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Mathematical problems of concentration difference
1. Hypothetical method: 85% mass is required:

40%: 10-6=4 (kg)

( 10x 67%- 10x 40%)(85%-40%)

=6.7÷0.45

=6 kg

2. 1500 x(40%-30%)-30%-25%

= 150÷5%

=3000 (grams)

3. The condition should be 10 kg orange juice, right?

(50x 20%)x( 1-20%)x( 1-20%)÷50x 100%

=6.4÷50x 100%

= 12.8%

4.(800 x 10%+500 x20 %)(800+500+200)x 100%

= 180÷ 1500 x 100%

= 12%

5. Water:

Sugar: 37.5 (1-10%) x10%.

(25x 6%)( 10%-6%)

=37.5÷0.9x0. 1

= 1.5÷4%

=25/6 (g)

=37.5 (gram)

6. Need 75% alcohol solution:

2400 times (60%-55%)-75%-55%

Need 55% alcohol solution: 2400-600= 1800 (g)

= 120÷0.2

=600 (grams)

7.650 times (20%- 15%)\( 15%-5%)

=32.5÷0. 1

=325 (gram)

8.430 times (20%-12%) (12%-10%)

=34.4÷0.02

= 1720 (g)