A scientist needs ten liters of a 20 percent acid solution for an experiment but she has only a five percent solution and a forty percent solution. To the nearest tenth of a liter about how many liters of the five percent and the forty percent solutions should she mix to get the solution she needs?

We need 5.7 liters of the 5% volume solution and 4.3 liters of the 40% volume solution.

Step-by-step explanation:

Let V₁ be the volume of the 5% solution required and V₂ be the volume of 40% solution required.

Since mass = concentration × volume, the combined mass of the 5% and 40% volume solution equal the mass of the 10 liter 20% volume solution.

So, 0.05V₁ + 0.4V₂ = 0.2 × 10

0.05V₁ + 0.4V₂ = 2 (1)

Also, the total volume of the solution must equal 10 liters. So,

V₁ + V₂ = 10 (2)

V₁ = 10 - V₂  (3)

substituting (3) into (1), we have

0.05(10 - V₂) + 0.4V₂ = 2

0.5 - 0.05V₂ + 0.4V₂ = 2

collecting like terms, we have

- 0.05V₂ + 0.4V₂ = 2 - 0.5

0.35V₂ = 1.5

dividing through by 0.35, we have

V₂ = 1.5/0.35

V₂ = 4.29

V₂ = 4.3 liters to the nearest tenth of a liter

Substituting V₂ into (3), we have,

V₁ = 10 - 4.3  

V₁ = 5.7 liters

So, we need 5.7 liters of the 5% volume solution and 4.3 liters of the 40% volume solution.

(a) the volume of the cone is 20/3 in3

hence, option a is correct.

529 and 531

step-by-step explanation: