Question

/\/49:При помоле ржи на каждые 3 части муки получается 1 часть отходов. Сколько центнеров ржи смололи, если муки получилось на 36 ц. больше, чем отходов? /\/50:а)Купили 1800 г. сухофруктов. Яблоки составляют 4 части, груши-3 части и сливы-2 части массы сухофруктов. Сколько граммов яблок, груш и слив в отдельности купили? б)Яблоки составляют 7 частей, груши-4 части, а сливы-5 частей массы сухофруктов. Сколько граммов яблок, груш и слив в отдельности содержится в 1600 г. сухофруктов? /\/51:Для компота взяли 6 частей яблок, 5 частей груш и 3 части слив. Оказалось, что груш и слив вместе взяли 2 кг. 400 г. . Определие массу взятых яблок; массу всех фруктов. /\/52:1)При изготовлении кофейного напитка «Ячменный» на 4 части ячменя берут 1 часть цикория. Сколько пачек напитка изготовлено, если каждая пачка весит 250 г. и на изготовление партии напитка израсходовано ячменя на 36 кг. больше, чем цикория. 2)При изготовлении кофейного напитка «Наша марка» на 7 частей кофе берут 6 частей цикория, 5 частей желудей и 2 части каштанов. Сколько пачек напитка изготовлено, если каждая пачка весит 200 г., а кофе и цикория вместе израсходовали 26 кг.? /\/53:1)Сплав содержит 1 часть свинца и 2 части олова. Во сколько раз в этом сплаве олова больше, чем свинца? 2)Сплав содержит олова в 3 раза больше, чем свинца. Сколько частей олова приходится на 1 часть свинца?

Answer 1

№50 а) 4x+3x+2x=1800 
9x=1800 
x=200 
1) 4*200=800 (г)-яблок
2) 3*200=600 (г)-груш
3)2*200=400 (г)-слив                                                                                                           б)  7x+4x+5x=1600                                                                                                   16x=1600                                                                                                                    x=100                                                                                                                   7*100=700 (г) - яблок                                                                                            4*100=400 (г) - груш
5*100=500 (г) - слив.                                                                                                №51. Всего 2 кг 400 г (2400 г) груш и слив вместе. 
5+3=8 частей груш и слив. 
2400:8=300 (г) – одна часть 
6*300=1800 (г) – купили яблок 
6+5+3=14 частей сухофруктов.                                                                             14*300=4200 (г) – купили фруктов                                                                           Ответ: купили 1800 г. яблок; 4200 г. фруктов.                                                         №52.1) 4х-х=36
3х=36
х=12 - Вес цикория
4*12=48  - Вес ячменя
Вес всей партии = 48+12 = 60кг 
Кол-во пачек = 60000\250 =  240
Ответ: 240                                                                                                                     2)7х+6х=26
13х=26
х=2кг
7*2=14кг-кофе
6*2=12кг-цикория
5*2=10кг-желудей
2*2=4кг-каштанов
14+12+10+4=40кг продуктов израсходовано всего.
200гр = 0,2кг
40:0,2=200 пачек                                                                                                   №53.1) 2/1=2 олова в 2 раза больше                                                                                                                    

Answer 2

Task 49:

To find the number of centners of rye that were ground, we need to determine the ratio of flour to waste. According to the problem, for every 3 parts of flour, 1 part of waste is obtained. It is also given that the amount of flour is 36 centners more than the amount of waste.

Let's assume that the amount of waste is x centners. Then the amount of flour would be x + 36 centners.

To find the value of x, we can set up the following equation:

3x = x + 36

Simplifying the equation:

2x = 36

x = 18

Therefore, the amount of waste is 18 centners, and the amount of flour is 18 + 36 = 54 centners.

Answer: The mill ground 54 centners of rye.

Task 50:

a) To find the amount of each type of fruit purchased, we need to determine the ratio of apples, pears, and plums in the total mass of dried fruits. According to the problem, apples constitute 4 parts, pears constitute 3 parts, and plums constitute 2 parts of the total mass of dried fruits.

Let's assume that the total mass of dried fruits is x grams. Then the amount of apples would be (4/9) * x grams, the amount of pears would be (3/9) * x grams, and the amount of plums would be (2/9) * x grams.

Answer: - The amount of apples purchased: (4/9) * 1800 grams = 800 grams. - The amount of pears purchased: (3/9) * 1800 grams = 600 grams. - The amount of plums purchased: (2/9) * 1800 grams = 400 grams.

b) To find the amount of each type of fruit contained in 1600 grams of dried fruits, we need to determine the ratio of apples, pears, and plums in the total mass of dried fruits. According to the problem, apples constitute 7 parts, pears constitute 4 parts, and plums constitute 5 parts of the total mass of dried fruits.

Let's assume that the total mass of dried fruits is y grams. Then the amount of apples would be (7/16) * y grams, the amount of pears would be (4/16) * y grams, and the amount of plums would be (5/16) * y grams.

We are given that the total mass of dried fruits is 1600 grams. Substituting this value into the equation, we can solve for y.

(7/16) * y + (4/16) * y + (5/16) * y = 1600

Simplifying the equation:

16y = 1600 * 16

y = 1600

Therefore, the amount of apples contained in 1600 grams of dried fruits is (7/16) * 1600 grams = 700 grams, the amount of pears is (4/16) * 1600 grams = 400 grams, and the amount of plums is (5/16) * 1600 grams = 500 grams.

Answer: - The amount of apples contained in 1600 grams of dried fruits: 700 grams. - The amount of pears contained in 1600 grams of dried fruits: 400 grams. - The amount of plums contained in 1600 grams of dried fruits: 500 grams.

Task 51:

To determine the mass of the taken apples and the total mass of all fruits, we need to determine the ratio of apples, pears, and plums in the mixture. According to the problem, the ratio is 6 parts of apples, 5 parts of pears, and 3 parts of plums. It is also given that the combined mass of pears and plums is 2.4 kilograms (or 2400 grams).

Let's assume that the mass of apples is x grams. Then the mass of pears would be (5/6) * x grams, and the mass of plums would be (3/6) * x grams.

We are given that the combined mass of pears and plums is 2400 grams. Substituting the values into the equation, we can solve for x.

(5/6) * x + (3/6) * x = 2400

Simplifying the equation:

8x/6 = 2400

8x = 2400 * 6

x = 1800

Therefore, the mass of the taken apples is 1800 grams, and the total mass of all fruits is 1800 + 2400 = 4200 grams.

Answer: - The mass of the taken apples: 1800 grams. - The total mass of all fruits: 4200 grams.

Task 52:

1) To find the number of packs of the "Barley" coffee drink, we need to determine the ratio of barley to chicory. According to the problem, the ratio is 4 parts of barley to 1 part of chicory. It is also given that each pack weighs 250 grams, and the amount of barley used in the production of the batch is 36 kilograms (or 36000 grams) more than the amount of chicory.

Let's assume that the amount of chicory used is x grams. Then the amount of barley used would be x + 36000 grams.

To find the value of x, we can set up the following equation:

4x = x + 36000

Simplifying the equation:

3x = 36000

x = 12000

Therefore, the amount of chicory used is 12000 grams, and the amount of barley used is 12000 + 36000 = 48000 grams.

To find the number of packs, we divide the total mass of the batch by the weight of each pack:

Number of packs = Total mass of the batch / Weight of each pack Number of packs = (12000 + 36000) grams / 250 grams Number of packs = 48000 grams / 250 grams Number of packs = 192 packs

Answer: The number of packs of the "Barley" coffee drink produced is 192.

2) To find the number of packs of the "Our Brand" coffee drink, we need to determine the ratio of coffee to chicory, chestnuts, and horse chestnuts. According to the problem, the ratio is 7 parts of coffee to 6 parts of chicory, 5 parts of chestnuts, and 2 parts of horse chestnuts. It is also given that each pack weighs 200 grams, and the total amount of coffee and chicory used in the production is 26 kilograms (or 26000 grams).

Let's assume that the amount of coffee used is x grams. Then the amount of chicory used would be (6/7) * x grams, the amount of chestnuts would be (5/7) * x grams, and the amount of horse chestnuts would be (2/7) * x grams.

We are given that the total amount of coffee and chicory used is 26000 grams. Substituting the values into the equation, we can solve for x.

x + (6/7) * x = 26000

Simplifying the equation:

13x/7 = 26000

13x = 26000 * 7

x = 14000

Therefore, the amount of coffee used is 14000 grams, the amount of chicory used is (6/7) * 14000 grams = 12000 grams, the amount of chestnuts used is (5/7) * 14000 grams = 10000 grams, and the amount of horse chestnuts used is (2/7) * 14000 grams = 4000 grams.

To find the number of packs, we divide the total mass of the batch by the weight of each pack:

Number of packs = Total mass of the batch / Weight of each pack Number of packs = (14000 + 12000 + 10000 + 4000) grams / 200 grams Number of packs = 40000 grams / 200 grams Number of packs = 200 packs

Answer: The number of packs of the "Our Brand" coffee drink produced is 200.

Task 53:

1) To determine the ratio of lead to tin in the alloy, we need to compare the parts of tin to the parts of lead. According to the problem, the alloy contains 1 part of lead and 2 parts of tin. Therefore, the amount of tin in the alloy is twice the amount of lead.

Answer: The amount of tin in the alloy is twice the amount of lead.

2) To determine the ratio of tin to lead in the alloy, we need to compare the parts of tin to the parts of lead. According to the problem, the alloy contains 3 times more tin than lead. Therefore, the ratio of tin to lead is 3:1.

Answer: The ratio of tin to lead in the alloy is 3:1.