 bicycle wheels, 21 tricycles, just one more.

The rest of the groups are not right.

So bicycle 11, tricycle 7

There are definitely more pairs in singles than doubles.

So it can only be.

Do the math. If it's 7,5

14 in singles and 20 in doubles Not right.

8,4 16 for singles and 16 for doubles is not true.

9,3 18 in singles, 12 in doubles, no.

10,2 20 for singles, 8 for doubles, right.

11,1 is also not true.

So, 10 tables for singles and 2 for doubles.

There are 11 bicycles and 7 tricyclesThere are 10 singles and 2 doubles

Bicycle 11 liang tricycle 7 liang

10 ren in singles and 2 ren in doubles

Bicycle 11 Tricycle 7

Singles 10 Doubles 2

Solution: Set up to make x three-pointers.

Throw the ball from within the 3-point line (9-x).

3x+2（9-x）=21

3x+18-2x=21

x=3A: Zhang Peng made 3 three-pointers in this game.

Scored 9 9 2 = 18 points.

A total of 21 points were scored 21 - 18 = 3

Scored 3 three-pointers.

Let x be the number of 3-point goals and y be the number of 2-point goals.

x+y=93x+2y=21

x=3y=6

Since a total of 9 is set to be made x from outside the three-point line, then 9-x are made from inside the three-point line.

3x+2(9-x)=21

3x+18-2x=21

x+18=21

x=3 for three three-pointers.

Set the number of shots from beyond the three-point line to x... The equation is y. 3x+2y=21

The equation x+y=9 is solved to y=6

Therefore, the answer is that Zhang Ming made 3 three-pointers in this game. Calculation method. 3x9 - 21) divided by 2 =3

This is a two-point hit.

Use 9-3 = 6 again

The algorithm previously assumed that all nine balls were three-pointers, and that the difference between the number of points that should be scored and the actual number of points was the number of points that were put in by the two points.

Divide by two points to get the number.

and then the number of three-pointers...

Unset x cast 2 points.

2x+3（9-x）=21

x=69-6=3 (pcs).

Answer... Have a nice day.

Scored six 2-pointers and 3 3-pointers.

Suppose x three-pointers 3x+2*(9-x)=21 and solve x=3, i.e. 3 three-pointers.

1) Solution: After X hours, the two stuffy orange boats met Zhitong.

21x+27x=384

48x = 384x = 8 kg).

1 out of 1}

5/5 km] A: This road is a kilometer long.

After the previous week passed, the proportion of the remaining part to the whole is 1-1 4-1 3=5 12, so the total length of this road is:

km).

1-25%-1 3=5 12 km A: This road is a full kilometer.

Set up this road for the whole field x kilometers.

x-25%x-1/3x=

Diameter = Yuan Perimeter Vulture =

Radius = half of the diameter = 12 2 = 6

Circle area = vulr =

Circle area = rectangular area.

The length of the rectangle = area and width =

3. Only under the condition of the same circle or equal circle).

4. Only the figure of the four axes of symmetry must be square) 5.

1. There are several axes of symmetry.

A: There are 2 axes of symmetry.

2. If the radius of a circle increases by 1cm, its diameter increases by 2cm. Answer: Yes.

3. The diameter of the semicircle is equal to half of the diameter of the circle. Answer: Wrong.

4. A figure with four axes of symmetry must be a square. Answer: Wrong.

5. The diameter of the great circle is 4 times the diameter of the small circle, and the radius of the great circle is 4 times the radius of the small circle. ( or ).

A: Yes.

What is the quotient of 20 of 120 divided by a quarter of 12/5? 120×20%÷（12/5×1/4）=40

65 of the first number is 26, what is its 75? 26÷65%×75%=30

Jingjing looked at a book, she had read 40 pages of the whole book, and there were still 90 pages left. What percentage more pages you haven't read than what you've read?

or (90- 90 (1-40%) 40%) 90 (1-40%) 40%) = 50%.

One-third of a number is one-fifth more than one-fifth of it14, find this number. 14÷（1/3-1/5）=105

A cup of sugar containing 10 sugar 100 grams of sugar water, if you add 60 grams of water, the sugar content is ( ).

A case of drinks, if you drink 50, there are 12 bottles left. If you drink 5 bottles, there are ( ) bottles left.

12 50%-5=19 bottles.

1, 5/22, 53

Three-two

Question 5: Solution: A + B = 282 75% * A = (5 9) * B.

A = 120 B = 162

7.Finding the perimeter and area is simpler.

The circumference of the circle c = pi x diameter or = 2 * pi x radius = 2 * the length of the wire required =

Area = 40 * 40 * square centimeters.

A has 120 copies, and B has 162 copies.

It needs to be squared centimeters.

This question is relatively computational, but the answer must be the longest saving method is the best, if it is a choice or fill-in-the-blank question, it is good to save it for three years. The specific process is as follows (tax included):

Save every year.

10,000 yuan).

10,000 yuan).

10,000 yuan).

10,000 yuan).

Saved every year and every year.

10,000 yuan).

10,000 yuan).

10,000 yuan).

Saved every year for two years.

10,000 yuan).

10,000 yuan).

10,000 yuan).

Save 10,000 yuan in three years).

10,000 yuan).

It's a bit of a struggle, think for yourself, it's actually quite simple.

Of course, it's every three years!

Because: save a year: 10,000 yuan).

250+10000) yuan).

10,000 yuan).

10,000 yuan).

Savings for two years: 10,000 yuan).

325+10000) yuan).

10,000 yuan).

One year and two years: 10,000 yuan).

250+10000) yuan).

10,000 yuan).

Save 10,000 yuan in three years).

10000 + 385 = 11155 (yuan).

I didn't count the money for taxes.

Answer: The three-year deposit method will withdraw the most money from the bank after three years.

Every three years.

Solution: 10,000 yuan a year).

250+10000) yuan).

10,000 yuan).

10,000 yuan).

Savings for two years: 10,000 yuan).

325+10000) yuan).

10,000 yuan).

One year and two years: 10,000 yuan).

250+10000) yuan).

10,000 yuan).

Save 10,000 yuan in three years).

10000 + 385 = 11155 (yuan).

So, A: The method of saving every three years is the most money withdrawn from the bank after three years.

1.Number of pages read 3/5 = number of pages in the storybook.

x - three-fifths x = 36

2/5 x = 36

x=902.Red chalk + two-thirds of red chalk = white chalk.

x + two-thirds x = 100

Five-thirds x five-thirds = 100 five-thirds.

x=60

2.Equivalence test: the number of boxes of red chalk (1 + 2/3) = the number of boxes of white chalk.

Equation test: (1 + 2/3) = 100