1. ## algebra linear

two pipes can fill a tank in two hours running simultaneously, If both pipes were open for 40 min. and the first is then shut off, it takes one hour and 30 min. more for the second pipe to fill the tank. how long for each pipe running to fill the tank?

2. ## Re: algebra linear

Let the rate of the water flowing from the first and second pipes be $\displaystyle x,y \hspace{2 mm} lit/hr$ respectively.

Given,
$\displaystyle 2(x+y) = \frac{2}{3}(x+y) + \frac{3}{2}y \implies y = 8x$
$\displaystyle \therefore x$ takes $\displaystyle 18$ hours to fill the tank and $\displaystyle y$ takes $\displaystyle \frac{4}{9}$ hours

3. ## Re: algebra linear

Originally Posted by vanpore

two pipes can fill a tank in two hours running simultaneously, If both pipes were open for 40 min. and the first is then shut off, it takes one hour and 30 min. more for the second pipe to fill the tank. how long for each pipe running to fill the tank?

I'll do it in vicious detail. Please don't mistake length for difficulty; I simply started writing and ended up writing A LOT. It's a straightforward problem.

The thing to notice is that usually, this kind of problem assumes equal "rates" for all the "pieces" (Ex: "It takes 5 men 80 hours to build a small barn. How long would it take 3 of them to build 4 barns identical to the first?"). But that's *not* the case here, as can be seen by noting that if it were, then "2 pipes fill it in 2 hours" is all the information you'd need to answer the question, and yet additional information was given ("both pipes were open for 40 min. and the first is then shut off, it takes one hour and 30 min"). It might work out that both pipes flow at the same rate, but that can't be assumed.

One thing to note about this kind of word problem: it's all about units and ratios, so keep track of the units, and let them guide you through the problem.

General:
To solve a word problem algebraically, you first go through a setup stage where you define useful variables, translate the problem's information into those variables, and identify what you're being asked to find. After that, you algebraically solve the problem. When you're done, you take your solution, check it (however best makes sense to you - depends on the problem), make sure it answers the word problem's question directly, make sure it "makes sense" ("Thus Bob bought 4.72 cars"??? "Thus Bob baked -2 pies"???), and write your answer (and on a quiz or test, circle it) in manner, with units, that clearly and directly answers the problem's question ("Thus Bob bought 6 cars.", "Thus Bob baked 5 pies.")

Setup Stage:
Let the flow of pipe #1 be $\displaystyle p_1 \ tanks/hour$, and the flow of pipe #2 be $\displaystyle p_2 \ tanks/hour$. Note that when you multiply $\displaystyle p_1 \ tanks/hour$ by a time, you get a result in units of "tanks".

Since I want to track units, I want to use the same units for the same "dimension" (time is a dimension, weight is a dimension, tanks is a dimension, etc.).
Since some of the times are in hours, and others in minutes, I'll convert everything upfront to avoid be sidetracked while working the problem. I get:

"two pipes can fill a tank in two hours running simultaneously, If both pipes were open for 2/3 hours. and the first is then shut off, it takes 3/2 hours more for the second pipe to fill the tank. how long for each pipe running to fill the tank?"

Now tranlsate the word problem into algebraic statements.

"two pipes can fill a tank in two hours running simultaneously" means $\displaystyle (p_1 \ tanks/hour + p_2 \ tanks/hour)(2 \ hours) = 1 \ tank$. Thus $\displaystyle 2(p_1 + p_2) = 1$ (units cancel!).

To translate into algebra the information "both pipes were open for 2/3 hours. and the first is then shut off, it takes 3/2 hours more for the second pipe to fill the tank", first make sure you know what it means "in essence", then try to translate each of it's parts, then put the whoel thing together.

1. In essence, it's saying "this amount + this other amount = 1 tank".

2. Translating the parts:

"both pipes were open for 2/3 hours" tells you an amount filled (unit of tanks) by the flow of both:

$\displaystyle = (p_1 \ tanks/hour + p_2 \ tanks/hour)(2/3 \ hours) = (2/3)( p_1 + p_2 ) \ tanks$.

"the first is then shut off, it takes 3/2 hours more for the second pipe" is again telling you an amount, in units of tanks. That amount is:

$\displaystyle = (p_2 \ tanks/hour)(3/2 \ hours) = (3/2) p_2 \ tanks$.

3. Putting those two parts together: "this amount + this other amount = 1 tank" becomes

$\displaystyle (2/3)( p_1 + p_2 ) \ tanks + (3/2) p_2 \ tanks = 1 \ tank$, so $\displaystyle (2/3)( p_1 + p_2 ) + (3/2) p_2 = 1$.

The last part of the setup stage is to make sure you know what you're supposed to find. The problem asks "how long for each pipe running to fill the tank?".

Thus it's requesting a TIME for each pipe - the time to fill 1 tank. Since the flow of pipe #1 is $\displaystyle p_1 \ tanks/hour$, then the time $\displaystyle t_1 \ hours$ it taks for pipe #1 to fill 1 tank satisfies:

$\displaystyle (p_1 \ tanks/hour) (t_1 \ hours) = 1 \ tank$, so $\displaystyle p_1 t_1 \ tanks = 1 \ tank$, so $\displaystyle p_1 t_1 = 1$, so $\displaystyle t_1 = 1/p_1$.

Similarly with $\displaystyle p_2$ and $\displaystyle t_2$, get that $\displaystyle t_2 = 1/p_2$.

So the answer to the question is to say "It would take pipe #1 $\displaystyle t_1$ hours to fill the tank, and pipe #2 $\displaystyle t_2$ hours to fill the tank." But $\displaystyle t_1 = 1/p_1$ and $\displaystyle t_2 = 1/p_2$, so we'll need to find $\displaystyle p_1$ and $\displaystyle p_2$ first. There's no sense in keeping those extra $\displaystyle t_1, t_2$variables around - we'll just find $\displaystyle p_1$ and $\displaystyle p_2$ and then remember that the final answer will be their inverses, in units of hours.

Form of our ultimate final answer: So find $\displaystyle p_1$ and $\displaystyle p_2$. Then the answer will be "It would take pipe #1 $\displaystyle (1/p_1)$ hours to fill the tank, and pipe #2 $\displaystyle (1/p_2)$ hours to fill the tank."

The setup stage is done.

Algebra Stage:
Find $\displaystyle p_1$ and $\displaystyle p_2$ such that $\displaystyle 2(p_1 + p_2) = 1$ and $\displaystyle \frac{2}{3}( p_1 + p_2 ) + \frac{3}{2}p_2 = 1$.

Simplify the second equation (collect like terms):

$\displaystyle \frac{2}{3}p_1 + (\frac{2}{3} + \frac{3}{2})p_2 = 1$. Have $\displaystyle \frac{2}{3} + \frac{3}{2} = \frac{4}{6} + \frac{9}{6} = \frac{13}{6}$. Thus

$\displaystyle \frac{2}{3}p_1 + \frac{13}{6}p_2 = 1$.

Finally really ready to go: Find $\displaystyle p_1$ and $\displaystyle p_2$ such that $\displaystyle 2(p_1 + p_2) = 1$ and $\displaystyle \frac{2}{3}p_1 + \frac{13}{6}p_2 = 1$.

Many ways to do this. I'll solve for $\displaystyle p_1$ in terms of $\displaystyle p_2$ in the first equation, and plug that into the second equation to find $\displaystyle p_2$, and thus also $\displaystyle p_1$.

$\displaystyle 2(p_1 + p_2) = 1 \Rightarrow p_1 + p_2 = \frac{1}{2} \Rightarrow p_1 = \frac{1}{2} - p_2$.

So $\displaystyle \frac{2}{3}p_1 + \frac{13}{6}p_2 = 1 \Rightarrow \frac{2}{3}(\frac{1}{2} - p_2) + \frac{13}{6}p_2 = 1 \Rightarrow (\frac{2}{3})(\frac{1}{2}) - (\frac{2}{3})p_2 + \frac{13}{6}p_2 = 1$.

$\displaystyle \Rightarrow \frac{1}{3} + (\frac{13}{6}- \frac{2}{3})p_2 = 1 \Rightarrow (\frac{13}{6}- \frac{4}{6})p_2 = 1 - \frac{1}{3} \Rightarrow \frac{9}{6}p_2 = \frac{2}{3}$

$\displaystyle \Rightarrow \frac{3}{2}p_2 = \frac{2}{3} \Rightarrow p_2 = (\frac{2}{3})(\frac{2}{3}) = \frac{4}{9}$. Then $\displaystyle p_1 = \frac{1}{2} - p_2 = \frac{1}{2} - (\frac{4}{9})$.

Arithmetic: $\displaystyle \frac{1}{2} - \frac{4}{9} = \frac{9}{18} - \frac{8}{18} = \frac{1}{18}$

Solved the algebra: $\displaystyle p_1 = \frac{1}{18}$ and $\displaystyle p_2 = \frac{4}{9}$.

Check:
$\displaystyle 2(p_1 + p_2) = 1$ and $\displaystyle \frac{2}{3}( p_1 + p_2 ) + \frac{3}{2}p_2 = 1$, so $\displaystyle 2(\frac{1}{18} + \frac{4}{9}) \overset{?}{=} 1, \ \frac{2}{3}( \frac{1}{18} + \frac{4}{9} ) + \frac{3}{2}(\frac{4}{9}) \overset{?}{=} 1$.

Both check out numerically.

Conceptual checks: Both are reasonable numbers for the flow - not negative, not ridiculous huge given the problem, etc.

Also (and this is the kind of "thinking" that comes with practice and experience), our result says that pipe#1 is flowing very slowly relative to pipe #2 - pipe #2 is doing almost all the work - so CRUDELY, it's kinda like pipe #1 has no flow, and so CRUDELY, like pipe#2 is filling the tank after, CRUDELY, 2 hours. Does that match our result? Yes, CRUDELY. The first condition is that both of them flowing together, which on our crude assumption is just pipe #2 flowing, fills a tank in two hours. Our second condition is that, since we're CRUDELY saying that pipe #1 has no flow, 2 hours and 10 minutes of the flow of pipe #2 fills a tank - and that's CRUDELY the same as what the first condition said. Since a flow of $\displaystyle \frac{4}{9}$ a tank/hour for 2 hours results in $\displaystyle \frac{8}{9}$ tank, when pipe #2 flows for 2 hours, and pipe #1 has no flow, we CRUDELY get a full tank, which CRUDELY matches both of the conditions given in the problem. Thus this is another check that the answer is reasonable.

Now all that's left is to answer the problem: Rmember it's NOT just $\displaystyle p_1$ and $\displaystyle p_2$, but rather the TIME needed for each pipe to indepedently fill the tank.
We already decided the answer would be "It would take pipe #1 $\displaystyle (1/p_1)$ hours to fill the tank, and pipe #2 $\displaystyle (1/p_2)$ hours to fill the tank."

Note that $\displaystyle (1/p_1)=18$ and $\displaystyle (1/p_2) = \frac{9}{4} = 2\frac{1}{4}$. Of course, a quarter of an hour is 15 minutes. Now can give the final answer:

ANSWER TO WORD PROBLEM: It would take pipe#1 18 hours to fill the tank, and pipe#2 2 hours and 15 minutes to fill the tank."

4. ## Re: algebra linear

Kalyanram: you made a typo? Should be 9/4, not 4/9
Johnsomeone: get a life

Make it a speed = distance * time problem; pick any distance, say 540 miles:

@(x+y)mph......................................... ...540 miles..........................................>[2 hours]

@(x+y)mph.....180 miles.....>[2/3 hour]@y mph..................360 miles....................>[3/2 hours]

x + y = 540 / 2 = 270 mph

y = 360 / (3/2) = 240 mph; so x = 270 - 240 = 30 mph

Time @x mph: 540 / 30 = 18 hours
Time @y mph: 540 / 240 = 2.25 hours

Assigning a "distance" is not a must of course, as the default is 1,
but makes it easier at the teaching stage...to illustrate...

5. ## Re: algebra linear

Wilmer... tell me about it : )