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Difficulty: HardAlgebraic Word Problems and Modeling

Two cyclists, Clara and Dan, start simultaneously from opposite ends of a 120120-mile trail and ride directly toward each other along the same path. Clara rides at a constant speed of cc miles per hour, and Dan rides at a constant speed of dd miles per hour. If they maintain their initial speeds, they will meet in 33 hours. However, if Clara increases her speed by 20%20\% and Dan increases his speed by 3313%33\frac{1}{3}\%, they will meet 3636 minutes earlier. What is Clara's original speed cc, in miles per hour?

  1. A
    15
  2. B
    20
  3. C
    22.5
  4. 25Answer
  5. E
    30

Answer

Clara's original speed cc is 25 miles per hour.
The correct answer is 25 miles per hour. Initial combined speed is c+d=1203=40c + d = \frac{120}{3} = 40 mph. Meeting 36 minutes earlier means a new duration of 2.4 hours, giving a new combined speed of 1202.4=50\frac{120}{2.4} = 50 mph. Setting up the equation 65c+43(40−c)=50\frac{6}{5}c + \frac{4}{3}(40 - c) = 50 yields c=25c = 25.

Step-by-Step Solution

1
Determine the initial combined rate of Clara and Dan.
Since they cover 120120 miles in 33 hours, their combined speed is c+d=1203=40c + d = \frac{120}{3} = 40 miles per hour.
Distance equals rate multiplied by time (D=r⋅tD = r \cdot t). When moving toward each other, combined rate is c+dc + d.
2
Calculate the new travel time and the new combined rate.
The new travel time is 3 hours−36 minutes=3−0.6=2.43\text{ hours} - 36\text{ minutes} = 3 - 0.6 = 2.4 hours (or 125\frac{12}{5} hours). The new combined rate is 1202.4=50\frac{120}{2.4} = 50 miles per hour.
36 minutes is equivalent to 3660=0.6\frac{36}{60} = 0.6 hours.
3
Express the modified individual rates algebraically.
Clara's new speed is 1.2c=65c1.2c = \frac{6}{5}c. Dan's new speed is (1+13)d=43d\left(1 + \frac{1}{3}\right)d = \frac{4}{3}d.
Increasing by 20%20\% multiplies rate by 1.21.2, and increasing by 3313%33\frac{1}{3}\% multiplies rate by 43\frac{4}{3}.
4
Formulate and solve the system of linear equations.
We have the system:
1) c+d=40  ⟹  d=40−cc + d = 40 \implies d = 40 - c
2) 65c+43d=50\frac{6}{5}c + \frac{4}{3}d = 50

Substituting d=40−cd = 40 - c into equation 2:
65c+43(40−c)=50\frac{6}{5}c + \frac{4}{3}(40 - c) = 50
Multiplying the entire equation by 1515 to eliminate denominators:
18c+20(40−c)=75018c + 20(40 - c) = 750
18c+800−20c=75018c + 800 - 20c = 750
−2c=−50  ⟹  c=25-2c = -50 \implies c = 25
Solving for cc yields Clara's original speed in miles per hour.

Key Concept

Distance-Rate-Time modeling using systems of linear equations with percentage changes in individual rates
Estimated Time:2m 30s
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Topics

Algebraic Word Problems and Modeling

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