# Problem on mixture. Please explain with steps.

• Oct 16th 2012, 08:37 AM
hisajesh
Problem on mixture. Please explain with steps.
A cask contains 3 parts wine and one part water. What part of the mixture must be drawn off and substituted by water so that the resulting mixture may be half wine and half water.
• Oct 16th 2012, 09:42 AM
HallsofIvy
Re: Problem on mixture. Please explain with steps.
Since there is no specific amount given, lets take the total liquid in the cask to be 1. Then the amount of water in the cask is 1/4 and the amount of wine is 3/4. Suppose you take out amount x (x< 1). The amount of water in that is x/4 and the amount of wine is 3x/4. That will leave 1- x/4 water and 1- 3x/4 wine. You want to add an amount of water to make this equal parts wine and water.

Since we "substitute" that by water, we simply replace the amount taken out, so that we are adding x water, then we will have 1- x/4+ x= 1+ 3x/4 water and 1- 3x/4 wine. To have equal amounts of water and wine, those must be equal: 1+3x/4= 1- 3x/4. Solve that for x.
• Oct 16th 2012, 01:33 PM
MaxJasper
Re: Problem on mixture. Please explain with steps.
Quote:

Originally Posted by HallsofIvy
To have equal amounts of water and wine, those must be equal: 1+3x/4= 1- 3x/4. Solve that for x.

This results in x=0.

I suspect we should have:

3/4-3x/4 = 1/4-x/4+x

which results in x = 1/3 (not sure yet)
• Oct 16th 2012, 01:36 PM
hisajesh
Re: Problem on mixture. Please explain with steps.
Solving that X gives 0.

Your 3rd line "That will leave 1- x/4 water and 1- 3x/4 wine. You want to add an amount of water to make this equal parts wine and water."
Will that not leave (1/4)-(x/4) water and (3/4)-(3x/4) wine and then adding x water gives water equation as (1/4)-(x/4)+x
and leaving wine the same amount (3/4)-(3/4)x. Now solving (1/4)-(x/4)+x and (3/4)-(3/4)x yields x=1/3 which is the correct answer.

I have another question. I started the problem by taking 4 parts as whole and removing x from 4 and going through this way
yields 4/3 as x.
You started the problem by taking 1 as whole and could arrive at 1/3 as x which is the correct anwer. kindly reflect what makes the difference.
• Oct 16th 2012, 01:48 PM
TheEmptySet
Re: Problem on mixture. Please explain with steps.
Quote:

Originally Posted by hisajesh
Solving that X gives 0.

Your 3rd line "That will leave 1- x/4 water and 1- 3x/4 wine. You want to add an amount of water to make this equal parts wine and water."
Will that not leave (1/4)-(x/4) water and (3/4)-(3x/4) wine and then adding x water gives water equation as (1/4)-(x/4)+x
and leaving wine the same amount (3/4)-(3/4)x. Now solving (1/4)-(x/4)+x and (3/4)-(3/4)x yields x=1/3 which is the correct answer.

I have another question. I started the problem by taking 4 parts as whole and removing x from 4 and going through this way
yields 4/3 as x.
You started the problem by taking 1 as whole and could arrive at 1/3 as x which is the correct anwer. kindly reflect what makes the difference.

Here is my 2 cents. I visualized the problem as this

At the begining the mixture has 75% wine and since it is well mixed what ever volume is removed with be 75% wine. At the end we want the percent to be 50%

This gives

$.75-.75x=.5 \iff -.75x =-.25 \iff x =\frac{25}{75}=\frac{1}{3}$

Also where abouts are you from in Utah. I went to school there for a while.
• Oct 16th 2012, 01:50 PM
Soroban
Re: Problem on mixture. Please explain with steps.
Hello, hisajesh!

Quote:

A cask contains 3 parts wine and one part water.
What part of the mixture must be drawn off and substituted by water
so that the resulting mixture may be half wine and half water?

Suppose the cask holds 60 liters of liquid.
It contains 45 liters of wine and 15 liters of water.

We remove $x$ liters of the mixture.
. . We remove $\tfrac{3}{4}x$ liters of wine.
. . We remove $\tfrac{1}{4}x$ liters of water.
The cask contains: $\begin{Bmatrix}45-\tfrac{3}{4}x\text{ liters of wine}\; \\ 15-\tfrac{1}{4}x\text{ liters of water} \end{Bmatrix}$

We add $x$ liters of water.
The cask contains: . $\begin{Bmatrix}45-\frac{3}{4}x\text{ liters of wine}\; \\ 15 + \frac{3}{4}x\text{ liters of water} \end{Bmatrix}$

But these two quantities are supposed to be equal.

There is our equaton! . . . $45-\tfrac{3}{4}x \;=\;15 + \tfrac{3}{4}x$
• Oct 16th 2012, 02:08 PM
MaxJasper
Re: Problem on mixture. Please explain with steps.
Quote:

Originally Posted by Soroban
There is our equaton! . . . $45-\tfrac{3}{4}x \;=\;15 + \tfrac{3}{4}$

adding missing x in above equations we have:

$45-\tfrac{3}{4}x \;=\;15 + \tfrac{3}{4}x$

resulting x=20 which is 1/3 original volume.
• Oct 16th 2012, 02:17 PM
hisajesh
Re: Problem on mixture. Please explain with steps.
@ everyone: Thank you for your flurry of answers!!!!!

I started the problem by taking 4 parts as whole and removing x from 4 and going through this way
yields 4/3 as x. If I start the problem by taking 1 as whole and could arrive at 1/3 as x which is the correct anwer. kindly reflect what makes the difference
• Oct 16th 2012, 02:20 PM
MaxJasper
Re: Problem on mixture. Please explain with steps.
Whatever you assume the original volume to be you always end up with 1/3 of it. Start with vol v and get v/3.
• Oct 16th 2012, 02:21 PM
TheEmptySet
Re: Problem on mixture. Please explain with steps.
Quote:

Originally Posted by hisajesh
@ everyone: Thank you for your flurry of answers!!!!!

I started the problem by taking 4 parts as whole and removing x from 4 and going through this way
yields 4/3 as x. If I start the problem by taking 1 as whole and could arrive at 1/3 as x which is the correct anwer. kindly reflect what makes the difference

You found the volume removed, but if you want the RATIO you need find

$\frac{\frac{4}{3}}{4}=\frac{1}{3}$
• Nov 9th 2012, 06:10 PM
Soroban
Re: Problem on mixture. Please explain with steps.
Hello, hisajesh!

Quote:

A cask contains 3 parts wine and one part water.
What part of the mixture must be drawn off and substituted by water
so that the resulting mixture may be half wine and half water?

Suppose the cask contains $W$ units of the wine/water blend.
. . It is 25% water; it contains $\tfrac{1}{4}W$ units of water.

We draw off $x$ units of the mixture.
. . We remove $\tfrac{1}{4}x$ units of water.

We add $x$ units of pure water.
. . This contains $x$ units of water.

The final amount of water is: . $\tfrac{1}{4}W - \tfrac{1}{4}x + x \:=\:\tfrac{1}{4}W + \tfrac{3}{4}x$ units.

But we know that the final mixture is $W$ units which is 50% water.
. . So it contains $\tfrac{1}{2}W$ units of water.

There is our equation! . . . . $\tfrac{1}{4}W + \tfrac{3}{4}x \:=\:\tfrac{1}{2}W$

Therefore: . $\tfrac{3}{4}x \:=\:\tfrac{1}{4}W \quad\Rightarrow\quad x \:=\:\tfrac{1}{3}W$

We should draw off one-third of the cask and replace it with water.