SAT Ratios, Rates & Proportions Question 48 | Free SAT Prep | Aniko

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Question 48·Hard·Ratios, Rates, Proportional Relationships, and Units

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An aquarium manager needs exactly 200 liters of seawater with a salinity of 42 parts per thousand (ppt). A 20-liter jug of pure freshwater (0 ppt) is accidentally emptied into an empty mixing tank. To reach the desired volume and salinity, the manager will add only a 35 ppt seawater solution and a 50 ppt seawater solution to the tank.

How many liters of the 50 ppt solution must the manager add?

Your answer

(Express the answer as an integer)

For mixture and concentration problems on the SAT, always define variables for the unknown amounts of each solution, then write two key equations: one for total volume (all parts must add to the final volume) and one for total amount of the substance (like salt), using concentration × volume. Solve the resulting linear system with substitution or elimination, and be especially careful to include components with 0 concentration in the volume equation (they still contribute volume even though they add no substance).

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Relate the unknown amounts to the total volume

Let $x$ be the liters of 35 ppt solution and $y$ be the liters of 50 ppt solution. You already have 20 liters of freshwater and need 200 liters total. How should $20$ , $x$ , and $y$ add up?

Think in terms of total salt, not just concentration

The concentration (ppt) times volume gives the total amount of salt contributed by that part. Write an equation where the total salt from the freshwater, 35 ppt solution, and 50 ppt solution equals the total salt in 200 liters at 42 ppt.

Use two equations together

You should now have one equation from the total volume and one from the total salt. Use substitution or elimination to solve this system for $x$ and $y$ , and remember that $y$ represents the liters of 50 ppt solution.

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Enter the volume equation

In the first expression line, type x + y = 180 to represent the relationship between the liters of 35 ppt solution ( $x$ ) and 50 ppt solution ( $y$ ).

Enter the salt equation

In a new line, type 35x + 50y = 42*200 to represent the total salt needed for 200 liters at 42 ppt.

Find the required amount of 50 ppt solution

Look at the intersection point of the two lines. The y-coordinate of this intersection gives the number of liters of the 50 ppt solution needed.

Step-by-step Explanation

Define variables and use the total volume

Let $x$ be the number of liters of the 35 ppt solution and $y$ be the number of liters of the 50 ppt solution.

You already have 20 liters of freshwater, and the final mixture must be 200 liters total. So the volumes must satisfy

$20 + x + y = 200$ , which simplifies to $x + y = 180$ .

Write an equation for total salt in the mixture

The total amount of salt in the final mixture comes from the salt in each component.

The 20 liters of freshwater contribute $0$ salt because they are 0 ppt.
The 35 ppt solution contributes $35x$ (in "ppt-liters" units).
The 50 ppt solution contributes $50y$ .

The final mixture is 200 liters at 42 ppt, so it contains $42 \cdot 200$ units of salt. This gives the equation

35x + 50y = 42 \cdot 200

You can compute $42 \cdot 200 = 8400$ , so the salt equation is $35x + 50y = 8400$ .

Solve the system of equations

Now solve the system

x + y = 180

35x + 50y = 8400

From $x + y = 180$, express $x$ in terms of $y$:

x = 180 - y

Substitute this into the salt equation:

35(180 - y) + 50y = 8400

Compute and simplify: $35 \cdot 180 = 6300$, so

6300 - 35y + 50y = 8400

6300 + 15y = 8400

15y = 8400 - 6300 = 2100

y = \dfrac{2100}{15} = 140 $$ Since $y$ is the liters of 50 ppt solution, the manager must add **140 liters** of the 50 ppt solution.

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Step-by-step Explanation

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Step-by-step Explanation