Question

Mr. Beecher and Mrs. Carter are teachers at the same school. They leave their houses at the same time in the morning to get to school. Mr. Beecher lives 8 miles away from school and rides his bicycle to work. Every minute, he gets 1 6 of a mile closer to school. Mrs. Carter lives 20 miles away from school and drives her car to work. Every minute, she gets 1 2 of a mile closer to school. After how many minutes will Mr. Beecher and Mrs. Carter first be the same distance away from school?

Answer 1

Let the school be at point x

 

Mr. Beecher lives 8 miles away from school

Therefore, his starting point is x – 8

 

Every minute, he gets 1/6 of a mile closer to school

Therefore, at time, t(in minutes), Mr. Beecher’s distance from school, Db = x – 8 + (1/6)t

Db = x – 8 + (1/6)t            … (Equation I)

 

Mrs. Carter lives 20 miles away from school

Therefore, her starting point is x – 20

 

Every minute, she gets 1/2 of a mile closer to school

Therefore, at time, t(in minutes), Mrs. Carter’s distance from school, Dc = x – 20 + (1/2)t

Dc = x – 20 + (1/2)t            … (Equation II)

 

After how many minutes will Mr. Beecher and Mrs. Carter first be the same distance away from school?

That’s when Db = Dc, given that t is same value in both equations

 

 Db = x – 8 + (1/6)t            … (Equation I)

Dc = x – 20 + (1/2)t            … (Equation II)

 

Db = Dc. Therefore,

x – 8 + (1/6)t = x – 20 + (1/2)t

 

Subtract x from both sides of the equation

x – x – 8 + (1/6)t = x – x – 20 + (1/2)t

– 8 + (1/6)t = – 20 + (1/2)t

(1/6)t – 8 = (1/2)t – 20

 

Add 20 to both sides of the equation

(1/6)t – 8 + 20 = (1/2)t – 20 + 20

(1/6)t + 16 = (1/2)t

 (1/2)t = (1/6)t + 16

 

Subtract (1/6)t from both sides of the equation

(1/2)t – (1/6)t = (1/6)t – (1/6)t + 16

1/3t = 16

 

Multiply both sides of the equation by 3

1/3t x 3 + 16 x 3

t = 48 minutes

 

 

Therefore, after 48 minutes, Mr. Beecher and Mrs. Carter will first be the same distance away from school.