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

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

Adenah Adenah | 6 days ago
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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Adenah Originally Answered: Algebra? word problem D; please help?
I would start with listing a few fractions that you know to be equivalent to 5/6. You can create fractions that are equivalent to 5/6 by multiplying by fractions that are equal to 1 ... 2/2,3/3,4/4,etc. 5/6 x 2/2=10/12 5/6 x 3/3 = 15/18 5/6 x 4/4 = 20/24 In the next step you are again using fractions that are equal to 1, as it states that the (same) number must be added to both numerator and denominator - so again we're talking about 2/2,3/3,5/5 ... I think it's easier to work backward. Look at fraction #1: 10/12. Is there a number that you can add to 3 to get 10 and that you can add to 5 to get 12? 3 + 7 =10, and 5 + 7 = 12. (This is not to say that 3/5 + 7/7 = 10/12.) 3/5 + 7/7 = 1 3/5 because 7/7 is equal to 1. Besides, you can't add fractions that have uncommon denominators ... however that's not what this question is asking. Algebraically, you would set your problem/solution up like this: Let x = any number Let y = any number For what values of x and y are the following equations equal? 5 (x) = 3 + y 6 (x) = 5 + y Let x = 1 5(1) = 3 + y, 5 = 3 + y, y = 2 6(1) = 5 + y, 6 = 5 + y, y = 1 ... x = 1 fails, there are no corresponding values of y that meet the criteria. Let x = 2 5(2) = 3 + y, 10 = 3 + y, y = 7 6(2) = 5 + y, 12 = 5 + y, y = 7 ... x = 2 satisfies the equation in the numerator and the denominator. Your final answer is 7. 7 can be added to the N & D of 3/5 to obtain a fraction equivalent to 5/6.
Adenah Originally Answered: Algebra? word problem D; please help?
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 ¯¯¯¯¯¯
Adenah Originally Answered: Algebra? word problem D; please help?
The best way to answer this is to take the derivative of the equation. By takin the derivative, the equation changes to v(t)=-32t+50 where v is the velocity. When the ball is at its maximum height, the velocity is zero, so set the equation equal to 0 and solve for t. Should get just over 1.5 seconds. when you get the time, plug it back into the original equation and it will give you your maximum height.

Teman Teman
Write out the equations as you read them. Let X = Bob's hose and Y = Jim's hose. You know that X+Y = 18 The problem then says that Bob's hose (X) takes 20% less time than Jim's (Y), so X is 0.80 times Y (because for however long Jim's hose takes, Bob's hose takes less time). So 0.80Y = X Then use substitution. Plug the value of X from the second equation into the first, then solve for Y. 0.80Y + Y = 18 Bob's hose (X) is the remainder after you find the length of time Jim's hose (Y) takes.
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Teman Originally Answered: Algebra word problem?
Your equation is a little ambiguous. To take it literally it would read; y = -1777(x) * 27153 where * is multiply, however it is more likely that the equation would be; y = -1777(x) + 27153. I will work through this problem with both of these equations. This will also hopefully give you the benefit of a greater understanding. For y = -1777(x) * 27153 In 2005, x = 5, therefore the equation will read: y = -1777(5) * 27153 Now to solve the equation: y = -8885 * 27153 y = -241254405 Therefore the number of gun deaths in 2005 is -241254405 (this is why I believe your equation is missing an operator) In 2007, x = 7, therefore the equation will read: y = -1777(7) * 27153 Now to solve the equation: y = -12439 * 27153 y = -337756167 Therefore the number of gun deaths in 2007 is -337756167 To predict when the number of deaths will be 11160 requires 11160 to be substituted for y: 11160 = -1777(x) * 27153 Now solve for s: Both sides now need to be divided by 27153 (or 27153 needs to swap sides and therefore swat operators): 0.4110043 = -1777(x) Again 0.411 needs to be divided by -1777: -0.00023129 = x Therefore -0.00023 years after 2000 the number of gun deaths was 11160 For y = -1777(x) + 27153: Again for 2005 , x = 5: y = -1777(5) + 27253 y = -8885 + 27153 y = 18268 Therefore the number of gun deaths in 2005 is 18268. In 2007, x = 7: y = -1777(7) + 27153 y = -12439 + 27153 y = 14714 Therefore the number of deaths in 2007 is 14714. To predict when the number of deaths will be 11160, y = 11160: 11160 = -1777(x) + 27153 27153 need to swap sides and indication (or 27153 needs to be subtracted from both sided) -15993 = -1777(x) -15993 needs to be divided by -1777: 9 = (x) Therefore 9 years after 2000 the number of deaths will be 11160; therefore in 2009 approx.11160 people will die due to guns. Sorry it's so long, but I hope this helps :) Remember that you won't be able to use Yahoo Answers in a test or an exam, so if you still have problems understanding you should ask your teacher to explain it more thoroughly or in a different way. :) Good luck =D
Teman Originally Answered: Algebra word problem?
So, the problem gives you an equation, with "x" being the years after 2000 and "y" being the number of gun deaths. For 2005, x = 5, right? Because "x" is the number of years after 2005. So, to solve that part, plug in 5 for "x" in the equation and solve for "y." That is, plug in -1777*(x) + 27153 into your calculator and you should get the number of gun deaths in 2005. Do the same thing for 2007 by plugging in 7 for "x." The last question is a bit trickier. This time, it tells you the number of gun deaths (y) as 11160. So, let's plug that into the equation. This time, we're solving for "x". We get 11160 = -1777x + 27153, right? With that, subtract 27153 from both sides to get -15993. Now, your equation is -15993 = -1777x. Now, we divide by -1777 to isolate "x". You should get x = 9. But you're not done yet! Remember that "x" is the number of years after 2000, which means that your year will be 2009. I hope that helps!

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