The Process and Answers of Junior High School 100 Calculations

Students, I advise you not to wait for the answer, but to make it up by yourself. I know I haven't seen anyone who can give an answer for such a long time. Actually, homework is not that difficult. Think more and you should be able to do it.

There are 100 calculation problems in the first semester. There is a process and an answer. Even if 50x = 60 (x-3) 50x = 60x-180180 =10xx =180% d% a.

Q: A calculation problem 1. 1 minus 65438+9 over 00 equals 0. 1, 1 minus 65438+99 over 000 equals 0.0 1 minus 65438+990 over 000. 1 minus 10000 part 9999 is equal to 0.0001.9/10+99/100+999/1000+999/9999.

2. 189÷( 189+ 189/ 190)

= 189÷ 189×( 1+ 1/ 190)

= 1/( 1+ 1/ 190)

= 1/( 19 1/ 190)

= 190/ 19 1

One hundred calculation questions (calculation process+answer) 1 in the last semester of the second day of junior high school. An engineering team wants to recruit two kinds of workers, A and B. 150. In addition, the wages of two kinds of workers are 600 and 1000 respectively. Now the number of hundreds of millions of workers is required to be no less than twice that of a worker. When asked how many workers A and B hired, they paid the least.

Solution: Let's recruit X workers of Class A .. and then recruit workers of Class B (150-X) with a monthly salary of Y yuan.

Yes: 150-x ≥ 2x.

The solution is x≤50.

y=600x+ 1000( 150-x)

= 150000-400x

Obviously, the bigger x, the smaller y. So when x = 50, y is the smallest.

That is, 50 workers from Party A are recruited, and 150-50 = 100 workers from Party B have the lowest salary.

Lao Zhang and Lao Li bought the same number of breeding rabbits. A year later, Lao Zhang raised more rabbits than he bought.

There are two more rabbits, and Lao Li has fewer rabbits 1 rabbit than Lao Zhang bought.

Lao Li keeps no more than two thirds of rabbits. How many kinds of rabbits did Lao Zhang buy at least a year ago?

Suppose Lao Zhang bought X rabbits a year ago.

x+2≤(2x- 1)*2/3

3x+6≤4x-2

-x≤-8

x≥8

Lao Zhang bought at least eight breeding rabbits a year ago.

The computer company sold a batch of computers, and sold 60 computers at the price of 5500 yuan/set in January.

The price was reduced from the second month, and then all these computers were sold at the price of 5000 yuan/set.

The total sales exceeded 550,000 yuan. How many computers are there in this batch?

Solution: suppose there are x computers in this batch.

5500*60+5000*(x-60)>550000

X> 104 (Taiwan Province)

X is an integer, so at least 105 computers can make the total sales exceed 550,000 yuan.

4. A popular science book has 98 pages in total. Xiao Wang hasn't finished reading it for a week (7 days), and Xiao Yong finished reading it in less than a week. On average, Xiao Yong reads three more pages than Xiao Wang. How many pages does little Wang Pingjun read every day?

If Wang Pingjun reads A pages every day, Xiao Yong reads +3 pages on average every day.

7a & lt98

7(a+3)>98

The solution is 1 1

5. An enterprise buys Phyllostachys pubescens in "Bamboo Sea in South Sichuan" for rough processing, which can process 8 tons per day and make a profit of 800 yuan per ton; If bamboo is processed, it can be processed 1 ton every day, and the profit per ton can reach 4000 yuan. Due to the limitation of conditions, only one method can be used for processing every day, and all bamboo products are required to be sold within one month (30 days). To this end, the factory director called a meeting of employees to discuss how to process and sell more economically.

A said: all bamboo should be sold after rough machining;

B said: 30 days to complete the processing, unprocessed bamboo directly sold;

C said: 30 days can take a few days for rough machining, and then a few days for fine machining before selling;

Which scheme should the director adopt to make the most profit?

This problem is a typical application problem of combining equations and inequalities. The specific solutions are as follows:

Assuming that rough machining takes X days, finishing takes Y days, and the total profit is Z, the equation is as follows:

8×800X+4000Y=Z ( 1)

X+Y=30 (2)

0 & lt=X，Y & lt=30 (3)

Substitute (2) into (1) to get192000-2400y = z.

So y=( 192000-z)/2400.

Namely 0

120000 & lt； = z<= 192000, that is, the maximum value of z is 192000.

At this time, y=0, that is, the rough machining profit in these 30 days is the largest, so scheme A should be adopted!

I wonder if you understand my answer!

6、

In order to further improve the conditions for running a school, the school plans to invest 6.5438+0.8 million yuan to buy LCD projection bulbs from an electronic mall. It is understood that this electronic shopping mall has three different types of light bulbs. The quotation is: 3000 yuan for A, 4200 yuan for B and 5000 yuan for C. ..

(1) If you buy 50 bulbs of two different models with 1.8 million yuan, how many different purchase schemes are there?

(2) The school decided to buy 50 bulbs of three different models with a price of 6.5438+0.8 million yuan, and requested to buy more than 5 bulbs of the second model, but not more than 654.38+00 bulbs. How many three different types of LCDs did the school actually buy?

Hello, I did this problem myself, for reference only. The answer is: (1) two formulas.

Case; (2) Buy 3 1 A, 10 B, 9 C, and solve the problem as follows:

We buy X pieces of A, and if the other one is B, it is (50-x) pieces. The formula can be listed as follows:

3000 x+4200 (50-x) = 180000 Solution: X is 25, so both parties are 25.

If you buy an X and the other model is a C,

Can be listed as follows:

3000 x+5000 (50-x) = 180000 Solution: X is 35, so I bought 35 A and 65438 C+05.

If you buy B X and the other one is c,

Can be listed as follows:

4200 x+5000 (50-x) = 180000, but the solution of x doesn't meet the question.

Therefore, there are two purchase schemes.

The second question:

If you buy X A and Y B, you buy C (50-x-y).

The formula is:

3000 X+4200Y+5000(50-X-Y)= 180000

Y=(-70000+2000X)/-800

Furthermore, if y is greater than 5 and less than or equal to 10, then

(-70000+2000x)/-800 should also meet this condition, and x = 3 1 or 32 can be obtained by solving the inequality. But when x is 32, y is not an integer, so x can only be 3 1. At this time, A buys 3 1 and B buys 65438+.

7. In recent years, the number of students enrolled in the first year of senior high school in Jiji has increased year by year, reaching 550 last year, including "Jiji class" students and ordinary students enrolled from the whole province. The maximum enrollment this year is 65,438+000 more than last year, of which 20% can be enrolled in regular classes and 65,438+00% can be enrolled in "gathering classes". How many students can you enroll at least this year?

For example, last year, X students enrolled in the "gathering intelligence class" and Y students enrolled in the ordinary class.

According to the conditions, X+y = 550.

10%X+20%Y≤ 100

If y = 550-x is substituted into the inequality, X≥ 1 10 can be obtained.

So (1+10%) x ≥110.

A: At least 1 10 students can be enrolled in the "intelligence gathering class" this year.

8. Xiaoming's father 1 month buys 6,000 shares at the share price 18 yuan. In the next two months, the stock continued to soar, and he bought 8000 shares several times. But since then, the stock price has continued to fall. When the stock price fell to 36 yuan, he began to sell one after another. When the share price fell to 30 yuan, he sold all his shares to make ends meet.

43.5~54

In two extreme cases, that is, when you think of 36 yuan, you buy it all at once.

And 30 yuan only throws it all at once, and gets two numbers respectively, and that's it.

9. It is understood that individual clothing sales can be profitable as long as they are 20% higher than the purchase price, but bosses often exceed the purchase price.

50%~60% price tag. If you are going to buy a dress with a price tag of 200 yuan, within what range should you make a counter-offer?

Let the cost price be x.

x( 1+50%)& lt； = 200 & lt=x( 1+60%)

1.5x & lt； = 200 & lt= 1.6x

125<= x & lt= 133.3

Bottom line range of profit price y:

125( 1+20%)& lt； = y & lt= 133.3( 1+20%)

125 * 1.2 & lt； = y & lt= 133.3* 1.2

150<= y & lt= 159.96=~ 160

If the counter-offer reaches the bottom line, the counter-offer range is between 150 yuan and 160 yuan.

10, pump the pool water with a type A pump that can pump 1. 1 ton of water per minute, and it will be finished in half an hour; If the pump is B-type, it is estimated that it will take 20 to 22 minutes to finish pumping. How many tons of water does type B pump pump per minute more than type A pump?

Let type b be x.

1. 1 * 30/20 & gt； X & gt 1. 1*30/22

1.65 >X & gt 1.5

At least 0.4 tons more water can be pumped and at most 0.55 tons more water can be pumped.

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Title: A plane flies from south to north at a speed of 300m/s and passes over a city at 2 pm. Plane B flies from west to east at a speed of 400m/s and passes over City A at 2: 20 pm. If two planes fly at the same altitude, when are the two planes 360 kilometers apart?

I don't have much time! Excuse me!

One hundred computing problems and processes

The solving process of several calculation problems and their answers 1, (y+2)2=9,

y+2= 3，

y 1= 1，y2=-5。

2、x2-7x=5x-36，

x2-7x-5x+36=0，

x2- 12x+36=0，

(x-6)(x-6)=0，

x=6。

3 、( 3x)2- 12=0

9x2- 12=0，

3x2-4=0，

3x2=4，

x2=4/3，

x 1=(2√3)/3

x2=-(2√3)/3。

4 、( y+2)2=9，

y+2= 3，

y 1= 1，y2=-5。

One hundred calculation problems. I'm glad to answer these questions for you.

0.4× 125×25×0.8

=(0.4×25)×( 125×0.8)

= 10× 100= 1000

1.25×(8+ 10)

= 1.25×8+ 1.25× 10

= 10+ 12.5=22.5

9 123-( 123+8.8)

=9 123- 123-8.8

=9000-8.8

=899 1.2

1.24×8.3+8.3× 1.76

=8.3×( 1.24+ 1.76)

=8.3×3=24.9

9999× 100 1

=9999×( 1000+ 1)

=9999× 1000+9999× 1

= 10008999

14.8×6.3-6.3×6.5+8.3×3.7

=( 14.8-6.5)×6.3+8.3×3.7

=8.3×6.3+8.3×3.7

8.3×(6.3+3.7)

=8.3× 10

=83

1.24+0.78+8.76

=( 1.24+8.76)+0.78

= 10+0.78

= 10.78

933- 157-43

=933-( 157+43)

=933-200

=733

482 1-998

=482 1- 1000+2

=3823

I32× 125×25

=4×8× 125×25

=(4×25)×(8× 125)

= 100× 1000

= 100000

9048÷268

=(2600+2600+2600+ 1248)÷26

=2600÷26+2600÷26+2600÷26+ 1248÷269

= 100+ 100+ 100+48

=348

288 1÷ 43

=( 1290+ 159 1)÷ 434

= 1290÷43+ 159 1÷43

=30+37

3.2×42.3×3.75- 12.5×0.423× 16

=3.2×42.3×3.75- 1.25×42.3× 1.6

=42.3×(3.2×3.75- 1.25× 1.6)

=42.3×(4×0.8×3.75- 1.25×4×0.4)

=42.3×(4×0.4×2×3.75- 1.25×4×0.4)

= 42.3×(4x 0.4x 7.5- 1.25 x4 0.4)

=42.3×[4×0.4×(7.5- 1.25)]

=42.3×[4×0.4×6.25]

=42.3×(4×2.5)

=4237

1.8+ 18÷ 1.5-0.5×0.3

= 1.8+ 12-0. 15

= 13.8-0. 15

= 13.65

6.5×8+3.5×8-47

=52+28-47

=80-47

(80-9.8) × 2/5- 1.32

=70.2X2/5- 1.32

=28.08- 1.32

=26.76

8×7 4÷[ 1÷(3.2-2.95)]

=8×4/7÷[ 1÷0.25]

=8×4/7÷4

=8/7

2700×(506-499)÷900

=2700×7÷900

= 18900÷900

=2 1

33.02-( 148.4-90.85)÷2.5

=33.02-57.55÷2.5

=33.02-23.02

= 10

( 1÷ 1- 1)÷5. 1

=( 1- 1)÷5. 1

=0÷5. 1

=0

18. 1+(3-0.299÷0.23)× 1

= 18. 1+ 1.7× 1

= 18. 1+ 1.7

= 19.8

Hope to adopt, O(∩_∩)O Thank you.

What does the title of a calculation problem (detailed process) mean in the first day of junior high school?

x+y/3

X+y is an integer.

Should we make it a whole, simplify it, or work out a fixed value?

The calculation process and answer analysis of a calculation problem are interchanged, and the work efficiency is improved to 9/8 of the original.

Exchange ethylene and propylene, the working efficiency is improved to 9/8 of the original,

At the same time, the total efficiency is improved to 9/8*9/8=8 1/64.

That is, the time to finish the work is 64/8 1, that is, (1-64/81) * 9 =17/9 hours in advance.

Senior one has 10 calculation problem ... (detailed process) correct answer (1) (30-3) * (30+3) = 30 2-3 2.

(2) (6-0. 1)*(6+0. 1)

(3) ( 100- 1)*( 100+ 1)

(4) ( 1000+5)*( 1000-5)

( 1) ( 100- 1)^2= 100^2-2* 100* 1+ 1^2

(2) ( 1000+2)^2