Expert: Sherry Wallin - 12/15/2008

Question
Robert can mow the lawn in one hour less time than Ed.  Together they can finish the job in 5 hours.  How long would it take for Robert to finish the lawn alone?

Can you show me the math?

Answer
Hi Walter~
    This is a standard type work = rate*time problem, however it doesn't work out nicely.

You are given that Robert and Ed can do the job together in 5 hours so think about it this way , that is 10 man hours so you should be looking for a number a little bit smaller than 10 (because Robert accomplishes more in less time) but much more than 5.
Recall that work = rate times time. So if you think about work = 1 whole job you really have rate times time is 1. Essentially this means that rate and time are reciprocals of each other.(remember to be a multiplicative inverse, the product of a number and it's inverse is 1).
Ok so we're given that Robert works a complete job in 1 hour less than Ed. Let t = time for Robert to mow the lawn by himself, then t+1 is the time for Ed to mow the lawn by himself. So since rate and time are inverses of each other we have Robert's rate as (1/t) and Ed's rate (1/(t+1)). Since the job together took 5 hours we have 5(Robert's rate + Ed's rate) = 1 job
5[1/t + 1/(t+1)] = 1. Clear the denominators by multiplying by the LCD
t(t+1)[5(1/t + 1/(t+1)]= t(t+1)*1
5(t+1) + 5t = t^2+t
10t + 5 = t^2+t
0 = t^2 -9t - 5
This is where you need to be creative and use the quadratic formula because this is not factorable over the integers or rational numbers.
{-(-9)+-sqrt[(-9)^2 -4(1)(-5)]}/2(1)
[9+-sqrt(81+20)]/2
[9+-sqrt(101)]/2
This is the value of t and we want it for Robert so you have (note: sqrt 101 is about 10)[9 +10]/2 is approximately equal to 19/2 or 9.5 hours which falls in our predicted range, doesn't it?

Hopefully I have explained the math necessary for you to do the problem.

Math Prof