A problem on quadratic equations .

[sankalpmittal]

Homework Statement

There are 2 beakers P and Q. P contains 50 litres of water and Q contains 40 litres of acid. x litres is taken from both the beakers, mixed and replaced. This process is repeated 2 times and after the second time, there is 71/8 litres of acid in P. What is the quantity of x ?

Homework Equations

Form the equation in the form of ax² + bx + c = 0 where a is not equal to 0 and a,b and c are constants.

The Attempt at a Solution

 

[SammyS]

What does it mean by: "... x litres is taken from both the beakers, mixed and replaced."

What is replaced? ... By what is it replaced?

 

[PeterO]

Homework Statement

There are 2 beakers P and Q. P contains 50 litres of water and Q contains 40 litres of acid. x litres is taken from both the beakers, mixed and replaced. This process is repeated 2 times and after the second time, there is 71/8 litres of acid in P. What is the quantity of x ?

Homework Equations

Form the equation in the form of ax² + bx + c = 0 where a is not equal to 0 and a,b and c are constants.

The Attempt at a Solution

Just checking, this reads as though the mixing is done three times in all - done once then repeated twice. Is that what is meant, or should it merely have stated "This process is done twice".

I suggest you choose a couple of values for x and see what is happening , to get a feel for the situation.

For example, what happens if the samples are 5 litres from each? 10 litres from each instead.

eg: when x is 10 the first process will transfer 5 litres of acid to the water and 5 litres of water to the acid, 1st repeat will transfer less than 5 litres to each, and so on.

 

[PeterO]

What does it mean by: "... x litres is taken from both the beakers, mixed and replaced."

What is replaced? ... By what is it replaced?

Sounds very much like half the mix is put back in each bucket, so that there is 50 and 40 litres of liquid in each bucket after each sample, mix and replacement

 

[sankalpmittal]

DETAILED EXPLANATION OF QUADRATIC EQUATION PROBLEM1. Suppose there is a beaker P. It contains 50 litres of water .
2. Also suppose there is beaker Q. It contains 40 litres of acid. Got it ?
3. Now "x" litres of acid is taken from beaker Q .
4. "x" litres of water is also taken from beaker P.
5. This x litres of water from beaker P and also the x litres of acid from beaker Q are mixed with each other. Got this statement ?
6. After they are mixed and form mixture , the same x litres are taken from mixed mixture and added to both the beakers P and Q.
7. Now this process from step 3 to 6 is repeated again .
8. Now the question asks what is the value of x ? How many litres ?I hope you get it . Please solve it .

Thanks in advance .

(I am 14 years , in class 10th . The above problem was from book "elementary algebra" by H.S. Hall and Knight . )

 

[sjb-2812]

OK, let's try one iteration of removal and replacement.

If x litres is removed from the acid container, how much is placed into the water container?

 

[sankalpmittal]

OK, let's try one iteration of removal and replacement.

If x litres is removed from the acid container, how much is placed into the water container?

umm umm

x litres removed from acid container ?? How much placed ??

I don't know where to start . Please solve it step by step .

 

[sjb-2812]

Remove x litres from Q. Add it to x litres from P. Mix. Replace x litres into P. How much acid is there in P?

 

[sankalpmittal]

Remove x litres from Q. Add it to x litres from P. Mix. Replace x litres into P. How much acid is there in P?

x litres from Q mixed with x litres from P becomes 2x , it is diffused uniformly .
Now x litres put back to Q , so
40 l of acid + x/2 l of water - x/2 l of acid .In P :
50 l of water - x/2 l of water + x/2 l of acid .

Right ?

That's not , what the question says ? Check my post #5 .

I am already at step 6.

Sorry my bad . I edited the post .

 

[sjb-2812]

That's not , what the question says ? Check my post #5 .

I am already at step 6.

 

[sjb-2812]

So if there is 40- (x/2) litres of acid in Q, what is its concentration?

 

[sankalpmittal]

So if there is 40- (x/2) litres of acid in Q, what is its concentration?

No idea . I can't solve this problem . Maybe you can .

In Q :
40 l of acid + x/2 l of water - x/2 l of acid .In P :
50 l of water - x/2 l of water + x/2 l of acid .

 

[SammyS]

We're here to help you solve problems. According to the rules of this site, we're not supposed to solve them for you.

 

[sankalpmittal]

We're here to help you solve problems. According to the rules of this site, we're not supposed to solve them for you.

See post #5 . Question corrected to extent .

In Q :
40 l of acid + x/2 l of water - x/2 l of acid .


In P :
50 l of water - x/2 l of water + x/2 l of acid .


After this what ?

 

[sjb-2812]

Can you define concentration?

 

[ehild]

So there is x/2 l of acid in P and 40-x/2 l acid in Q after the first step. You take out x litres from both bakers again. How much acid is in the x litres taken from P and how much acid is in the x litres taken from Q?

ehild

 

[sankalpmittal]

So there is x/2 l of acid in P and 40-x/2 l acid in Q after the first step. You take out x litres from both bakers again. How much acid is in the x litres taken from P and how much acid is in the x litres taken from Q?

ehild

Umm


we again take out x l from both . So in
P:

How will I form quadratic equation ? I don't know . Please just suggest only one step so that i can get it .

 

[ehild]

The quadratic comes at the end. Answer my question. There is x/2 l of acid in 50 l mixture. You take x litres again. How much acid is in that x litres?

If it is easier with number, let be 5 l acid in 50 l mixture. You take out 10 l. How much acid is in that 10 litres?

ehild

 

[PeterO]

See post #5 . Question corrected to extent .

In Q :
40 l of acid + x/2 l of water - x/2 l of acid .


In P :
50 l of water - x/2 l of water + x/2 l of acid .


After this what ?

Try some actual numbers [ and hopefully not happen upon the correct answer ] just to see how it works. To dive straight into x is really quite advanced - especially if this is the first such problem you have done.

For example let the sample be 40 litres from each - all of the smaller [acid] container. [ie x = 40]

After mixing and replacing, you now have 20 litres of acid in each.
The 40 litre container is 20 litres of water plus 20 litres of acid, the 50 litre beaker is 30 / 20.

Now you again take 40 litres from each [1st repeat]

You get 20 lites of water plus 20 litres of acid from the 40l beaker [everything again] and 40l from the 50 litre pot which is 60% water / 40% acid so 24 litres of water - 16 litres of acid.

mix together and the 80 litres is 44 litres water - 36 litres acid

So you now return 22l water - 18l acid to each beaker

The 50 litre beaker already had the remaining 6 l water and 4 l acid to make a new mix of 28 l water and 22 l acid. The 40 litre container is just re-filled with the 22-18 mix.

Now you do it again - you calculate this time.

You will then end up with a certain amount of acid /water in each beaker.

If you keep track of what has been happening to the original 40, you can then do those same things to the general sample size, x.

For example, after the first sample - mix - replace cycle, there is x/2 litres of acid in the water and x/2 litres of water in the acid.

Work through a couple of times with numbers if necessary, and you should see how this works. For example let x = 6,10,14 or 22 - each of those is 2 times a prime. We know we are going to halve the first time, so we may as well choose an even number to make that work. From there on you will be doing something to a prime number, so should be able to keep track of what is happening.

[For example if we chose 8, it would become 4 after the first operation. If it became 2 after the second, we wouldn't know if it had been halved or "square rooted"]

Peter

 

[sankalpmittal]

Try some actual numbers [ and hopefully not happen upon the correct answer ] just to see how it works. To dive straight into x is really quite advanced - especially if this is the first such problem you have done.

For example let the sample be 40 litres from each - all of the smaller [acid] container. [ie x = 40]

After mixing and replacing, you now have 20 litres of acid in each.
The 40 litre container is 20 litres of water plus 20 litres of acid, the 50 litre beaker is 30 / 20.

Now you again take 40 litres from each [1st repeat]

You get 20 lites of water plus 20 litres of acid from the 40l beaker [everything again] and 40l from the 50 litre pot which is 60% water / 40% acid so 24 litres of water - 16 litres of acid.

mix together and the 80 litres is 44 litres water - 36 litres acid

So you now return 22l water - 18l acid to each beaker

The 50 litre beaker already had the remaining 6 l water and 4 l acid to make a new mix of 28 l water and 22 l acid. The 40 litre container is just re-filled with the 22-18 mix.

Now you do it again - you calculate this time.

You will then end up with a certain amount of acid /water in each beaker.

If you keep track of what has been happening to the original 40, you can then do those same things to the general sample size, x.

For example, after the first sample - mix - replace cycle, there is x/2 litres of acid in the water and x/2 litres of water in the acid.

Work through a couple of times with numbers if necessary, and you should see how this works. For example let x = 6,10,14 or 22 - each of those is 2 times a prime. We know we are going to halve the first time, so we may as well choose an even number to make that work. From there on you will be doing something to a prime number, so should be able to keep track of what is happening.

[For example if we chose 8, it would become 4 after the first operation. If it became 2 after the second, we wouldn't know if it had been halved or "square rooted"]

Peter

Hmm a bit complicated though !

The quadratic comes at the end. Answer my question. There is x/2 l of acid in 50 l mixture. You take x litres again. How much acid is in that x litres?

If it is easier with number, let be 5 l acid in 50 l mixture. You take out 10 l. How much acid is in that 10 litres?

ehild

Ummmm

50 l -> x/2 l of acid
so x l -> x²/100 l of acid ?

 

[ehild]

Hmm a bit complicated though !
Ummmm

50 l -> x/2 l of acid
so x l -> x²/100 l of acid ?

Why are you confused? It is right. Let's go ahead.

1. How much acid remained in beaker P?

2. What is the amount of acid in the x l sample taken from the 40 l beaker?

3. You mix the two x l samples and halve the mixture. What is the amount of acid in one sample?

4. You return that x l of mixture to beaker P. How much will be the acid in it altogether in terms of x?

But the amount of acid is given after the second step: it is 71/8 l. So you have the equation for x.

ehild

 

[sankalpmittal]

Why are you confused? It is right. Let's go ahead.

1. How much acid remained in beaker P?

2. What is the amount of acid in the x l sample taken from the 40 l beaker?

3. You mix the two x l samples and halve the mixture. What is the amount of acid in one sample?

4. You return that x l of mixture to beaker P. How much will be the acid in it altogether in terms of x?

But the amount of acid is given after the second step: it is 71/8 l. So you have the equation for x.

ehild

In beaker Q :
40 l -> 40-x/2 l
x l -> (80x-x²)/80 l of acid

we mix those two :

x²/100 + 80x-x²/80

In one sample therefore :

(4x²+400x-5x²)/200 l of acid .
=(400x-x²)/200 l of acid .
Now ?

 

[ehild]

Do not be so desperate! You do it very well... Just some little mistake: the acid content in the new x l sample is x²/100 + (80x-x²)/80. Do not forget the parentheses! And they add up to (400x-x²)/400 in the 2x l sample. Taking the half,
one sample contains (400x-x²)/800 l acid. You add it to beaker P. But remember: when taken out x l from it, you removed x²/100 l acid. So the acid content will be the present amount + the acid content of the sample. How much is it?

ehild

 

[PeterO]

In beaker Q :
40 l -> 40-x/2 l
x l -> (80x-x²)/80 l of acid

we mix those two :

x²/100 + 80x-x²/80

In one sample therefore :

(4x²+400x-5x²)/200 l of acid .
=(400x-x²)/200 l of acid .
Now ?

"sankalpmittal" when this tread is read from post #1 to the end, you are not solving anything.
I would advise no further help from anyone until a full - even if incorrect - solution is offered by OP and not just a lot of frown/confused emoticons.

 

[ehild]

PeterO,

I lead Sankalpmittal through the problem by asking questions and him answering them, almost correctly up to now. He is 14 years old, someone has to show him once how to solve such problems. Let me do it in peace. Please, do not discourage him. This is not a simple problem, you can not expect somebody at his age to solve it by himself.

ehild

 

[sankalpmittal]

PeterO,

I lead Sankalpmittal through the problem by asking questions and him answering them, almost correctly up to now. He is 14 years old, someone has to show him once how to solve such problems. Let me do it in peace. Please, do not discourage him. This is not a simple problem, you can not expect somebody at his age to solve it by himself.

ehild

Even my school math teacher failed to solve this question. He doesn't even tried it and said that this question is weird and I can't solve it . I asked 3 or 4 high school level teachers , they can't solve it . I am trying in my copy . These questions from Hall and Knight book are rather difficult not like normal text school book questions . ehild please solve it once so that I can understand its technique of solving .

 

[ehild]

Sankalpmittal, you were almost at the end. You should not have given up.

Well, you determined correctly the amount of acid both in P and in Q after the first step of mixing. Taken out x l from both beakers and mixed, then returned x l back to each beakers, the amount of acid was x/2 l in P and 40-x/2 l in Q.
In the next step, x l is removed again from both bakers. The sample from P contains x²/100 l acid, and that from Q contains (40-x/2)x/40=(80x-x²)/80 l acid. The two samples altogether contain

x²/100+ (80x-x²)80= (400-x²)/400 l acid.

This mixture is halved and one half is returned to beaker P, the other to Beaker Q. The amount of acid in each half is

(400-x²)/800 l.

(You were here, making only a little mistake.)

As x²/100 l acid has been removed from beaker P at the beginning of the second mixing step, it contains

(x/2-x²/100) l

acid in the (50-x) l solution. Now x l mixture is returned, with (400x-x²)/800 l acid content. So the total amount of acid becomes

x/2-x^2/100+(400x-x^2)/800=\frac{400x-8x^2+400x-x^2}{800}=\frac{800x-9x^2}{800}.

This amount is given in the problem, it is 71/8 l.

So your quadratic equation is

\frac{800x-9x^2}{800}=\frac{71}{8}

Can you continue from here? Have you learned solving quadratic equations?

You will get two roots, but one is physically meaningless. Let me see what you get for x. I am not allowed to give full solution.

ehild.

 

[Ray Vickson]

See post #5 . Question corrected to extent .

In Q :
40 l of acid + x/2 l of water - x/2 l of acid .


In P :
50 l of water - x/2 l of water + x/2 l of acid .


After this what ?

Answer this: how do you *know* that after the first mixing we are putting back x/2 l of acid and x/2 l of water in each vat? In other words, if we mix x l of water and x l of acid to get 2x l of mixture, how do we know that x l of the mix has x/2 l of acid and x/2 l of water? If you can answer that, you should be able to similarly decide how much acid there is in x l of a mixture from (40 - x/2) of water and x/2 of acid, etc.

RGV

 

[PeterO]

PeterO,

I lead Sankalpmittal through the problem by asking questions and him answering them, almost correctly up to now. He is 14 years old, someone has to show him once how to solve such problems. Let me do it in peace. Please, do not discourage him. This is not a simple problem, you can not expect somebody at his age to solve it by himself.

ehild

You could have supplied a clear/concise solution to "similar" problem with different sized beakers as a guide to how such a question is solved.

 

[SammyS]

You could have supplied a clear/concise solution to "similar" problem with different sized beakers as a guide to how such a question is solved.

Yes, ehild could have approached this in anyone of many ways. He chose to try to lead sankalpmittal through the problem in such a way so that sankalpmittal might actually discover some of the steps himself.

Frankly, I admire the patience shown by ehild in this endeavor.

 

[ehild]

You could have supplied a clear/concise solution to "similar" problem with different sized beakers as a guide to how such a question is solved.

People of different age need different explanation. If you need help to find a general solution to similar problems on your level of knowledge, I show you:

Assume you have two beakers P and Q. The volumes are V_p and V_q, both beakers are filled with a mixture of two fluids, "1" and "2". The concentrations of one of the components in the vessels are p and q, respectively. Assume that mixing of the components preserves the volume, that is, the volume of a mixture of v1 volume of component 1 and v2 volume of component 2 is v1+v2.

Taking x volume from both beakers and mixing them, the concentration of this sample is 0.5(p+q)

Returning x l of this mixture to beaker P, the new concentration p' becomes:

p'=[p(V_p-x)+0.5(p+q)x]/V_p=p+0.5(x/V_p)(q-p)

Returning x l of the sample mixture to Q, the new concentration q' becomes

q'=[q(V_q-x)+0.5(p+q)x]/V_q=q-0.5(x/V_q)(q-p)

In case of subsequent mixing steps you have the recurrence relations between concentrations:

p(n+1)=p(n)+0.5(x/V_p)(q(n)-p(n))

q(n-1)=q(n)-0.5(x/V_q)(q(n)-p(n))

I hope this is enough help for you to solve the original problem.

ehild

 

[PeterO]

People of different age need different explanation. If you need help to find a general solution to similar problems on your level of knowledge, I show you:

Assume you have to beakers P and Q. The volumes are V_p and V_q, both beakers are filled with a mixture of two fluids, "1" and "2". The concentrations of one of the components in the vessels are p and q, respectively. Assume that mixing of the components preserves the volume, that is, the volume of a mixture of v1 volume of component 1 and v2 volume of component 2 is v1+v2.

Taking x volume from both beakers and mixing them, the concentration of this sample is 0.5(p+q)

Returning x l of this mixture to beaker P, the new concentration p' becomes:

p'=[p(V_p-x)+0.5(p+q)x]/V_p=p+0.5(x/V_p)(q-p)

Returning x l of the sample mixture to Q, the new concentration q' becomes

q'=[q(V_q-x)+0.5(p+q)x]/V_q=q-0.5(x/V_q)(q-p)

In case of subsequent mixing steps you have the recurrence relations between concentrations:

p(n+1)=p(n)+0.5(x/V_p)(q(n)-p(n))

q(n-1)=q(n)-0.5(x/V_q)(q(n)-p(n))

I hope this is enough help for you to solve the original problem.

ehild

Yeah, right; I am sure that is much simpler to follow.

 

[sankalpmittal]

Sankalpmittal, you were almost at the end. You should not have given up.

Well, you determined correctly the amount of acid both in P and in Q after the first step of mixing. Taken out x l from both beakers and mixed, then returned x l back to each beakers, the amount of acid was x/2 l in P and 40-x/2 l in Q.
In the next step, x l is removed again from both bakers. The sample from P contains x²/100 l acid, and that from Q contains (40-x/2)x/40=(80x-x²)/80 l acid. The two samples altogether contain

x²/100+ (80x-x²)80= (400-x²)/400 l acid.

This mixture is halved and one half is returned to beaker P, the other to Beaker Q. The amount of acid in each half is

(400-x²)/800 l.

(You were here, making only a little mistake.)

As x²/100 l acid has been removed from beaker P at the beginning of the second mixing step, it contains

(x/2-x²/100) l

acid in the (50-x) l solution. Now x l mixture is returned, with (400x-x²)/800 l acid content. So the total amount of acid becomes

x/2-x^2/100+(400x-x^2)/800=\frac{400x-8x^2+400x-x^2}{800}=\frac{800x-9x^2}{800}.

This amount is given in the problem, it is 71/8 l.

So your quadratic equation is

\frac{800x-9x^2}{800}=\frac{71}{8}

Can you continue from here? Have you learned solving quadratic equations?

You will get two roots, but one is physically meaningless. Let me see what you get for x. I am not allowed to give full solution.

ehild.

9x²-800x+7100 = 0
9x²-90x-710x+7100=0
9x(x-10)-710(x-10)=0
(x-10)(9x-710)=0

These are the two roots of equation . The second root isn't possible because the value exceeds 50 . So the appropriate answer is that x is equal to 10 l .


Yes , I have learned quadratic equations in class 8th . This question was rather different though .

People of different age need different explanation. If you need help to find a general solution to similar problems on your level of knowledge, I show you:

Assume you have two beakers P and Q. The volumes are V_p and V_q, both beakers are filled with a mixture of two fluids, "1" and "2". The concentrations of one of the components in the vessels are p and q, respectively. Assume that mixing of the components preserves the volume, that is, the volume of a mixture of v1 volume of component 1 and v2 volume of component 2 is v1+v2.

Taking x volume from both beakers and mixing them, the concentration of this sample is 0.5(p+q)

Returning x l of this mixture to beaker P, the new concentration p' becomes:

p'=[p(V_p-x)+0.5(p+q)x]/V_p=p+0.5(x/V_p)(q-p)

Returning x l of the sample mixture to Q, the new concentration q' becomes

q'=[q(V_q-x)+0.5(p+q)x]/V_q=q-0.5(x/V_q)(q-p)

In case of subsequent mixing steps you have the recurrence relations between concentrations:

p(n+1)=p(n)+0.5(x/V_p)(q(n)-p(n))

q(n-1)=q(n)-0.5(x/V_q)(q(n)-p(n))

I hope this is enough help for you to solve the original problem.

ehild

Hm. This part is something which I am unable to understand. Your previous train of thoughts was easier to follow.


Thanks !

 

[ehild]

Well done! And you are right, it was a difficult problem, something for people over 18... Now you can teach your teachers how to solve such problems of mixing

The other method was for PeterO, he felt it easier than the one I showed to you.

ehild