Mixture Problems (Algebra)

How many ounces of a 30% acid solution should be mixed with 40 ounces of a 2% acid solution to make a 5% acid solution?

• 1.2 ounces

• 12 ounces

• 2 ounces

• 4.8 ounces

Tags

CCSS.HSA.CED.A.1

A 40 lb. bag of gravel that was 80% small stones by weight was mixed with another bag that was 60% small stones by weight to produce a new mixture that was 67% small stones by weight.

What was the weight of the other bag? Round to the nearest tenth of a lb.

• 62.8 lbs.

• 81.6 lbs.

• 52.9 lbs.

• 74.3 lbs.

Tags

CCSS.8.EE.C.7B

A chemist has 60 g of solution that is 70% acid.  how much water should be added to make a solution that is 40% acid?

• 40 g water
• 50 g water
• 45 g water
• 15 g water

Tags

CCSS.8.EE.C.7B

How many kilograms of pure alcohol must be added to 40 kg of a 70% alcohol solution to produce an 85% alcohol solution?

• 30 kg alcohol
• 40 kg alcohol
• 50 kg alcohol
• 35 kg alcohol

Tags

CCSS.HSA.CED.A.1

Which equation could help you find how many ounces of 30% acid solution should be mixed with 50 ounces of 2% acid solution to make 3% acid solution?

•  $\frac{0.02\cdot50+0.30x}{50+x}=\frac{3}{100}$  

•  $\frac{0.30x}{50+x}=\frac{3}{100}$  

•  $\frac{0.30x}{50+0.02x}=\frac{3}{100}$  

•  $\frac{0.02\cdot50+0.30x}{50+3}=\frac{30}{100}$  

Tags

CCSS.HSA.CED.A.1

A chemist has 80 g of solution that is 65% acid. Which equation below could help you find how much water should be added to make a solution that is 45% acid?

• $\frac{0.65\cdot80}{80+x}=\frac{45}{100}$

• $\frac{0.65\cdot80+x}{80}=\frac{45}{100}$

• $\frac{0.65\cdot80+x}{80+x}=\frac{45}{100}$

• $\frac{80+0.65x}{80}=\frac{45}{100}$

A chemist has 80 g of solution that is 65% acid. Which equation below could help you find how pure acid should be added to make a solution that is 85% acid?

•  $0.65\cdot80+x=0.85\left(80+x\right)$ 

•  $0.65x+80=0.85\left(80+x\right)$ 

•  $80x+0.65\cdot80=0.85\left(80+x\right)$ 

•  $0.85\cdot80+x=0.65\left(80+x\right)$ 

A chemist has 80 g of solution that is 65% acid. How much pure acid should she add to this solution in order for it to become 90% acid?

• 200 g.

• 106.7 g.

• 100 g.

• 86.7 g.

Tags

CCSS.8.EE.C.7B

What amount of 20% salt solution should be mixed with some 8% salt solution to make 500 ml of a 15% salt solution?

• 291.7 ml

• 208.3 ml

• 216.9 ml

• 242.8 ml

Tags

CCSS.8.EE.C.7B

Which equation could help you find how much 20% salt solution should be mixed with some 8% salt solution to make 500 ml of a 15% salt solution?

•  $0.2x+0.08\left(500-x\right)=500\cdot0.15$  

•  $0.2x+0.8\left(500-x\right)=500\cdot0.15$  

•  $0.2x+0.08\left(x-500\right)=500\cdot0.15$  

•  $0.2x+0.8\left(x-500\right)=500\cdot0.15$