T & W q12

S

Saumyojit

Senior Member
Joined
Jan 21, 2020
Messages
1,032
Two pipes A and B fill up a half tank in 1.2 hours. The tank was initially empty. Pipe B was kept open for half the time required by pipe A to fill the tank by itself . Then pipe A was kept open for as much time as was required by the pipe B to fill up 1/3 of the tank by itself. It was found that the tank was 5/6 full. The least time in which any of the pipes can the fill the tank fully is ?



Work done per minute by pipe A = Wa
Time taken by pipe A to fill up a tank for x minutes = Wa * x = Volume ( V )
x = V/ Wa


Work done per minute by pipe B = Wb
pipe B kept open for x/2 min = Wb * x/2


Work done by pipe B to fill up in one-third of tank in y min = Wb *y min = V / 3

y = V / 3Wb


=> Wb = V/ 3y = (Wa x) / 3y


=> Wa * y min = kept open

( Wa + Wb ) * 72 min = V/2 ---(i)


=> (Wb * x/2 ) + ( Wa *y ) + 5/6 = 1

Substituting y = V / 3Wb .
( ( Wa x ) /3y * V/ 2*Wa ) + ( Wa * V/ 3Wb ) + 5/6 = 1



Then?
 
Saumyojit said:
Two pipes A and B fill up a half tank in 1.2 hours. The tank was initially empty. Pipe B was kept open for half the time required by pipe A to fill the tank by itself . Then pipe A was kept open for as much time as was required by the pipe B to fill up 1/3 of the tank by itself. It was found that the tank was 5/6 full. The least time in which any of the pipes can the fill the tank fully is ?



Work done per minute by pipe A = Wa
Time taken by pipe A to fill up a tank for x minutes = Wa * x = Volume ( V )
x = V/ Wa


Work done per minute by pipe B = Wb
pipe B kept open for x/2 min = Wb * x/2


Work done by pipe B to fill up in one-third of tank in y min = Wb *y min = V / 3

y = V / 3Wb


=> Wb = V/ 3y = (Wa x) / 3y


=> Wa * y min = kept open

( Wa + Wb ) * 72 min = V/2 ---(i)


=> (Wb * x/2 ) + ( Wa *y ) + 5/6 = 1

Substituting y = V / 3Wb .
( ( Wa x ) /3y * V/ 2*Wa ) + ( Wa * V/ 3Wb ) + 5/6 = 1



Then?
Click to expand...
If I am not mistaken you have 2 equations and 5 variables: Wa, Wb, V, x, y. How do you expect to solve this system?
I would suggest taking another look at this post as an example of how to set up your equations:

t & w q11

Q. Mini and Vinay are quiz masters preparing for a quiz. In 'x' minutes, Mini makes 'y' questions more than Vinay. If it were possible to reduce the time needed by each to make a question by 2 mins , then in 'x' minutes Mini would make '2y' questions more than Vinay. How many questions does...
www.freemathhelp.com
 
Saumyojit said:
Work done per minute by pipe A = Wa
Time taken by pipe A to fill up a tank for x minutes = Wa * x = Volume ( V )
x = V/ Wa


Work done per minute by pipe B = Wb
pipe B kept open for x/2 min = Wb * x/2


Work done by pipe B to fill up in one-third of tank in y min = Wb *y min = V / 3

y = V / 3Wb


=> Wb = V/ 3y = (Wa x) / 3y


=> Wa * y min = kept open

( Wa + Wb ) * 72 min = V/2 ---(i)


=> (Wb * x/2 ) + ( Wa *y ) + 5/6 = 1

Substituting y = V / 3Wb .
( ( Wa x ) /3y * V/ 2*Wa ) + ( Wa * V/ 3Wb ) + 5/6 = 1



Then?
Click to expand...
Wht is the problem with this steps ?
 
Saumyojit said:
Two pipes A and B fill up a half tank in 1.2 hours. The tank was initially empty. Pipe B was kept open for half the time required by pipe A to fill the tank by itself . Then pipe A was kept open for as much time as was required by the pipe B to fill up 1/3 of the tank by itself. It was found that the tank was 5/6 full. The least time in which any of the pipes can the fill the tank fully is ?
Click to expand...
I would use only two variables. More only complicates things. Your equations aren't worth trying to solve, or even check if they can be rescued.

I let A = time for A alone to fill the tank; B = time for B alone to fill tank.

Write two equations, and show them to us; don't try solving until we agree you have interpreted the problem correctly.

The first time I tried solving, I got a cubic equation, which is not good. The next time, I solved it easily using a trick based on the special form of one of the equations, but got two solutions, which explains the unusual question at the end.
 
Saumyojit said:
Wa (per min work) * A mins = V

Wb * B mins = V
Click to expand...
That's still 5 variables! You don't need them all. There is no mention of actual volumes, so you don't need V; and the rate for A, in tanks per hour, is just 1/A, where A is the time taken to fill one tank (hours per tank). All you've done so far is to write two relationships among your excessive variables, not the equations that represent the actual problem.

Note that time in the problem is in hours, not minutes. So we define the variables as

A = time (in hours) for A alone to fill the tank​
B = time (in hours) for B alone to fill tank​

And I'll give you the first equation:
Saumyojit said:
Two pipes A and B fill up a half tank in 1.2 hours.
Click to expand...
Amount filled in 1.2 hours, at rate 1/A + 1/B = 1 tank:
1.2(1/A + 1/B) = 1​

or, if you prefer,

Rate together = 1 tank/1.2 hour:
1/A + 1/B = 1/1.2​

Do you understand these relationships? What you need to do is to first carefully define your variables, including units, and check that they are consistent and not redundant. You never fully did this. What you said was:
Saumyojit said:
Work done per minute by pipe A = Wa
Time taken by pipe A to fill up a tank for x minutes = Wa * x = Volume ( V )
x = V/ Wa

Work done per minute by pipe B = Wb
pipe B kept open for x/2 min = Wb * x/2
Click to expand...
You defined Wa and Wb, partly, but didn't state the units for "work", which you seem to be taking as "volume" (in unstated units) rather than "tankfuls", which is the usual way to handle these problems (so that V = 1). If you persisted, you would presumably find that the (unknown constant) V disappeared in your final answer, so it isn't really wrong, but it is a huge waste to use it.

You didn't define x and y, except that they are (inappropriately) in minutes. Presumably you intend x to be "time required by pipe A to fill the tank by itself", and you've recognized that this is V/Wa, but have missed the fact that therefore you don't need x as a separate variable. Again, yes, you could continue, eventually replacing x with V/Wa, and get reasonable equations; but you are making things much too hard for yourself.

(Your line "Time taken by pipe A to fill up a tank for x minutes =..." is nonsense as written; you presumably meant something like "Time taken by pipe A to fill up a tank = x minutes; therefore ...".)

This is why you are getting so tangled up in equations. Making the effort at the start to keep things simple would prevent your wasted time (and ours).
 
Dr.Peterson said:
Right. That's what I meant, and that's what I did when I previously solved it. I just wrote the equations too fast, without looking at my work.

Let's pretend I was giving @Saumyojit a chance to catch and fix an error ...
Click to expand...
Wa ( Per hr work by pipe A ) = 1/ A

Wb ( Per hr work by pipe B ) = 1/B


1.2(1/A + 1/B) = 1 /2

1.2 ( A + B ) = AB / 2

Wb * A/2 = kept open

Wb * C hrs = 1/ 3


Wa * C = kept open


Wb * A/2 + Wa * C = 1/6



1/3c * A/2 + C/A = 1/6

A/ 6c + C/A = 1/6

Then?
 
I defined two variables for you; that's all you need. Where did C come from? How do you define it?

It is possible to introduce intermediate variables to keep equations simple, but you have to do that carefully, so that you can soon replace such variables with expressions in the actual variables.

I can't follow what you're doing here, which is presumably intended to write an equation representing
Saumyojit said:
Pipe B was kept open for half the time required by pipe A to fill the tank by itself . Then pipe A was kept open for as much time as was required by the pipe B to fill up 1/3 of the tank by itself.
Click to expand...
I would start by writing an expression for "half the time required by pipe A to fill the tank by itself" (which apparently is what you mean by C), and then for how much of the tank is filled by pipe B alone in that time.

Then I'd write an expression for "as much time as was required by the pipe B to fill up 1/3 of the tank by itself", using only our two variable B.

One thing you need to do to keep track of what you are doing is to use words! Say what each expression or equation means.
 
Saumyojit said:
Wb * C hrs = 1/ 3
Click to expand...
C is time taken by pipe B to fill 1/3 of tank .


Pipe B was kept open for half the time which is ( A/2 hrs ) required by pipe A to fill the tank by itself

Wb * A/2 = filled some amount of tank




pipe A was kept open for as much time as was required by the pipe B to fill up 1/3 of the tank by itself.
Yeah so Wa * C hrs= Filled Some Amount of tank
 
Saumyojit said:
C is time taken by pipe B to fill 1/3 of tank .
Click to expand...
What is C, in terms of B?

What are Wa and Wb in terms of A and B?

You don't need these extra variables; if you use them, they should almost immediately be replaced, so only two variables remain.
 
Where did the 1/6 come from? Should that be 5/6?

What you did at the end didn't eliminate a variable, so in itself it is not helpful.

The goal is to eliminate a variable, right? So try to solve one equation for one variable in terms of the other. That should be a familiar idea, not something you have to ask about.

On the other hand, I think I recall that when I solved this, I observed that the one equation is homogeneous (every term having the same degree), so by dividing it by B^2, I could obtain a quadratic equation in A/B and solve for that, which simplified the rest of the work.
 
Write the two equations you have in A and B only. That's what we're working on, so we need to see them in one place.

Then decide whether you can solve one for either variable. Do it, and show us what you did. If you really can't do anything, then I guess you are stuck.
 
Saumyojit said:
That is B
Click to expand...
So the equation is: (assuming correct work)

A * B = 2.4 * (A + B)

A = 2.4 * B / (B - 2.4) .... Continue...

I would suggest - you start with a clean slate and start over.
 
Last edited by a moderator:
You must log in or register to reply here.
Top