1. Ratio and Proportions Question

Q: Two equal glasses filled with mixture of milk and water in the proportions of 2:1 and 1:1 respectively are emptied into a third glass. What is the proportion of milk and water in the third glass?

This reads quite simple and I should be able to do it but I'm stumped. Can someone give me a hint? I assume cross multiplying or adding the ratios is not the way to go... Trying to understand the why to this...

2. Originally Posted by dumluck
Q: Two equal glasses filled with mixture of milk and water in the proportions of 2:1 and 1:1 respectively are emptied into a third glass. What is the proportion of milk and water in the third glass?

This reads quite simple and I should be able to do it but I'm stumped. Can someone give me a hint? I assume cross multiplying or adding the ratios is not the way to go... Trying to understand the why to this...
glass 1 has 2/3 of a glass milk and 1/3 of a glass water

glass 2 has 1/2 of a glass of milk and 1/2 of a glass water.

So the mixture in glass 3 (assuming it lage enough to contain the contents of both glass 1 and glass 2 and that glasses 1 and 2 are of equal volumes) has (2/3)+(1/2) of a glass of milk and (1/3)+(1/2) of a glass of water.

So the ratio of milk to water in the mixture is ((2/3)+(1/2)) : ((1/3)+(1/2))

CB

3. I remember seeing a way to solve such problems that involves drawing a diagram consisting of crossing lines, but I can't find it now. If anybody could give a link, I would appreciate it.

4. Thanks Emakarov. I would assume there is an easier method than the above. The answer is 7:5 if that helps. Just struggling to get to it.

5. Originally Posted by dumluck
Thanks Emakarov. I would assume there is an easier method than the above. The answer is 7:5 if that helps. Just struggling to get to it.
Well, if we draw the 3 glasses, one glass twice as tall as the other two with same radius,
with a partition to prevent the milk and water from mixing.
Suppose the water is under the partition with the milk on top,
then the water is at 1/3 level in one glass and at 1/2 level in the other.
Then, in glass 3, the water level will be halfway between these two levels.

1/3 = 2/6.
1/2 = 3/6

Halfway is (2.5)/6 = 5/12.

So the milk occupies 7/12 of the third glass if we have a partition.

6. Thanks Archie,
That seems overly complex and how does it come to the ratio 7:5. It must be a lot simplier. The question doesn't state that the 3rd glass is a particular size so I would assume it''s large enough to take the combination of 2:1 and 1:1. Is this not the additions of the fractions 2/1 and 1/1? Which still does not give the correct answer so it's a confusing one.

7. Ok.. I think I got it.

2 + 1 = 3 so let's assume the glasses are 30mls.
2:1 would give 20/10 , the second would have 15/15
adding them gives us 35/25, divide by 5 and we get 7/5.

Thanks to all.

8. Originally Posted by dumluck
Thanks Archie,
That seems overly complex and how does it come to the ratio 7:5. It must be a lot simplier. The question doesn't state that the 3rd glass is a particular size so I would assume it''s large enough to take the combination of 2:1 and 1:1. Is this not the additions of the fractions 2/1 and 1/1? Which still does not give the correct answer so it's a confusing one.
Yes, I was "doing something with lines",
not the best explanation though!

CB's method works it through for the general case.

9. Originally Posted by Archie Meade
Yes, I was "doing something with lines",
not the best explanation though!

CB's method works it through for the general case.
It's no problem, I very much appreciate the help .

10. OK, I found the so-called "cross rule" here and here (the last link is in Russian).

It is especially suitable when the concentration of the resulting mix is given and one has to find the proportion in which to mix the two solutions. Here $w_1$ and $w_2$ are the concentrations of the initial solutions, $\displaystyle m_1$ and $\displaystyle m_2$ are their masses, and $w_3$ is the concentration of the mixture. We assume that $w_1\ge w_2$, so $w_1\ge w_3\ge w_2$. One finds $w_1-w_3$ and $w_3-w_2$. Then $m_1/m_2=(w_3-w_2)/(w_1-w_3)$. In particular, one can take $w_3-w_2$ grams of the first solution and $w_1-w_3$ grams of the second one.

In this thread's problem, suppose we know that we need to get milk's concentration to be 7/12. Then we have

i.e., the proportion of the initial solutions is 1:1.

If we want to find the mixture concentration x, then we have the following diagram.

So, (x - 1/2)/(2/3 - x) = 1/1, i.e., x - 1/2 = 2/3 - x, from where x = 7/12.

The reason the diagram works is the following. The mass of the substance in the first solution, second solution and their mixture is $w_1m_1$, $w_2m_2$ and $w_3(m_1+m_2)$, respectively, so $w_1m_1+w_2m_2=w_3(m_1+m_2)$. From here we get $m_1/m_2=(w_3-w_2)/(w_1-w_3)$.

,
,
,
,
,
,
,

,

,

,

,

,

,

,

www.two equal glass field with alcohol and water in the proportion 2ratio1 and 3 ratio1 are empited into a third glass the proportion of alcohol nand water in third glass will be give answer

Click on a term to search for related topics.