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Biking rates

by Andrew
(New York)










































Rudolph bikes a constant rate and stops for a five-minute break at the end of every mile. Jennifer bikes at a constant rate which is three-quarters the rate that Rudolph bikes, but Jennifer takes a five-minute break at the end of every two miles. Jennifer and Rudolph begin biking at the same time and arrive at the 50-mile mark at exactly the same time. How many minutes has it taken them?

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Dec 12, 2011
Biking Rates – Word Problem
by: Staff


Question:

by Andrew
(New York)


Rudolph bikes a constant rate and stops for a five-minute break at the end of every mile. Jennifer bikes at a constant rate which is three-quarters the rate that Rudolph bikes, but Jennifer takes a five-minute break at the end of every two miles. Jennifer and Rudolph begin biking at the same time and arrive at the 50-mile mark at exactly the same time. How many minutes has it taken them?


Answer:

Rudolph: bikes at rate R

Rudolph takes 5 minute breaks every mile. During the 50 mile trip, Rudolph takes 49 breaks (he arrives at the 50 mile stop at the beginning of his 50th mile break). 49 breaks × 5 min = 245 minutes.



Jennifer: bikes at rate J

J = (¾)×R

Jennifer takes 5 minute breaks every 2 miles. During the 50 mile trip, Jennifer takes 24 breaks (Jennifer arrives at the 50 mile stop at the beginning of her 25th mile break). 24 breaks × 5 min = 120 minutes.



Distance = 50 miles

Time = t in minutes, the time for both bikers to travel 50 miles is the same


R(t - 245) = 50 miles

J(t - 120) = 50 miles

J = (¾)×R

(¾)×R×(t - 120) = 50 miles


You now have two equations with two unknowns.

1st equation: R(t - 245) = 50 miles

2nd equation: (¾)×R×(t - 120) = 50 miles



Solve the 1st equation for R

R(t - 245) = 50 miles

R(t - 245)/(t-245) = 50/(t-245)

R*[(t - 245)/(t-245)] = 50/(t-245)

R*(1) = 50/(t-245)

R = 50/(t-245)




Substitute 50/(t-245) for R in the 2nd equation

(¾) × R × (t - 120) = 50 miles

(¾) × [50/(t-245)] × (t - 120) = 50 miles

.75 × [50/(t-245)] × (t - 120) = 50 miles


Solve for t

(¾) × [50/(t-245)] × (t - 120) = 50


Multiply each side of the equation by (t-245)

(t-245) × (¾) × [50/(t-245)] × (t - 120) = (t-245) × 50

(¾) × 50 × [(t-245) / (t-245)] × (t - 120) = (t-245) × 50

(¾) × 50 × 1 × (t - 120) = (t-245) × 50

(¾) × 50 × (t - 120) = (t-245) × 50


Divide each side of the equation by 50

(¾) × 50 × (t - 120) / 50 = (t - 245) × 50 / 50

(¾) × (t - 120) × (50/ 50) = (t - 245) × (50 / 50)

(¾) × (t - 120) × (1) = (t-245) × (1)

(¾) × (t - 120) = t - 245


Using the distributive law, expand the expression on the left hand side of the equation.

(¾) × t - (¾) × (120) = (t - 245)


Add 245 to each side of the equation

(¾)t - 90 + 245= t - 245 + 245

(¾)t + 155= t + 0

(¾)t + 155= t


Subtract (¾)t from each side of the equation

(¾)t + 155 - (¾)t = t - (¾)t

(¾)t - (¾)t + 155 = t - (¾)t

0 + 155 = t - (¾)t

155 = t - (¾)t

4 * 155 = 4 * (1/4)t


Multiply each side of the equation by 4

4 * 155 = t * (4/4)

4 * 155 = t * (1)

620 = t

t = 620 minutes


the final answer is: t = 620 minutes (10 hours, 20 minutes)






Thanks for writing.

Staff
www.solving-math-problems.com


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