Ask for the 6th grade math saline mixing problem exercises

40 of the salt water" is called "alpha salt water", 10 of the "salt water" is called "B salt water", and 20 of the salt water is called "propylene salt water".

What is the weight ratio of alpha saline and beta saline?

Answer: 30 10 ):40 30 ) 2:1 What is the weight ratio of brine and propylene brine after mixing A and B?

Answer: (25 20): 30 25) 1:1

959.Dissolve 12 grams of table salt in 600 grams of water, and the table salt is water ( ) 96020 grams of salt dissolved in 1000 grams of water, salt accounts for ( ) 961

8.Add 25 grams of salt to 100 grams of water, then the salt content of the brine is ( ) 962To prepare a brine, put 5 grams of salt in 120 grams of water, the weight ratio of salt to brine is ( ).

963.Add 50 grams of salt to 200 grams of water, and the salt content of the brine is ( ) 964Add 40 kg of salt to 200 kg of water, and the salt content of the brine is ( ) 965

The salt content of brine is that ( ) grams of water are required to prepare 200 grams of brine.

966.Add 200 grams of sugar to 2 kg of water, and the sugar accounts for 967 of the sugar water10 grams of sugar dissolved in 100 grams of water, sugar accounts for 968 of the sugar water

Dissolve 5 grams of salt into 50 grams of water, and the salt accounts for 969 of the brine5 kg of salt is dissolved in 100 kg of water, and the salt content of brine is ( ) 970The salinity of a brine is 10%, and the ratio of salt to water is ( ).

971.Xiaoli prepares a glass of sugar water for her mother every day, and among the following four days, ( The sugar water is the sweetest.

a.On the first day, the ratio of sugar to water was 1:9.

b.The next day, 20 grams of sugar is mixed with 200 grams of sugar water.

c.On the third day, add 20 grams of sugar to 200 grams of water.

d.On the fourth day, the sugar content was 12%.

972.Put 10 grams of salt in 100 grams of water, and the weight of the salt accounts for the weight of the salt water ( ).

500 grams of existing 5% brine, after adding some 8% brine, the brine becomes ( ).

A, unchanged B, lightened C, thickened.

How much water do you need to add to a cup of 20% saline and weighing 2 kg to turn it into brine with a salt content of 15?

100 grams of salt water, the salt content is 10%, add 5 grams of salt, what is the salt content? (Retain three significant digits).

The ratio of one cup of sugar to water is 1; 5.。After drinking half a cup of sugar water and filling it with water, what is the ratio of sugar to water in the cup?

The key to solving the brine problem lies in the following two points:

1. Memorize and understand the concentration calculation formula

Salt content Weight of salt The weight of salt water or spring belt.

The weight of salt water The weight of salt The weight of water.

2. Distinguish between the amount of change and the amount of unchanged in the brine.

There are three quantities of salt, water, and brine in salt water, and the amount that remains unchanged is the key to solving the problem.

For example, if salt is added to the brine, the salt changes, and the brine changes accordingly, and the water does not change.

When water is added to the brine, the water changes, the brine changes accordingly, and the salt does not change.

The brine evaporates and the water changes, the brine also changes, and the salt does not change.

For example, if there are two bottles of brine with the same weight, the ratio of salt to water in the first bottle is 1:3, and the ratio of salt to water in the second bottle is 3:2. Nanshan Shuangshan Luyu True Volume).

Analysis: The amount of salt and water in this question is not the same, but the weight of the two bottles of salt water is the same, so the weight of salt water is the key to solving the problem.

First bottle of salt: 1:3 water, 1 part salt, 3 parts water, 4 parts brine.

Second bottle of salt: 3:2 for water, 3 parts salt, 2 parts water, 5 parts for brine.

Because the weight of the two bottles of brine is the same, the number of portions of the two bottles of brine can be changed to the same by finding the minimum common multiple of the first of the forest, and the least common multiple of 4 and 5 is 20, so it can be turned into 20 parts, so.

First bottle of salt: water 1:3=5:15, 5 parts salt, 15 parts water, 20 parts salt.

Second bottle of salt: water 3:2=12:8, 12 parts salt, 8 parts water, 20 parts salt water.

After mixing, there are 5 + 12 = 17 parts of salt, 15 + 8 = 23 parts of water, and 20 + 20 = 40 parts of brine.

So Salt: Water 17:23

Method Formula:The quality of the salt The mass of the brine x 100% = the concentration of the brine.

Mass of salt: (brine quality - mass of salt) = mass ratio of salt to water.

For example:

Calculate how much salt and how much water is in the original brine.

The brine weighs 25 kg and contains 20% salt.

So the salt content is 25*20%=5kg, and the water content is 25-5=20kg.

From the amount of salt content now, it is calculated as the amount of salt water and the amount of water.

Salt water 5 8% = now water.

Subtracting the original water content from the current water content is equal to the added water.

So add water. Example 2: "40 brine" is called "alpha saline", "10 brine" is called "B saline", and "20 brine" is called "propylene saline".

What is the weight ratio of alpha saline and beta saline? Answer: 30 10): 40 30) 2:1.

What is the weight ratio of brine and propylene brine after mixing A and B? Answer: (25 20): 30 25) 1:1

Salinity = Salinity (Salinity + Moisture).

60 grams of salt water with 20 percent salt content, 60 * grams of salt, 60 * grams of water, the water content is unchanged, when it becomes 25 percent of the salt water, the water content is, so the total amount of salt water after the change is 48 grams, that is, the amount of salt added is 64-60 = 4 grams.

Water does not change 60*20%=12 (grams).

60-12 = 48 (grams).

48 75% = 64 (grams).

64-60 = 4 (grams).

Equation: Solution: Let x grams of salt be added.

12+x)/(60+x)=25%

4(12+x)=60+x

48+4x=60+x

3x=12x=4

First, calculate the mass of the total brine: 60 20% = 300 grams in, 300 grams 25% = 75 grams.

75 g - 60 g = 15 g.

So, add another 15 grams of salt water.

According to the first condition, it is solved: 60x20%=12g (salt), so the salt and water are 12g and 48g respectively.

Add xg of salt. According to the second condition, the equation is set: (12+x) divided by (60+x)=.

x=4。

Pour water once: pour out 40g of salt water and leave 60g, pour in water and restore 100g, so the concentration becomes 60% of the original.

After three times, the concentration is 80%*60% 3=

Option 1: Add salt.

10% salt.

8 liters of salt water in 8 liters of water content 8 * (1-10%) = liters.

There are liters of salt water containing 20% salt.

Salt needs to be added 9-8 = 1 liter.

Scheme 2: Evaporate water.

Salt content 8 * 10% = liters.

There are liters of salt water containing 20% salt.

Need to evaporate water 8-4 = 4 liters.

20 grams of salt are put in 500 grams of water to make a kind of brine, and after using 150 grams now, 200 grams of water and 10 grams of salt are added to the remaining brine. Answer: 20 grams of sugar in 500 grams of water.

It was made into a kind of sugar water.

Now drink 150 grams.

In the remaining sugar water, another 200 grams of water and 10 grams of white sugar were added, and the sugar content was 20 500 = 4%.

Now the mass of sugar in the sugar water (500-150) * 4% + 10 = 24 Now the sugar content is 24 (200 + 500-150) = the current water is sweet.

Set the original salt x grams, water 9x grams.

x+5):(9x)=1:5

x = 25 4 = grams.

In the original brine, water accounts for 100 9 (1+9) 90 grams, salt is 10 grams, to make the water-salt ratio is 6:1, 90 6 1 15 grams of salt are needed, 15-10 5 grams, 5 grams of salt are added.