A problem on quadratic equations .

[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