a scientist needs 10 liters of a 20% acid solution for an experiment, but she has only a 5% solution and a 40% solution. to the nearest tenth of a liter, about how many liters of the 5% and the 40% solutions should she mix to get the solution she needs? write and solve an equation to match the situation.

To prepare 10 liters of a 20% acid solution by mixing only a 5% acid solution and a 40% acid solution, let:

• xxx = liters of the 5% acid solution
• yyy = liters of the 40% acid solution

Since the total volume needed is 10 liters, the first equation is:

x+y=10x+y=10x+y=10

The total amount of pure acid in the final mixture must be 20% of 10 liters, which is 2 liters. The amount of acid from each solution is the concentration multiplied by the volume, so the second equation is:

0.05x+0.40y=20.05x+0.40y=20.05x+0.40y=2

Solving the system: From the first equation:

y=10−xy=10-xy=10−x

Substitute into the second equation:

0.05x+0.40(10−x)=20.05x+0.40(10-x)=20.05x+0.40(10−x)=2

0.05x+4−0.40x=20.05x+4-0.40x=20.05x+4−0.40x=2

−0.35x+4=2-0.35x+4=2−0.35x+4=2

−0.35x=−2-0.35x=-2−0.35x=−2

x=20.35≈5.71 litersx=\frac{2}{0.35}\approx 5.71\text{ liters}x=0.352​≈5.71 liters

Then,

y=10−5.71=4.29 litersy=10-5.71=4.29\text{ liters}y=10−5.71=4.29 liters

Answer:
The scientist should mix approximately 5.7 liters of the 5% acid solution and 4.3 liters of the 40% acid solution to get 10 liters of a 20% acid solution