# Ratio and Proportions Question

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• Jan 13th 2011, 02:23 AM
dumluck
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...
• Jan 13th 2011, 03:12 AM
CaptainBlack
Quote:

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
• Jan 13th 2011, 03:21 AM
emakarov
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.
• Jan 13th 2011, 03:24 AM
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.
• Jan 13th 2011, 03:34 AM
Archie Meade
Quote:

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.
• Jan 13th 2011, 03:58 AM
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.
• Jan 13th 2011, 04:04 AM
dumluck
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.
• Jan 13th 2011, 04:08 AM
Archie Meade
Quote:

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.
• Jan 13th 2011, 04:09 AM
dumluck
Quote:

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 :).
• Jan 13th 2011, 07:03 AM
emakarov
OK, I found the so-called "cross rule" here and here (the last link is in Russian).

http://lh3.ggpht.com/_SNa3egOo9rk/TS...cross-rule.png

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 \$\displaystyle w_1\$ and \$\displaystyle w_2\$ are the concentrations of the initial solutions, \$\displaystyle \displaystyle m_1\$ and \$\displaystyle \displaystyle m_2\$ are their masses, and \$\displaystyle w_3\$ is the concentration of the mixture. We assume that \$\displaystyle w_1\ge w_2\$, so \$\displaystyle w_1\ge w_3\ge w_2\$. One finds \$\displaystyle w_1-w_3\$ and \$\displaystyle w_3-w_2\$. Then \$\displaystyle m_1/m_2=(w_3-w_2)/(w_1-w_3)\$. In particular, one can take \$\displaystyle w_3-w_2\$ grams of the first solution and \$\displaystyle 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

http://lh4.ggpht.com/_SNa3egOo9rk/TS...ross-rule1.png

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.

http://lh5.ggpht.com/_SNa3egOo9rk/TS...ross-rule2.png

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 \$\displaystyle w_1m_1\$, \$\displaystyle w_2m_2\$ and \$\displaystyle w_3(m_1+m_2)\$, respectively, so \$\displaystyle w_1m_1+w_2m_2=w_3(m_1+m_2)\$. From here we get \$\displaystyle m_1/m_2=(w_3-w_2)/(w_1-w_3)\$.