Speed/dist./time problem (need to find average speed)

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Homework Help Overview

The problem involves calculating the average speed of a cyclist over two days, where the distance traveled each day is known, and the average speed on the second day is double that of the first day. The total time for the journey is also provided.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the formulation of an equation to model the problem, with attempts to solve it using different methods, including a chart method and algebraic manipulation. Questions arise regarding the least common denominator (LCD) and the steps taken in the algebraic process.

Discussion Status

Some participants have provided algebraic solutions and alternative approaches to the problem, while others are seeking clarification on the reasoning behind the steps taken. There is an ongoing exploration of the mathematical concepts involved, particularly regarding the manipulation of fractions and equations.

Contextual Notes

Participants express confusion about the algebraic steps and the concept of finding the LCD, indicating a need for clearer understanding of these mathematical principles. The original poster's initial attempt yielded an implausible speed, prompting further discussion.

bigmac
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Homework Statement


Andrew won a two day bicycle race. He traveled 60km each day and his average speed on the second day was double that of the first day. If Andrew rode for a total of 6 hours, what was his average speed each day?

a) state an equation that models this problem.
b) Solve the problem.

Homework Equations



Speed = distance / time

The Attempt at a Solution



I tried to do it using the chart method but its not working out...I get x = 0.0333 km/h...the speed can't be that slow :S.

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(60/x) + (60/2x) = 6 >> (120/2x) + (60/2x) = 6 >> 180/2x = 6

>> 180 = 12 x , finally x =180/12 = 15 km/h .. what about this one :)
 
yeah that works but, how did you take 2x as the LCD? don't you just multiply the denominators to get the LCD? And when you take 2x as a LCD, you cross multiply the left side right? I get 120x + 60x / 2x = 6...how did you get 120/2x + 60/2x...
 
hmm, in the equation you have x and 2x in the denominators and you want to add them, in order to proceed the denominatos should be the same, and the shortest way to get them equal is by multiplying with 2 ( in the x denominator) ..

if it is still not clear please ask again, it is important that you be able to grasp this one..
 
bigmac said:
yeah that works but, how did you take 2x as the LCD? don't you just multiply the denominators to get the LCD? And when you take 2x as a LCD, you cross multiply the left side right? I get 120x + 60x / 2x = 6...how did you get 120/2x + 60/2x...

it seems that confuses you .. another way to solve it by taking 1/x as a common factor :

(60/x) + (60/2x) = 6 >> 1/x (60 + 60/2) = 6 >> 1/x (60 +30) = 6 >> 1/x (90) = 6

>> 1/x = 6/90 >> x = 90/6 = 15 km/h .. how about this one?
 

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