<p>Jan can mor her lawn in 60 minutes by herself if she hires wally to do it, he takes 30 minutes,while peter can do it in 40 minutes. If Jan starts mowing the lawn for 15 minutes, then decides to hire Wally and Peter to help her finish, how long will it take (in hours), from when Jan started to when the three finished together ??</p>
<p>1)1/6
2)5/12
3)7/12
4)3/4
5)5/6</p>
<p>I tried to solve this,but I think I got something wrong while solving.</p>
<p>Jan = 1;</p>
<p>Wally = 60/30 = 2;
Peter = 60/40 = 3/2;
JanNew = 15/60 = 1/4 part of the work ;</p>
<p>The three should be together is 1 + 2 + 3/2 = 9/2;</p>
<p>but I couldn't really figure out how can then how long it takes.</p>
<p>You were almost there. Remember the d = rt formulas. </p>
<ol>
<li><p>Jan mows 1/4 of lawn in 15 minutes – or 3/12 hour.</p></li>
<li><p>She leaves 3/4 of lawn for the combined work at the rate of 9/2 you calculated.</p>
<br>
<br></li>
<li><p>Time to mow entire lawn is 3/12 + 2/12 or 5/12.</p></li>
</ol>
<p>You can also make up numbers to help. Gives you a little more arithmetic, but no algebra.</p>
<p>Say the lawn is 120 sq ft. (Not entirely a random choice BTW.) That means Wally mows 120/30=4 sq ft/min and Peter can mow 120/40 = 3 sq ft / min. And Jan mows at 120/60 = 2 per minute.</p>
<p>Now, Jan mows for 15 out of her usual 60 minutes, so she mows 1/4 of the lawn, leaving 90 ft to mow. The time left will be 90/(2 + 3 + 4) = 10 min…Total time with the original 15 minutes included: 25 min or 25/60 = 5/12 hr.</p>