The concentration question is not a fractional problem

Solution: There is a type of concentration ratio problem in the fraction application problem, which is the concentration problem. We know that when you put sugar in water, you can get sugar water, in which the sugar is called solute, water is called solute, and sugar water is called solvent.

If the amount of sugar does not change, the more sugar added, the sweeter the sugar water will be, which means that the sweetness of the sugar water is determined by the ratio of the mass of sugar (solute) to sugar water (solution = sugar + water). This ratio is called the sugar content or sugar content of the sugar water, which is the concentration. Therefore, the concentration is the ratio of the solute mass to the solvent mass, which is generally expressed as a percentage (or a fraction), and is calculated as follows:

Concentration = solute mass Solution mass 100 = solute mass (solute mass + solute mass) 100

Example: If you add 400 grams of water to 80 grams of salt solution, your concentration will be reduced by 10How many grams of this solution is there and what is its concentration?

Analysis: The original solution was 80 grams, and the total mass of the solution became 480 grams after adding 400 grams of water; If the concentration of the original solution is x, then the concentration of the new solution becomes (x-10); But the solute does not change in the old and new solutions. According to the formula "solute = solution concentration", because the solute always does not change from the old and new solutions, the equation is columned.

Solution: Let the concentration of the original solution be x, then the concentration after adding water is (x-10), and the equation is:

80 times x = (80 + 400) times ((x-10), solve this equation to get: x = 12, 80 times 12 = g), g).

Answer: This solution has a concentration of 12 grams of water.

I think it's exciting, and I don't need to perfect it.

In the concentration problem, there is also the problem of adding salt or water multiple times. The solution to this kind of problem is the same as the problem of adding salt or water once, except that it is necessary to determine the invariant many times, deal with the conditions step by step, and solve the problem step by step.

Example 1A glass of pure milk, drink 20%. Fill with water and stir well, and after drinking 25% more, the pure milk in the cup accounts for ()% of the volume of the cup

Analysis: If you start drinking 20% of pure milk, you will have 80% of pure milk left. Fill up with water and drink 25%, which is 25% of the water and 25% of the pure milk.

In other words, what is 75% of 80% of the remaining pure milk? This kind of question reflects the quantitative relationship of "dividing and re-dividing" in the fraction and percentage application questions.

Example 2Pour 40g of brine from a 100g cup filled with 80% saline and fill up with water. After stirring, pour out another 40g of salt water, and then top up with water. At this point, the concentration of salt water in the cup is ()%

Analysis: The first 40g of brine contains both salt and water, and the key is to find out the pure salt left in the cup each time. Then fill up with water, and the total amount of salt water is always 100g.

Each time the remaining pure salt in the cup is removed with 100g, the concentration of each new brine is obtained.

100 40) 80% 60 Salt left after pouring 40g of brine for the first time.

48÷（40＋60）＝48%……The concentration of new brine after filling with clean water.

100 40) 48% 60 The remaining salt after pouring out 40g of brine for the second time.

The concentration of the new brine after filling it with water again.

Example 3A and B have two identical cups, cup A has only half a cup of water, and cup B is filled with 30% salt water. Pour half of cup B of brine into cup A, shake well, and then pour half of cup of brine into cup B.

At this time, the concentration of the brine in cup A is ()% and the concentration of the brine in cup B ()%.

Analysis: The key to pouring two cups over each other is to find the pure salt left in the cup after each pour. And the total amount of brine after each pour is a full glass.

The unit "1" is definite. The quantitative relationship of such problems is complex, how can it be accurately analyzed? With the help of the list method, we can solve the problem step by step and solve the problem.

Set 20% alcohol 3x

Then 30% alcohol x

Then 20%*3x+30%*x+(45-4x)*45%=35%*45x=5 liters.

20% alcohol 15 liters.

30% alcohol 5 liters.

45% alcohol 25 liters.

Twenty percent 15 liters.

Thirty-five percent liters.

Forty-five percent 25 liters.

Set 23 percent liters.

35-20)*3x+(35-30)*x+=(45+35)*(45-4x)x=5

A and B walked in the opposite direction from A and B and met for a few hours. Five hours later, car B arrived at place A, and car A not only walked the whole distance, but also traveled an additional 45 kilometers, which accounted for exactly 25% of the distance between A and B. Q: How many kilometers does car A travel per hour?

I want to drip ......Hee-hee.

The basic idea is that the quality of the solute remains unchanged before and after mixing or dilution. For example, in the first question, if the hail is mixed with a concentration of 80% and a concentration of 40% can be mixed according to x:1 to obtain a potion with a concentration of 50%, then x 80%+1 40%=(x+1) 50%, and the solution is x=1 3, so it is 1; 3.

2.Let the first 45% of the brine be 100 grams, add x grams of water to dilute into a concentration of 36% brine, then 100 45% (100+x)=36%, and get x=25, and then add the same amount of water, the mass of the solution becomes 150 grams, and the concentration of the brine will become 100 45% 150=30%.

Solution x=520

4.Let the mass of brine containing 9% salt be x, (130 5% + x 9%), and the solution is x=70, and the prepared brine is 130+70=200 grams.

300 + 250 + 200) * 50% = 375 (g) 200 + 150 + 200) * 80% = 440 (g) 440-200 = 240 (g).

375-240 = 135 (grams).

300-200) + (250-150) = 200 (grams) then 200 grams of sulfuric acid for each of the two kinds of sulfuric acid A and B, then containing pure sulfuric acid is a disadvantage:

135*2=270 (grams).

200 + 200) - (200 + 150) = 50 (g) 270-240 = 30 (g) hail so that the concentration of B sulfuric acid is: 30|50=60%

The concentration of methyl sulfuric acid is: 440-(60%*150+200)=150 (grams).

The quality of B salt water is X times that of A. then the ratio of A and B is 1:x.

20%+x*80%)/1+x)=45%

Solution. x=5/7

1:x=1:(5/7)=7:5

The ratio of A and B brine after mixing the bending macro is 7:5

Let the concentration of methyl sulfuric acid be x, and the concentration of acetic sulfuric acid shall be y.

The concentration is equal to the solution of the bridge chain solution, so the solute is equal to the concentration * of the solution.

300x+250y)/(300+250+200)=50%,(200x+150y+200)/(200+150+200)=80%

Solution: x=75%, y=60%.

1.How many grams of water should be poured into container A with 300 grams of salt water with a concentration of 8% and container B with a concentration of 120 grams of salt water in container B.

300* Let x be the mass of water, x = 180g

2.Add 5 kg of salt to a vat of brine with 10% salt, and after dissolving, the concentration of brine in the drums increases to 20%, how many kg of salt is in the drum?

Cup A is filled with 40 kg of brine containing 20% salt, cup B is filled with 60 kg of brine containing 4% salt, now take out some brine from cup A and put it into cup C, and then take some brine from cup B and put it into cup D, and then pour all the salt water of cup D into cup A, pour all the brine in cup C into cup B, and as a result, cup A and cup B become two cups of brine with the same salt concentration, if it is known that the weight of brine taken out from cup B and poured into cup D is 6 times the weight of the brine taken out from cup A and poured into cup C, Try to determine how many kilograms of salt water is taken out of cup A and poured into cup C? Dissolve the saline taken from cup A and poured into cup C x kg.

20%（40-x）+4%*6x]/(40-x+6x)=[(60-6x)4%+20%x]/(60-6x+x)

20（40-x）+4*6x]/(40-x+6x)=[(60-6x)4+20x]/(60-6x+x)

800-20x+24x)/(40+5x)=(240-24x+20x)/(60-5x)

800+4x)/(40+5x)=(240-4x)/(60-5x)

200+x)/(8+x)=(60-x)/(12-x)

60-x)(8+x)=(200+x)(12-x)

480+60x-8x-x^2=2400+12x-200x-x^2

240x=1920

x = 8 A 8 kg of saline taken from cup A and poured into cup C.

Some data is different.

Set up x kg of brine taken from cup A and poured into cup C.

The solution is x=8 (kg).

[x=8A: The amount of salt water taken from cup A and poured into cup C is 8 kg.

1 question. 40×

2 questions. 40 (1-8 Nazhong) = (40+x) (1-20 state resistance) 40 (1-8 ) 1-20 Dongji Mountain) -40 = 6 (kg).

Solution: Add x grams of 38% syrup.

Then 8000*20%+38%x=(8000+x)*25% according to the sugar phase, then the equiseries equation is called.

Xie Sun Kaide x = 40000 13