T & W q 13

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Saumyojit

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A tank of 425 litres capacity has been filled with water through two pipes, the​

first pipe having been opened five hours longer than the second. If the first pipe​

were open as long as the second, and the second pipe was open as long as the​

first pipe was open, then the first pipe would deliver half the amount of water​

delivered by the second pipe; if the two pipes were open simultaneously, the​

tank would be filled up in 17 hours. How long was the second pipe open?​



Wa & Wb --> per hour work

x hrs is time for pipe B was opened .

y is the amount filled up by pipe B in x hrs



(Wa + Wb) * 17 hrs = 425


Wa ( x +5) + Wb x = 425

Wb x = y

Wb ( x +5) = (y +k) & Wa * x = (y + k ) /2

Wb ( x +5) = 2 Wa x

Wb x + 5Wb = 2 Wa x

Wb x = 2Wa x - 5 Wb




Will i substitute Wb x in Wa ( x +5) + Wb x = 425

Wa x + Wa 5 + Wb x = 425


Wa x + Wa 5 + 2Wa x - 5 Wb = 425


Wb = ( (3Wa x + 5Wa ) - 425 ) / 5

then?
 
Saumyojit said:

A tank of 425 litres capacity has been filled with water through two pipes, the​

first pipe having been opened five hours longer than the second. If the first pipe​

were open as long as the second, and the second pipe was open as long as the​

first pipe was open, then the first pipe would deliver half the amount of water​

delivered by the second pipe; if the two pipes were open simultaneously, the​

tank would be filled up in 17 hours. How long was the second pipe open?​



Wa & Wb --> per hour work

x hrs is time for pipe B was opened .

y is the amount filled up by pipe B in x hrs



(Wa + Wb) * 17 hrs = 425


Wa ( x +5) + Wb x = 425

Wb x = y

Wb ( x +5) = (y +k) & Wa * x = (y + k ) /2

Wb ( x +5) = 2 Wa x

Wb x + 5Wb = 2 Wa x

Wb x = 2Wa x - 5 Wb




Will i substitute Wb x in Wa ( x +5) + Wb x = 425

Wa x + Wa 5 + Wb x = 425


Wa x + Wa 5 + 2Wa x - 5 Wb = 425


Wb = ( (3Wa x + 5Wa ) - 425 ) / 5

then?
Click to expand...
When starting a new problem do you forget everything you learned while solving previous problems?
 
Saumyojit said:

A tank of 425 litres capacity has been filled with water through two pipes, the​

first pipe having been opened five hours longer than the second. If the first pipe​

were open as long as the second, and the second pipe was open as long as the​

first pipe was open, then the first pipe would deliver half the amount of water​

delivered by the second pipe; if the two pipes were open simultaneously, the​

tank would be filled up in 17 hours. How long was the second pipe open?​



Wa & Wb --> per hour work

x hrs is time for pipe B was opened .

y is the amount filled up by pipe B in x hrs



(Wa + Wb) * 17 hrs = 425


Wa ( x +5) + Wb x = 425

Wb x = y

Wb ( x +5) = (y +k) & Wa * x = (y + k ) /2

Wb ( x +5) = 2 Wa x

Wb x + 5Wb = 2 Wa x

Wb x = 2Wa x - 5 Wb




Will i substitute Wb x in Wa ( x +5) + Wb x = 425

Wa x + Wa 5 + Wb x = 425


Wa x + Wa 5 + 2Wa x - 5 Wb = 425


Wb = ( (3Wa x + 5Wa ) - 425 ) / 5

then?
Click to expand...
I can't force you to use variables representing times rather than rates, as we did when I helped you with a similar problem; and I can't force you to use as few variables as possible, since others here like to start out with a variable for everything. Different styles are valid (though when you ask for help and take up a lot of people's time, they have a right to expect you to have learned from it, and improved your approach to the next problem).

But I can strongly recommend, at least, that you state all your equations first, before starting to solve them. We need to see definitions of n variables, and n equations, in order to determine whether you have a solvable problem. At the end here, you have three variables; I am not about to comb through your mess of random work to determine whether you have three equations in those three variables.

As one small detail, when you define a variable as "per hour work" you should state the units, which here evidently is "litres per hour". That just illustrates the sort of sloppiness that gets you in trouble. An orderly beginning prevents a lot of struggles.

And when you are asking someone to help you, you need to explain your work in words, so we don't have to try to figure it all out. You'll have helpers much more ready to help if you help them help you!
 
Dr.Peterson said:
per hour work" you should state the units, which here evidently is "litres per hour
Click to expand...
yeah .

Simultaneous work done by both pipes for 17 hrs​

(Wa + Wb) * 17 hrs = 425lit


Assuming x hrs is time for pipe B was opened . The first pipe having been opened five hours longer than the second to fill 425 liters .​

Total work done by both pipes during their respective hrs
Wa ( x hr+5 hr ) + Wb x = 425 lit


y is the amount filled up by pipe B in x hrs where work rate is Wb litre per hr .

Wb x = y



the second pipe was open as long as the first pipe was open​

Wb ( x +5) = (y +k) ; y+ k liters is the amt filled up




first pipe was open as long as the second and first pipe would deliver half the amount of water delivered by the second pipe​

Wa * x = (y + k ) /2
2 Wa x = y + k

We know , Wb ( x +5) = (y +k)

subst. 2 Wax in it gives

Wb ( x +5) = 2 Wa x

Wb x + 5Wb = 2 Wa x

Wb x = 2Wa x - 5 Wb

Right?
 
I am going to give you a gift: doing the work I asked you to do, since you at least did part of it.

Here are your definitions of four variables, slightly improved:

Wa = litres per hour for pipe A​
Wb = litres per hour for pipe B​
x = hrs pipe B was opened in first scenario.​
y = litres delivered by pipe B in x hrs​

The last definition immediately implies one of your equations:
Wb x = y​

But in the course of your work, you introduce a fifth variable, k, that isn't clearly defined, except that
"y + k liters is the amt filled up [i.e. delivered by the second pipe in scenario 2]"​

Here are the statements in the problem, each followed by the equation you wrote for it:

Scenario 1: A tank of 425 litres capacity has been filled with water through two pipes, the first pipe having been opened five hours longer than the second.​
Wa ( x + 5) + Wb x = 425​
​
Scenario 2: If the first pipe were open as long as the second [was actually open], and the second pipe was open as long as the first pipe was [actually] open, then the first pipe would deliver half the amount of water delivered by the second pipe;​
[I've added words to clarify this, which is very awkward]​
Wb ( x + 5) = (y + k)​
Wa * x = (y + k) /2​
​
Scenario 3: if the two pipes were open simultaneously, the tank would be filled up in 17 hours.​
(Wa + Wb) * 17 = 425​
​
How long was the second pipe open?​
Solve for x.​

We do appear to have five equations for the five unknowns; I'm not looking at your solving work yet at all, just trying to do for you what I asked you to do, which was to state all your equations first, before starting to solve them. This allows me to check that they all make sense. They do, now that I have put them in order.

Now you need to solve. If I were doing this, I would be trying to eliminate variables, one by one. Clearly we can eliminate y by replacing it with Wb * x; and we can eliminate "y+k" by combining the two equations containing it. This reduces us to three variables, with these equations:

Wa ( x + 5) + Wb x = 425​
Wa * x = Wb ( x + 5) /2​
(Wa + Wb) * 17 = 425 (which quickly becomes Wa + Wb = 25)​

Now, I think you've done these things; what you didn't do is to gather the three remaining equations so you can see where you stand.

Since we are solving for x, I would probably first try to eliminate Wa or Wb using the last equation, and then the other.

Do you see yet how an orderly approach can make a complicated problem manageable? With random equations written all over the place, you are bound for trouble. Please learn this, if nothing else!
 
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