A simple calculation problem using the law of distribution of multiplication

A simple calculation problem using the law of distribution of multiplication

1）2 × 68 ＋ 68 × 98
2）71 × 512 ＋ 512 × 29
3）92 × 879 ＋ 879 × 8
4）2 × 967 ＋ 967 × 98
5）88 × 386 ＋ 386 × 12
6）65 × 300 ＋ 35 × 300
7）35 × 744 ＋ 65 × 744
8）55 × 111 ＋ 45 × 111
9）64 × 199 ＋ 36 × 199
10）5 × 422 ＋ 95 × 422
11）357 × 95 ＋ 643 × 95
12）1053 × 539 ＋－53 × 539
13）264 × 906 ＋ 736 × 906
14）353 × 995 ＋ 647 × 995
15）221 × 413 ＋ 779 × 413
16）964 × 945 ＋ 945 × 36
17）661 × 389 ＋ 389 × 339
18）871 × 756 ＋ 756 × 129
19）960 × 844 ＋ 844 × 40
20）828 × 263 ＋ 263 × 172
21）1394 × 594 － 594 × 394
22）2090 × 37 － 37 × 1090
23）1301 × 405 － 405 × 301
24）1390 × 493 － 493 × 390
25）1258 × 911 － 911 × 258
26）1930 × 438 － 930 × 438
27）1627 × 882 － 627 × 882
28）1837 × 249 － 837 × 249
29）1395 × 524 － 395 × 524
30）1263 × 943 － 263 × 943
31）195 × 671 － 95 × 671
32）165 × 115 － 65 × 115
33）185 × 482 － 85 × 482
34）154 × 9 － 54 × 9
35）141 × 427 － 41 × 427
36）84 × 907 ＋ 907 × 42 ＋ 907 × 74
37）90 × 272 ＋ 272 × 97 ＋ 272 × 13
38）80 × 614 ＋ 614 × 93 ＋ 614 × 27
39）33 × 267 ＋ 267 × 69 ＋ 267 × 98
40）34 × 101 ＋ 101 × 71 ＋ 101 × 95
41）84 × 907 ＋ 907 × 42 ＋ 907 × 74
42）76 × 491 ＋ 491 × 95 ＋ 491 × 29
43）39 × 508 ＋ 508 × 75 ＋ 508 × 86
44）29 × 850 ＋ 850 × 70 ＋ 850
45）16 × 788 ＋ 788 × 19 ＋ 788 × 65
46）17 × 623 ＋ 623 × 21 ＋ 623 × 62
47）32 × 960 ＋ 960 × 83 ＋ 960 × 85
48）3 × 2 ＋ 2 × 79 ＋ 2 × 18
49）37 × 97 ＋ 97 × 74 ＋ 97 × 89
50）56 × 546 ＋ 546 × 60 ＋ 546 × 84
51）36 × 363 ＋ 363 × 29 ＋ 363 × 35
52）5 × 27 ＋ 27 × 5 ＋ 27 × 90
53）6 × 10 ＋ 10 × 7 ＋ 10 × 87
54）85 ＋ 85 × 48 ＋ 85 × 51
55）71 × 89 ＋ 89 × 47 ＋ 89 × 82
56）6 × 99 ＋ 99 × 39 ＋ 99 × 55
57）77 × 78 ＋ 78 × 92 ＋ 78 × 31
58）58 × 60 ＋ 60 × 61 ＋ 60 × 81
59）80 × 24 ＋ 24 × 91 ＋ 24 × 29
60）89 × 91 ＋ 91 × 6 ＋ 91 × 5
61）79 × 25 ＋ 25 × 2 ＋ 25 × 19
62）65 × 19 ＋ 19 × 52 ＋ 19 × 83
63）13 × 25 ＋ 25 × 78 ＋ 25 × 9
64）7 × 100 ＋ 100 × 21 ＋ 100 × 72
65）77 × 4 ＋ 4 × 19 ＋ 4 × 4
66）13 × 14 ＋ 14 × 11 ＋ 14 × 76
67）31 × 59 ＋ 59 × 98 ＋ 59 × 71
68）52 × 38 ＋ 38 × 71 ＋ 38 × 77
69）3 × 12 ＋ 12 × 70 ＋ 12 × 27
70）60 × 20 ＋ 20 × 74 ＋ 20 × 66
71）23 × 43 ＋ 43 × 78 ＋ 43 × 99
72）87 × 32 ＋ 32 × 76 ＋ 32 × 37
73）20 × 76 ＋ 76 × 75 ＋ 76 × 5
74）25 × 100 ＋ 100 × 24 ＋ 100 × 51
75）5 × 85 ＋ 85 × 14 ＋ 85 × 81
76）82 × 98 ＋ 98 × 71 ＋ 98 × 47
77）7 × 50 ＋ 50 × 36 ＋ 50 × 57
78）61 × 89 ＋ 89 × 27 ＋ 89 × 12
79）57 × 36 ＋ 36 × 3 ＋ 36 × 40
80）20 × 56 ＋ 56 × 87 ＋ 56 × 93
81）59 × 6 ＋ 6 × 26 ＋ 6 × 15
82）82 × 10 ＋ 10 × 63 ＋ 10 × 55
83）70 × 100 ＋ 100 × 91 ＋ 100 × 39
84）45 × 45 ＋ 45 × 34 ＋ 45 × 21
85）90 × 82 ＋ 82 × 9 ＋ 82 × 1
86）43 × 53 ＋ 53 × 87 ＋ 53 × 70
87）73 × 22 ＋ 22 × 35 ＋ 22 × 92
88）34 × 74 ＋ 74 × 5 ＋ 74 × 61
89）46 × 98 ＋ 98 × 10 ＋ 98 × 44
90）20 × 68 ＋ 68 × 99 ＋ 68 × 81
91）67 × 98 ＋ 98 × 31 ＋ 98 × 2
92）14 × 57 ＋ 57 × 29 ＋ 57 × 57
93）67 × 100 ＋ 100 × 42 ＋ 100 × 91
94）11 × 34 ＋ 34 × 75 ＋ 34 × 14
95）75 × 26 ＋ 26 × 94 ＋ 26 × 31
96）33 × 9 ＋ 9 × 52 ＋ 9 × 15
97）41 × 5 ＋ 5 × 33 ＋ 5 × 26
98）12 × 5 ＋ 5 × 10 ＋ 5 × 78
99）65 × 65 ＋ 65 × 24 ＋ 65 × 11
100）74 × 33 ＋ 33 × 40 ＋ 33 × 86

About percentage
Xiaohong bought 55 strawberries and Xiaolan 187 strawberries. The proportion of strawberries in their hands is 27%: 73%. How many strawberries can Xiaohong buy up to 80%?

The final number of strawberries is unclear
Is it the total quantity?
Let's buy X more and play 80%
（55+x）/（55+x+187）=0.8
The solution is x = 693
So we need to buy 693 more, up to 80% of the total
Hope to help you, do not understand, welcome to ask

Percentage of calculation questions
What's the percentage of 84g flour in 210g flour?

84/210=0.4=40%

Proportion problem
The original number ratio of English reference books and mathematics reference books is 7:3. Later, she bought back 14 of the two reference books. Now the ratio of the two reference books is 7:4. How many of the two books do you have? There should be a detailed explanation and formula,

The teacher bought back 14 English reference books and 14 mathematics reference books,
That is (7x + 14): (3x + 14) = 7:4,
It's solved, x = 6,
So the teacher has 56 English reference books and 32 mathematics reference books

80% of the boys in a school are 20% more than the girls, and the number of boys accounts for 20% of the total number of students______ %．

Suppose the number of male students is 1; 1 × 80% = 0.8; 0.8 △ 120%, = 0.8 △ 120%, = 23; 1 △ 23, = 1 △ 53, = 60%; answer: the number of male students accounts for 60% of the total number of students

Two percentage application questions
It used to take 12 hours to get from Wuhan to Beijing by train. Now it only takes 8 hours after the speed of the float car is raised. How much higher is the speed of the train?
A garment factory used to burn 8 tons of coal every day, but now it only burns 6 tons of coal every day. How much has it saved?

Speed increased by (1 / 8-1 / 12) / (1 / 12) × 100 ^ = 50%
Saving (8-6) △ 8 × 100% = 25%

Percentage application
Xiao Li bought 500 shares at the price of 10 yuan. In the process of stock trading, he had to pay stamp duty and commission at 0.2% and 0.35% of the purchase price. Finally, he made a net profit of 9395 yuan. How many yuan did Xiao Li sell all the shares?
How many kilograms of water should be evaporated from 40 kilograms of salt water containing 16% salt to make 20% salt water?
The weight of banana is 48% of that of apple. The weight of pear is 7 / 12 of that of apple. It is known that the imported banana is 28.8kg. The price of banana is 3 yuan per catty, that of pear is 1 / 3 of that of banana, and that of apple is 4 times of that of pear, In the process of sale, there is a 5% loss due to decay and other reasons. If all the fruits are sold, how much money does the store earn (the result is rounded to retain the whole yuan)?
The yield per square meter of peanut planted in an agricultural park in our city the year before last was 2kg, and the oil yield was 40%. After the new peanut varieties were planted last year, the yield per square meter was 2 / 3 times that of the year before last, and the oil yield reached 50%
(1) How many kilogrammes of peanut oil can be extracted from every square meter of peanut produced in the year before last year and the year before last?
(2) This agricultural park squeezes all the peanuts produced in that year into peanut oil every year. Last year, 400 square meters of peanuts were planted, and the peanut oil extracted is 25% more than that of the previous year. How many square meters of peanuts were planted in the previous year?
(3) Under the condition of (2), it is known that the planting cost of peanuts in the previous year and last year was 10 yuan per square meter. Last year, the price of peanut oil was 30 yuan per kilogram, equivalent to 6 / 5 of the price of peanut oil in the previous year. Moreover, all the peanut oil squeezed in that year was sold at the end of each year, How much of the profit from the sales of peanut oil last year accounted for in the profit from the sales of peanut oil last year (profit from the sales of peanut oil = money from the sales of peanut oil - cost of planting peanut)?
Thank you for your good answer!

1. Let the selling price be x and solve the equation
X*5000-(10*5000*(1+0.0035)-X*5000*(0.002+0.0035)=9395
X*5000-X*5000*0.0055=9395+(10*5000*(1+0.0035)
5000X-5000X*0.0055=59570
X(5000-5000*0.0055)=59570
X=59570/4972.50
X=11.98
Selling price: 11.98 yuan
2. It should evaporate x kg of water
16％×40=﹙40－x﹚×20％
x=8
4. (1) the year before last: 2 × 40% = 0.8 (kg / m2),
last year:
=1.5 (kggm2),
A: the peanut planted the year before last can produce 0.8kg peanut oil per square meter. The peanut planted last year can produce 1.5kg peanut oil per square meter
(2) Peanut oil pressed last year: 1.5 × 400 = 600 (kg),
Species of the year before last: 600 (1 + 25%) / 0.8,
=600÷1.25÷0.8,
=480÷0.8,
=600 square meters;
A: 600 square meters of peanuts were planted the year before last
(3) Last year's cost: 10 × 400 = 4000 yuan,
Last year's income: 30 × 600 = 18000 yuan,
Previous year's income: 18000
=18000×
=15000 yuan,
The cost of the previous year: 10 × 600 = 6000 yuan,
The profit from sales of peanuts in the previous year accounted for 1% of the profit from sales of peanut oil in the previous year
（15000-6000）÷（18000-4000）,
=9000÷14000,
=
；
A: the profit from selling peanuts the year before last accounted for 20% of the profit from selling peanut oil last year
．

Application of percentage of quadratic equation of one variable
There are 20 pieces of clothes to be treated in Yuyu shopping mall. The original price of each piece is 50 yuan. The shopping mall decided to reduce the price. After selling 10 pieces, in order to speed up the recovery of funds, it decided to reduce the price again by the same margin. As a result, they sold out quickly and recovered 855 yuan in total
Calculate the percentage of the two price cuts
Calculate the unit price of the last 10 pieces of clothes

Let the percentage of the two price cuts be X
50(1-x)*10+50(1-x)^2*10=855
(1-x)^2+(1-x)-1.71=0
1-x = (- 1 + 2.8) / 2 = 0.9, (negative root omitted)
x=0.1
That is, the price reduction rate is 10%
(2) The price of the last ten pieces is: 50 * (1-10%) ^ 2 = 40.5 yuan

How to solve the problem of positive and negative proportion

The first thing is to make clear whether it is a positive proportion or an inverse proportion. There must be two variables constituting the proportional relationship. A constant maintains the proportional relationship: positive proportion: Generally speaking, the quotient of the two variables is certain. For example, the width b of your house must be 7 meters (constant). The longer the length a is, the larger the area s is. The area is the length

Use a 72cm iron wire to make a cube frame with equal edge length. What is the total edge length of the cube?
The ratio of length, width and height of a cuboid is 5:4:2, in which the length is 20cm. Find the cuboid: 1 the total length of all edges; 2 the surface area; 3 the volume
A boy has finished 160 pages of a story book and he finds that there is 20% of the book left .How many pages are in the book?
Simple warehouse with length, width and height of 20m, 4m and 3M respectively. It can put TV boxes with length, width and height of 1m, 0.8m and 0.8m at most?
Now there is a 30 cm long and 20 cm wide rectangular cardboard. Can we make a 5 cm high uncovered rectangular box? If so, try to find out the volume of the box?

1. There are 12 edges in the cube, 72 ／ 12 = 6cm, each edge is 6cm long, the sum of edge length is: 6 × 12 = 72cm 2, set the height as 2x, width as 4x, length as 5x. 5x = 20, x = 20 ／ 5 = 4, 2x = 8, 4X = 16, so the cuboid is 8cm high, 16cm wide, 20cm long, the sum of edge length = 8cm × 4 + 16cm × 4 + 20cm × 4 = 176cm surface area