The growth rate is a hot topic in the senior high school entrance examination in recent years. Only by mastering the essential connotation of the growth rate problem can we change constantly in the senior high school entrance examination.

The essence of growth rate is; The increase amount is the percentage of the initial amount, and the increase amount is the final amount minus.

Starting quantity.

Let the initial quantity be q and the final quantity be p? If the growth rate is x, increase it once to p=q( 1+x)? L continuously increases to p=q( 1+x)2.

If x > 0, it means growth; If x < 0, it means decreasing.

1. Average growth rate?

Example 1:? An orchard planted 200 fruit trees this year, and now plans to expand the planting area, so that the planting amount in the next two years will increase in the same proportion as that in the previous year, so that the total planting amount in the three years (including this year) will be 1400. Find this percentage.

Analysis: If the growth rate is X, the planting amount will be 200( 1+x) next year and 200( 1+x)2 in the following year.

Then the total planting amount in three years is 200+200 (1+x)+200 (1+x) 2.

Solution: if the growth rate is x, it is obtained according to the meaning of the question.

200+200( 1+x)+200( 1+x)2 = 1400

Let 1+x=y, then 200+200y+200y2= 1400.

Get a solution? y 1=2？ y2=-3

That is, 1+x=2 or 1+x=-3.

X 1= 1 x2=-4

So this percentage is 100%

Example 2: The sales of a commercial building in February was 6.5438+0 million yuan, and the sales in March decreased by 20%. Since April, the business measures of the commercial building have been continuously improved, and the sales volume has steadily increased. In May, the sales reached 6.5438+0.352 million yuan. Try to find the average growth rate in April and May.

Analysis: The sales in January and March were calculated as 100 (1-20%) ten thousand yuan. Let the average growth rate in April and May be X, then the sales in April will be 100 (1-20%) ten thousand yuan. In May, the sales volume was100 (1-20%) (1+) =100 (1-20%) (1+x) 20,000 yuan.

Solution: Let the average growth rate in April and May be, and get the equation from the meaning of the question.

100( 1-20%)( 1+x)2 = 135.2

( 1+x)2= 1.69

That is 1+x = 1.3.

Therefore, x 1=0.3 and x2 =-2.3.

X2=-2.3 = 30%, because X2=-2.3 is not practical and is discarded.

That is, the average growth rate in April and May is.

Special exercises:

1. In order to develop education, a county has increased its investment in education, with an investment of 30 million yuan in 2007 and an estimated investment of 50 million yuan in 2009. Suppose the average annual growth rate of education funds is, according to the meaning of the question, the following equation is correct (? )?

A.3000( 1+x)2=5000？ B.3000x2=5000

C.3000( 1+x%)2=5000？ d . 3000( 1+x)+3000( 1+x)2 = 5000

2. The output of an electric bicycle factory in March was 1 1,000 vehicles. Due to the increase in market demand, the output in May increased to 65,438+0,265,438+00 vehicles, so the monthly average growth rate of the factory in April and May was _ _ _ _ _.

3. After the price of a commodity has been reduced twice in a row, the price of each commodity has dropped from 55 yuan to 35 yuan. Let the average percentage of each price reduction be x, then the correct one in the following equation is ().

a . 55( 1+x)2 = 35 b . 35( 1+x)2 = 55 c . 55( 1-x)2 = 35d . 35( 1-x)2 = 55

The retail price of a commodity after two price reductions is the price before the price reduction.

Then the average price will decrease every time ()

A. 10% B. 19% C.9.5%？ D.20%

Second, the change of growth rate.

Example 3: The import volume of a commodity in country K in February this year decreased by 20% compared with the end of last year. Due to the rising price of this commodity, the cost of importing this commodity has increased by 30% compared with the end of last year. The increase in February is 5% more than that in 65438+ 10. Find the increase in the price of this commodity in 65438+ 10.

Analysis: If the import volume of the commodity was A at the end of last year, the price of the commodity was B, and the price growth rate in June this year was X, then the import volume of the commodity in February this year was (1-20%) A, the price in June was 1+X, and the price in February was B (/kloc-0).

Solution: What is the price growth rate of 65438+ 10 in June this year? According to the meaning of the question

( 1-20%)a×b( 1+x)( 1+x+5%)= ab( 1+30%)

After simplification, 8 (1+x) 2+0.4 (1+x)-13 = 0.

Let 1+x=y, then 8y2+0.4y- 13=0.

Get a solution? y 1= 1.25？ y2=- 1.3

That is, 1+x= 1.25 or 1+x=-2.3.

∴ x 1=0.25=25%？ X2=-2.3 (truncated)

Therefore, the price of this commodity in June 5438+ 10 increased by 25% compared with the end of last year.

Comments: This issue is a question of changing the growth rate. If the value before growth is a, the first growth rate is x, and the second growth rate is m more than the first growth rate, then the second growth rate is (x+m), and the result after growth is b. The method of listing equations according to the meaning of the question can be summarized as formula A (1+x) (1+x+).

Special exercise: 1 (a senior high school entrance examination in Shaanxi province) A shopping mall continuously lowered the price of a household appliance in the first quarter, in which the decline in March was 2 percentage points more than that in February (1 percentage point = 1%). As a result, the number of sales units in March increased by four times compared with that in June, and the sales revenue increased by 296%.

Third, the related growth rate.

Example 3. As shown in the figure, a kindergarten has a wall with a length of 16 meters, and it is planned to enclose a rectangular lawn ABCD. The area is120m2, and the fence is 32m long. Find the length of BC side of rectangular lawn.

2. Under the condition of 1, if the restriction condition of16m wall length is removed, the area of ABCD in rectangular lawn will be expanded to 216m2, in which the growth rate of BC is 2.5 times that of AB, and the growth rate of AB side will be calculated.

Solution: 1 Let the length of AB be x? The length of BC side is (32-2x) meters.

X(32-2x)= 120

Solution: x 1= 10? x2=6

When x= 10, AB is 10 m and BC is 12 m.

When x=6, AB is 6m and BC is 20m > 16m.

Therefore, the length of BC side of rectangular lawn is12m.

2. If the growth rate of AB side is y and that of BC side is 2.5y,

According to the meaning of the question,10 (1+y) ×12 (1+2.5y) = 216.

Y 1 = 0.2 = 20%？ Y2=- 1.6 (truncated)

So the growth rate at the AB end is 20%

Special exercises:

1 (Nanjing senior high school entrance examination) A farm planted 10 mu of pumpkin with an yield of 2000kg per mu. According to the market demand, this farm has expanded its planting area this year, and all the new varieties of high-yield pumpkins are planted. It is known that the growth rate of pumpkin planting area is twice the yield per mu. This year, the total output of pumpkin is 60,000, and the growth rate of pumpkin yield per mu is sought.