6th grade percentage math problem concentration problem

1) 300*13%+700*7%=88g- 7%*700=

Answer: The same concentration of A and B is 210 grams from A and B respectively.

2) 15%*200=30g

10%*100=10g

30-10)/(200/100-1)=20g20-10=10g

A: 10 grams of salt should be added.

3)14/ 35% =40kg

800-40 =760kg

Answer: 40 kg of pesticides with a concentration of 35% and 760 kg of water can be used to prepare 800 kg of pesticides with a concentration of 800 kg.

4) 30*15%=liters.

20%-15%) = 90 liters.

90 + 30 = 120 liters.

120*25%=30 liters.

90*20%=18 liters.

30-18) (1-25%) = 16 liters.

A: Add another 16 litres of alcohol and the concentration becomes 25%.

5) 150+450=600g

150* 4% = 6g

Answer: The concentration of brine in container B is:

6) Suppose there is 1 part salt in the brine.

1 3% = 100 3 servings.

1 2% = 50 servings.

50-100 3 = 50 3 servings.

Answer: The concentration of the brine after adding the same amount of water for the third time is.

7) 100-40=60g for the first time

60*80%=48g

2nd 100-40=60g

3rd 100-40=60g

60*A: The concentration of the brine in the cup is.

8) 8%*300=24g

300-120=180g

180/60%=300g

300-120=180g

A: Each container should be poured with 180 grams of water.

Hopefully LZ can give a few more points.

One. Set x grams from A and B respectively.

13 (300 x) 7 x = 300 times.

x = 210 two (200·15% x) (200 x) (100·10% x) (100 x).

x = three pesticides 800 kg water 800 40 760 kg.

Fourth, the original alcohol solution is X liters.

20％x÷（x＋30）=15％

x = 90 90 (1 20 ) 30] (1 25%) 90 30 16 liters.

Suppose x grams of alcohol solution A and 270-x grams of alcohol solution B are taken, which are derived from the title.

95 x+80 (270-x)=85 270 gives x=90

When x=90, 270-x=180

Answer: 90 grams and 180 grams should be taken from the two alcohol solutions of A and B.

If you can set a type to take xg, then type B is (270-x)g

95%x+80%(270-x)=270*85%The solution yields x=90g

B: (270-90)*80%=180g