Question

After training for a​ year, a cyclist finds that she has increased her average rate by 7 miles per hour and can travel 392 miles in 1 less hour than she did a year ago.

a. What was her old average​ rate? What is her new average​ rate?
b. What was her old time to travel 392 ​miles? What is her new time for 392 ​miles?

Answer 1

a.

Original average rate = 49 miles per hour

New average rate = 56 miles per hour

b.

Original time = 8 hours

New time = 7 hours

Explanation:

Original conditions:

Distance = 392 miles

Time = t

Average speed = a

After 1 year of training:

Distance = 392 miles

Time = t - 1

Average speed = a + 7

average speed = distance/time

Original:

a = 392/t Eq. 1

After 12 year of training:

a + 7 = 392/(t - 1)

(a + 7)(t -1) = 392 Eq. 2

Equations 1 and 2 form a system of equations.

a = 392/t

(a + 7)(t -1) = 392

Substitute 392/t for a in equation 2.

(392/t + 7)(t - 1) = 392

392 - 392/t + 7t - 7 = 392

-392/t + 7t - 7 = 0

-392 + 7t² - 7t = 0

7t² - 7t - 392 = 0

392 = 8 × 49

(7t + 49)(t - 8) = 0

7t + 49 = 0 or t - 8 = 0

7t = -49 or t = 8

t = -7 or t = 8

Discard the negative solution.

a = 392/t

a = 392/8 = 49

a.

Original average rate = a = 49 miles per hour

New average rate = a + 7 = 56 miles per hour

b.

Original time = t = 8 hours

New time = t - 1 = 7 hours