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Difficulty: MediumSolving Linear Inequalities

A chemical solution in a laboratory has an initial volume of 5050 milliliters and evaporates at a rate of 72\frac{7}{2} milliliters per hour. A second chemical solution has an initial volume of 2020 milliliters and evaporates at a rate of 54\frac{5}{4} milliliters per hour. After how many hours, tt, will the volume of the first solution be at most the volume of the second solution?

  1. t≥403t \ge \frac{40}{3}Answer
  2. B
    t≤403t \le \frac{40}{3}
  3. C
    t \ge 90
  4. D
    t \le 90
  5. E
    t≤−403t \le -\frac{40}{3}

Answer

The correct solution is the set of values t≥403t \ge \frac{40}{3}.
Translating the problem statement gives the inequality 50−72t≤20−54t50 - \frac{7}{2}t \le 20 - \frac{5}{4}t. Subtracting 50 from both sides and adding 54t\frac{5}{4}t to both sides results in −94t≤−30-\frac{9}{4}t \le -30. Multiplying both sides by −49-\frac{4}{9} and reversing the inequality sign because of the negative multiplier yields the correct solution, t≥403t \ge \frac{40}{3}.

Step-by-Step Solution

1
Write the inequality representing the physical situation.
50−72t≤20−54t50 - \frac{7}{2}t \le 20 - \frac{5}{4}t
The volume of the first solution after tt hours is 50−72t50 - \frac{7}{2}t, and the volume of the second solution is 20−54t20 - \frac{5}{4}t. The phrase 'at most' means the first volume must be less than or equal to the second volume.
2
Isolate the variable terms on the left side and the constant terms on the right side.
−72t+54t≤20−50-\frac{7}{2}t + \frac{5}{4}t \le 20 - 50
Grouping like terms makes it possible to simplify both sides of the inequality.
3
Find a common denominator of 4 to combine the fractions on the left side, and simplify the constant terms on the right side.
−94t≤−30-\frac{9}{4}t \le -30
Converting −72t-\frac{7}{2}t to −144t-\frac{14}{4}t allows us to add it to 54t\frac{5}{4}t, resulting in −94t-\frac{9}{4}t.
4
Multiply both sides of the inequality by −49-\frac{4}{9} to solve for tt, reversing the inequality sign because we are multiplying by a negative number.
t≥403t \ge \frac{40}{3}
Multiplying by the reciprocal of the coefficient isolates tt. The inequality sign must be reversed (≤{\le} to ≥{\ge}) because we are multiplying by a negative value.

Key Concept

Solving Linear Inequalities
Estimated Time:1m 30s
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Topics

Solving Linear Inequalities

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