How to solve the problem of math concentration in the sixth grade of primary school

The following are the quantity relationships commonly used in concentration problems:

1. Solution quality, solute quality, solvent quality.

2. Concentration solute mass solution mass 100

3. Solute mass Solute (the percentage of solution occupied by solute) The mass of the solution.

4. Solvent quality Solvent (percentage of solvent in solution) Solution quality.

5. The concentration of the mixed solution: the mass of the solution a, the concentration of the solution a, the mass of the solution b, the concentration of the solution b, the concentration of the solution (the mass of the solution a, the mass of the solution b) 100

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For example, if you prepare a brine and add 20 grams of salt to 480 grams of water, what is the concentration of this brine? The concentration of one brine is 15%, how many grams of salt is contained in 300 grams of this brine? How many grams of water do I need to make 120 grams of salt to make a 15% concentration of brine?

Answer: 1) The concentration of saline: 20 (480 20) 100%=4%2) Salt content: 300 15%=45 (g).

3) Solution: 120 15% = 800 (g) Solvent: 800 120 = 680 (g).

Distinguish between the three concepts of salt, water, and salt water.

The concentration is salt water * 100%.

If you really can't, use the solution equation, list the equation and there are points, and you only learned to solve this kind of equation in the first year, but it is not difficult.

The amount of solute contained in a unit solution is called the concentration of that solution. The higher the solute content, the greater the concentration. The concentration can be calculated in grams of solutes, gram molecules, or gram equivalents of solutes in a given solution. It is generally expressed as a percentage of the weight of the solute contained in a unit solution.

If you think it's good, remember me

Generally: medicine (medicine + water) * 100% or medicine liquid.

Math concentration problem? You give an example.

You go and taste the milk, I think it's quite strong.

Concentration Problem 6th Grade Math Problem Solving Skills are as follows:

Solute, solution, solvent, the three elements of the concentration formula:

Solubility: A substance that is dissolved. For example, the solute in sugar water is sugar, and the solute in brine is salt.

Solvent: Something (mostly liquid) used to dissolve solutes. For example, water in sugar water, water in salt water.

Solution: A mixture of solute and solvent. For example, sugar water, salt water.

The relationship between the three: the mass of the solution = the quality of the solute + the quality of the solvent.

Concentration: The ratio of solute mass to solution mass, often expressed as a percentage.

Basic formula: concentration = solute mass solution mass 100% = solute mass (solute mass + solvent mass) 100%.

Deformation formula: solute mass = solution mass concentration, solution mass = solute mass concentration tung defeat good.

When doing the problem, it is necessary to pay attention to distinguish which solute concentration is being asked for in the question. For example, if sugar and salt are put in water at the same time, if you ask for the concentration of sugar, the sugar is the solute; If you let find the concentration of salt, then salt is the solute.

Basic question type: directly test the use of formulas.

Skills: Review the questions carefully, find the solutes, solvents, and solutions, and directly apply the formula to answer.

Example 1: What is the concentration of salt obtained by putting 50 grams of local lead into 150 grams of water?

Idea: Salt is soluble in water, 50 grams of salt is the solute, 150 grams of water is the solvent, solution = mass of salt + mass of water = 50 + 150 = 200 grams.

Solution: Concentration = 50 (50 + 150) 100% = 25%.

A: The concentration of this brine is 25%.

Example 2: How many grams of water do you need to add to prepare 15 grams of salt to make brine with a salt content of 5%?

Idea: Directly use the deformation formula of the concentration formula to find the mass of the solution, that is, the quality of the brine. Then subtract the amount of salt.

Solution: Mass of brine = solute mass Concentration = 15 5% = 300 (g).

The quality of water = the quality of brine - the quality of salt = 300-15 = 285 (grams).

A: Add 285 grams of water.

Advanced question type: existing solution, change the concentration.

Example 3: Add 100 grams of brine to 400 grams of brine with a salt content of 5%, what is the salt content of the brine at this time?

Idea: Add 100 grams of water, the quality of the salt as a solute remains unchanged, and it can be obtained by 400 5%; The mass of the solution, which was originally 400 grams, is now (400 + 100) grams. Then use the concentration formula to solve it.

Solution: salt content = 400 5% (400 + 100) 100% = 4%.

A: At this time, the salt content of the brine is 4%.

Example 4: If there is 20 kg of brine with 15% salt, how many kilograms of salt do you need to add to make the concentration of brine 20%?

Idea: The mass of the solute and the mass of the solution increase after the addition of salt, and it is easier to understand by using the equation to answer. If x kg of salt needs to be added, then the mass of salt is (20 15%+x) kg, and the mass of the solution is (20+x) kg.

Solution: Suppose you need to add x kg of salt.

20×15%+x）÷（20+x）=20%

Solution: x = A: Kilograms of salt need to be added.

The percentage of concentration in the solution is called the percentage concentration, referred to as the concentration, and the concentration problem is a percentage problem. The main quantitative relationship of the concentration problem is: weight of the solution = weight of the solute + weight of the solvent, concentration = mass of solute 100% of the weight of the solution.

Let's analyze it through some typical examples.

<> example 1: When salt is added to the brine, the weight of the salt changes, but the weight of the water does not change. To solve this similar problem, first find the invariant in the problem, find the invariant, and then find the changed amount of solution (saline).

Example 2 solves the problem of water becoming salt unchanged, the key is that the salt in the original brine does not change after the water evaporates (or a certain amount of water is added). Generally, the amount of salt is found first, and then the current amount of salt water is found according to the current concentration, and the difference between the original salt water and the current salt water is the evaporated water.

Example 3: The weight of the mixed salt is the sum of the weight of the original salt in each solution, the weight of the mixed brine is the sum of the weight of each of the original brine, and the weight of the mixed salt divided by the weight of the mixed brine is the concentration of the mixed brine.

Example 4 sets the unknown according to the equality of the mixed solution, and then according to the equation of the equality of the solute before and after mixing.

Suppose: take x grams from 5% saline, then take 5000-x grams from 15% saline.

x g 5% brine contains 5% salt x, 5000-x g 15% brine contains 15% salt (5000-x), and the total salt is 5000*12%.

5%x+(5000-x)15%=5000*12%0,05x+

x=1500 5000-x=3500 Therefore, 1500 grams should be taken from 5% brine and 3500 grams from 15% brine to prepare.

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1.The concentration of salt water is 700 grams, indicating that it contains grams of salt;

Evaporation will not evaporate the salt, so the resulting concentration is that the salt contained in the brine is still grams, and the filial piety of the Huai people divides the quality of the brine by grams, so 700-500 grams of water are evaporated.

2.1200 grams of saline solution with a concentration of 20% contains 20% * 1200 = 240 grams of salt, after adding 800 grams of water, the quality of salt is still 240 grams, and the total mass of brine is 1200 + 800 = 2000 grams, so the concentration of panicle closure is 240 divided by 2000 = 12%.

In the future, you should pay attention to the invariants in these processes, such as evaporation and adding water, which will not change the quality of the solvent in the solution.

There are 180 grams of sugar water with a concentration of 2 in container A, which means that there are 180*2% grams of sugar.

There are several grams of sugar water with a concentration of 9 in the B container, assuming that there are n grams, then there is 9n % of sugar 9n% from B to take out 240 grams of sugar water and transport it into A, at this time there is sugar water 180 + 240 420 in A, there is sugar, and the concentration of sugar water in A at this time is.

After Qinghong tossed just now, B has sugar water n-240, and sugar 9n%-240*9%=9n%.

The concentration of sugar water in B is (9n% Answer: N-240) and then pour water into B so that there is the same concentration of sugar water in both containers.

At this time, the concentration of sugar water is 6% and 9n% after pouring water

What is the concentration of sugar water in the two containers right now? 6%, of course. Feel.

There is something wrong with this question.

Concentration = mass of salt (mass of salt + water) solution:

Mass of salt = grams.

then the mass of the water is grams.

x = grams. It is necessary to laugh and drain to evaporate the gram of water.

Mass of salt = 20% * 1200 = 240 g picoliter bush.

Concentration = 240 (Burning Sakura 1200 + 800) = 12%.

40% sugar water becomes 30%.

The concentration was reduced by 40%-30%=10%.

The water added, which originally had a concentration of 0, becomes 30% sugar water.

The concentration increased by 30%.

The ratio of 40% sugar water to 5 kg of water added, is:

It turns out that 40% of the sugar water has: 5 3 = 15 kg.

After adding 5 kg of water, a total of 15 + 5 = 20 kg of water is included: 20 (1-30%) = 14 kg.

When the concentration changes to 50%, the water content remains unchanged and the total weight is:

14 (1-50%) = 28 kg.

Add sugar: 28-20 = 8 kg.

Let the weight of the raw sugar water be a kilogram, I will change the % to a decimal and the sugar to be added will be b kilogram a = 15

b+6)/(b+20)=b=8

A concentration: 180 (180 + 120) = 60% B concentration: 150 (150 + 350) = 30% Using the cross method, 42% sugar water is prepared, and the quantity ratio of A and B is required

42%-30%): 60%-42%) = 2:3 Species: 500 (2+3) 2 = 200 grams.

Type B: 500-200 = 300 grams.

Proficient in the cross method, it is very convenient to do concentration and mixing problems.

The concentration of A is: 180 (180+120)=60% The concentration of B is: 150 (150+350)=30%, assuming that the sugar water is: x, then B is 500-x;

Depending on the title, there is the following equation:

60%x+30%(500-x)=500*42%Solve the equation to get x = 200 A: 200 B: 500-200=300

Suppose you take x grams of sugar water.

42% of 500 grams of sugar water is: 210 grams of sugar, 290 grams of water when added to the concentration of sugar water 180 (180 + 120) = 60% = (

Lift x = 200 grams.

B is 500-200 = 300 grams.

It can be seen that the concentration of A is 180 300 = and the concentration of B is 150 500 = If A x grams are taken, then B is 500-x grams.

Columnable equations. x=200, so A is 200 grams and B is 300 grams.