Multiple choice

At $10:00$, Angela starts from her home and runs at a constant pace to Kathleens house, which is exactly $2$ miles away. Immediately, she and Kathleen turn around and walk back to Angelas house exactly $4$ miles an hour slower than Angela ran. When they arrive at Angelas house, the time is $10:45$. At what speed did Angela run?

  1. $6$ miles per hour
  2. $6.5$ miles per hour
  3. $7$ miles per hour
  4. $7.5$ miles per hour
  5. $8$ miles per hour
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Let speed be v. Time to house = 2/v. Return speed = v - 4. Return time = 2/(v-4). Total time = 45 min = 0.75 hours. 2/v + 2/(v-4) = 0.75. 2(v-4 + v) / (v(v-4)) = 0.75. 2(2v-4) = 0.75(v^2 - 4v). 4v - 8 = 0.75v^2 - 3v. 0.75v^2 - 7v + 8 = 0. Multiply by 4/3: v^2 - 9.33v + 10.66 = 0. Roots are 8 and 1.33. Since speed must be > 4, v = 8.

AI explanation

The total time taken from start to return home is 45 minutes, which is 0.75 of an hour. Let Angela's running speed be x miles per hour; then the walking speed back is (x - 4) miles per hour. Using the formula time = distance/speed, the total time equation is 2/x + 2/(x - 4) = 0.75. Multiplying by 4x(x - 4) yields 8(x - 4) + 8x = 3x(x - 4), which simplifies to 3x^2 - 28x + 32 = 0. Factoring this gives (3x - 4)(x - 8) = 0, and since x must be greater than 4, Angela ran at 8 miles per hour.

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