1. ## Work Problem

Dave can mow his lawn in 20 minutes less time than with a power mower that with his hand mower. One day, the power mower broke 15 minutes after he started mowing and he took 25 minutes more time to complete the job with the hand mower. How many minutes does Dave take to mow the lawn with the power mower?

The book tells me to use this:

rate of work x time worked = part of job completed

Who can clarify this question for me?

2. Originally Posted by magentarita
Dave can mow his lawn in 20 minutes less time than with a power mower that with his hand mower. One day, the power mower broke 15 minutes after he started mowing and he took 25 minutes more time to complete the job with the hand mower. How many minutes does Dave take to mow the lawn with the power mower?

The book tells me to use this:

rate of work x time worked = part of job completed

Who can clarify this question for me?

1. Let t denote the time (measured in minutes) you need to mow the lawn by hand. Let L denote the area of the lawn.

Then (t-20) is the time you need when you use a power mower.

2. In one minute the hand mower has done $\displaystyle \dfrac Lt$ of the lawn and
in one minute the power mower has done $\displaystyle \dfrac L{t-20}$ of the lawn. Therefore t > 20.

3.
$\displaystyle \underbrace{\dfrac L{t-20} \cdot 15}_{work\ of\ power\ mower} + \underbrace{\dfrac Lt \cdot 25}_{work\ of\ hand\ mower} = L$

Divide through by L. Multiply by the denominators to get rid of the fractions:

$\displaystyle 15t+25t-500 = t^2-20t~\implies~t^2-60t+500=0$

Solve for t.

I've got t = 10 or t = 50. So the only possible solution is t = 50. (see #2.)

3. ## ok

Originally Posted by earboth
1. Let t denote the time (measured in minutes) you need to mow the lawn by hand. Let L denote the area of the lawn.

Then (t-20) is the time you need when you use a power mower.

2. In one minute the hand mower has done $\displaystyle \dfrac Lt$ of the lawn and
in one minute the power mower has done $\displaystyle \dfrac L{t-20}$ of the lawn. Therefore t > 20.

3.
$\displaystyle \underbrace{\dfrac L{t-20} \cdot 15}_{work\ of\ power\ mower} + \underbrace{\dfrac Lt \cdot 25}_{work\ of\ hand\ mower} = L$

Divide through by L. Multiply by the denominators to get rid of the fractions:

$\displaystyle 15t+25t-500 = t^2-20t~\implies~t^2-60t+500=0$

Solve for t.

I've got t = 10 or t = 50. So the only possible solution is t = 50. (see #2.)
The answer in the book is 30 minutes.

How do I get 30 minutes?

Thanks