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# Help with a algebra word(sharing a job problem) problem please?

Topic: Problem solving algebra word problems
June 24, 2019 / By Becca
Question: Next door neighbors Bob and Jim use hoses from both houses to fill bob's swimming pool. They know it takes 18 h using both hoses. They also know that bob's hose, used alone, takes 20% less time than Jim's hose alone. How much time is required to fill the pool by each hose alone? (sharing a job problem)

## Best Answers: Help with a algebra word(sharing a job problem) problem please?

This is a bit complicated, but once you get your thinking straight, it's not hard. The swimming pool has a volume V. Both hoses together fill the pool in time t (which is 18 hours) at a volumetric rate of r[both]. This rate is the sum of Bob's hose's rate (r[Bob]) and Jim's hose's rate (r[Jim]). That is, r[both] = r[Bob] + r[Jim] and V = r[both]t = r[both] (18 h) What else do we know? We know that Bob's hose by itself takes 20% less time to fill the pool than Jim's hose by itself. So if Jim's hose could fill a barrel in 5 minutes, Bob's hose would fill that same barrel in 4 minutes. (This is just an example; we don't really care about "how big" our fictional barrel is, just that Bob's hose takes 4 minutes and Jim's takes 5 minutes -- that is, Bob's hose is 20% faster.) Since V = rt for the barrel, we see that V[barrel] = r[Jim] t[Jim, barrel] and V[barrel] = r[Bob] t[Bob, barrel] where t[Jim, barrel] and t[Bob, barrel] are the times it takes Jim and Bob, respectively, to fill the barrel alone -- that is, 5 minutes and 4 minutes. Since both equations are equal to V[barrel], we can set them equal to each other: v[barrel] = r[Bob] t[Bob, barrel] = r[Jim] t[Jim, barrel] r[Bob] (4 minutes) = r[Jim] (5 minutes) r[Bob] = (5/4) r[Jim] Now that we know how Bob's and Jim's hose rate compare to each other, we can solve the original problem. V[swimming pool] = r[total] (18 h) What's r[total]? It's just the sum of Jim's and Bob's hose rates: r[total] = r[Bob] + r[Jim] But we already found out that r[Bob] = (5/4) r[Jim], so let's substitute that in: r[total] = (5/4) r[Jim] + r[Jim] = (9/4) r[Jim] and our equation for filling the pool becomes: V[swimming pool] = (9/4) r[Jim] (18 h) How long does it take Jim's hose alone to fill the pool? Obviously, it takes V[swimming pool] = r[Jim] t[Jim] Now we have two equations that are both equal to V[swimming pool]. So they're equal to each other: (9/4) r[Jim] (18h) = r[Jim] t[Jim] Solve for t[Jim] (note that r[Jim] divides out from both sides): t[Jim] = (9/4) (18 h) = 40.5 h for Jim's hose alone. Now find Bob's time. You can repeat the steps above for Bob, or you can work off of the knowledge you already have. Let's do the latter. To fill the pool volume for Bob: V[swimming pool] = r[Bob] t[Bob] But since we also know that V[swimming pool] = r[Jim] t[Jim] we can tell that r[Bob] t[Bob] = r[Jim] t[Jim] Solve for t[Bob]: t[Bob] = t[Jim] (r[Jim] / r[Bob]) We found out a long time ago that r[Bob] = (5/4) r[Jim], so therefore r[Jim] / r[Bob] = 4/5 and t[Jim] = 40.5 h Putting those into the above equation, we find that t[Bob] = (40.5 h) (4/5) = 32.4 h So to fill the swimming pool, Jim's hose alone takes 40.5 h and Bob's hose alone takes 32.4 h.
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If you multiply both numerator and denominator of a fraction by the same number, the value of the fraction remains the same: 3x/5x = (3/5)(x/x) 3x/5x = 3/5(1) x = 1 ¯¯¯¯¯¯