Solution: Method 1:
Known person:
A desktop needs: 1/50 cubic meters of wood;
A table leg needs: 1/300 cubic meters of wood;
A desktop with four legs.
Suppose: * * made X tables, and a square table has four legs.
Therefore, the column equation: x/50+4x/300= 10.
Solution: x = 300
So: to produce 300 desktops, use 6 cubic meters of wood; 1200 table legs, using 4 cubic meters of wood; * * * Made 300 square tables.
Method 2:
Because the material of the desktop is six times that of the legs, six legs are equal to a desktop, that is, ten legs are equal to a square table. 10 cubic meter of wood can make 3000 legs, 3000/ 10 = 300, so we have to make 300 square tables.
Method 3:
In the same way, we can also consider making a desktop.
2. Sales problem
The market demand D (1000 pieces) and unit price p (yuan/piece) of a commodity obey the demand relationship: one third D+P- 17 =0. What is the market demand when the unit price is 4 yuan?
Solution: one third D+P- one third seventeen =0.
D/3？ + P？ - 17/3 ? =0
When the unit price is 4, that is, P=4, substitute the above formula.
D/3？ = ? 17/3 ? - ? 4 ? = ? 5/3
D=5
The market demand is 5000 pieces.
3. A computer company has brand A computers of type A, type B and type C, and brand B computers of type D and type E, and it is hoped that middle school will choose one computer from each of the two brands. Their prices are 6000 yuan for Type A, 4000 yuan for Type B and 2500 yuan for Type C respectively. D5000 and E2000 Hope Primary School plan to buy two sets of solutions from the company at a cost of 654.38+ten thousand yuan: 36 computers of different models, of which brand A computer is Type A computer. How many computers did the school buy?
A brand computer buys a type computer, and B brand computer may have two possibilities: D type and E type. Suppose the school buys a B-brand D computer.
Assuming that X units of type A are purchased, the equation of 36-X units of type D is:
6000X + (36-X)× 5000 = 100000
X = -80？ Is a negative number and does not hold water.
Therefore, the school bought an E-type computer, and the equation is:
6000X + (36-X)× 2000 = 100000
The solution is X = 7.
So the school bought seven A-type computers.
4. Geometric problems
In the right triangle ABC, angle C = 90 and angle A = 30. The bisector of angle C and the bisector of angle B intersect at point E, connecting AE. What is the angle AEB?
Solution: The vertical lines EF, EG and EH of BC, AB and AC passing through E are F, G and H respectively.
∵CE and BE are angular bisectors, ∴ EF=EG=EH, AE bisector ∠BAH,
∴∠BAE=75，
∴ ∠AEB= 180 -60 -75 =45
5. Installment payment problem
A supermarket has introduced the following preferential schemes: (1) no discount for one-time shopping 100 yuan; (2) 10% discount for one-time shopping over 100 yuan but not exceeding 300 yuan; (3) 20% discount for one-time shopping in 300 yuan. Wang Bo spent 80 yuan and 252 yuan on two purchases. How much should Wang Bo pay if he buys the same goods as the first two times?
Solution: Wang Bo bought it twice, and the preferential payment was the first time in 80 yuan.
The second time it exceeds 100 yuan but does not exceed 300 yuan, it is 252/0.9=280 yuan.
Exceeding 300 yuan is 252/0.8=3 15 yuan.
Therefore, the total non-preferential payment for two purchases in Wang Bo is 80+280=360 yuan, that is, 80+3 15=395 yuan.
360 yuan, if you buy the same goods as the last two times at one time, the payable amount is 360*0.8=288 yuan.
When it is 395 yuan, if you buy the same goods as the last two times at one time, the payable amount is 395*0.8=3 16 yuan.
Find a standard questions.
Solution: From the first picture and the second picture, we know that the sum of four numbers should be19;
There are two13+1=14 in the upper left corner of the first picture; Two 2+3 = 5 in the lower right corner;
The top left corner of the second picture is1+4 = 5; 5+9 in the lower right corner =14;
That is, one of the two angles is 14 and the other is 5;
So from these two conditions:
The number on the third picture should be 2, so 3+2 = 5; The following is 9, with 9+5 =14; Of course it adds up to 19!
(Here, if the top left corner is 14, that is, the top left corner is filled with 1 1, then the bottom left corner is filled with 0, but it depends on the picture.
Integer greater than 0)
Similarly,
In the fourth picture, fill in 6 on the left, so 6+8= 14, and fill in 4 on the right, so 4+1= 5;
You can't fill in 13 on the right here, so 1+ 13= 14, because it's unnecessary to fill in it on the left, because it's already 8> and it's five o'clock.
In the fifth picture, just fill in 3 above and 7 below. This is also the case.
7. Mathematical thinking problems (overall introduction, classified discussion)
It's just a walk, it took one minute for a person to walk, and a car left for half an hour, and finally arrived 10 minutes later than this person. It takes x points to set up a car to complete this section, so what is the reciprocal relationship between A and X?
Solution: If a person gets an A after walking, he walks 1/a every minute.
It takes x minutes for the car to walk this road, and the car covers the whole journey of 1/x every minute.
A car leaves half an hour after people leave and finally arrives 10 minutes later than people.
Namely a-30=x+ 10,
Arrive late 10 minutes, which means people arrive first. 10 minutes later, the bus will arrive.
8. Geometric space imagination
The vertebral body is cut by a plane parallel to the bottom surface. If the cross-sectional area is the same as the bottom area, what is the ratio of the height of the vertebral body to the upper and lower parts?
Solution: A vertebral body is cut by a plane parallel to the bottom. If the cross-sectional area is half of the bottom area,
The cross section is similar to the bottom surface.
The ratio of the corresponding edge to the bottom of the section = 1: root number 2.
Then the ratio of the height of vertebral body to the height of small and large vertebral bodies is 1: root number 2.
The ratio of the heights of the upper and lower parts is 1: (root number 2)- 1.
9.
As shown in the figure, given AF//BE//CD and AB//ED, can it be judged that ∠A and ∠D are equal according to these conditions? Why?
Solution: ∫AF//BE
∴∠A and ∠ Abe are complementary.
Again: AB//ED
∴∠ABE=∠BED (inner corner)
∵BE//CD
∴∠BED and ∞∠EDC are complementary.
So < a and < d