<p>James can clean a certain road in 50 minutes. If Mary can clean the same road in 20 minutes, how long will it take them to the nearest minute, to clean the road if they work together?</p>
<p>A) 12
B) 13
C) 14
D) 16
E) 18</p>
<p>Does anyone know how to solve this algebraicly?</p>
<p>Set up a system of equations.</p>
<p>The standard way is to note that James can clean 1/50 of the road in 1 min and Mary can clean 1/20 in one minute. Together, they can complete 1/50 + 1/20 = 7/100 of the road in one minute. The amount of time needed is 100/7 minutes, roughly 14 min, C.</p>
<p>Actually, here’s what I did (it may be a little convoluted at first and somebody may find an easier way. </p>
<p>Take a road of length x ft. James can clean it in 50 minutes. That means his rate is x/50 ft/min. According to similar reasoning, Mary’s rate is x/20 ft/min. The length of road James can clean in y minutes is given by (x/50)y ft. Similarly, the length of road Mary can clean is given by (x/20)y ft. If you add these two, then the sum has to equal the total length of road, namely x. Therefore:
(x/50)y + (x/20)y = x. </p>
<p>Factoring out xy,
xy(1/50 + 1/20) = x </p>
<p>Simplifying…
y(7/100) = 1 </p>
<p>Simplifying…
y = 14.29 minutes, or 14 minutes.</p>
<p>I got up to where you said “7/100 of the road in one minute” but then I got lost when you said “the amount of time needed is 100/7 minutes”. Sorry, I’m confused.</p>
<p>A simple ratio: If you can clean up 7/100 of the road in one minute, how many minutes will it take to clean the entire road? 100/7.</p>
<p>Analogous to Distance = rate*time, 1 = (7/100)(100/7).</p>
<p>This is the method I use:</p>
<p>t= time t/50 + t/20 = 1</p>
<p>multiply whole equation by 100</p>
<p>2t + 5t = 100 —> 7t = 100 ----> t = 14.3 = C</p>
<p>@rspence Sweet I understand it now, thank you!
I like riadapaki95’s strategy too.</p>
<p>I don’t think they test work/rate problems of this type on the SAT. I am 99% sure. Has anyone seen a similar problem on the Official SAT tests of this type?</p>
<p>^^ You may be 99% sure, but I can assure you there are problems like this on the SAT.</p>
<p>@SATQuantum, I believe I’ve seen these types of problems. They’re really easy once you know the solution.</p>