Hard 600-700 ratios problem. please advise

S:A:W = 2:50:100

Altered
S:A = 4:50 (ratio doubled)
S:W = 1:100 (halved)
=>S:W = 4:400 (as we have Soap:Alcohol in altered is 4:50, we need 4 for Soap here )
So new ratio S':A':W' = 4:50:400
We can rephrase Soap = 4y, Alcohol = 50y, Water=400y

Now 50y = 100 cubic centimeters
=> 400y = 800 cubic centimeters

IMO E

Hey rtaha2412:

I like this question and a pretty healthy takeaway that it brings up: When you're dealing with three-part problems (ratios, inequalities, etc.) it's often best to break them into two, two-part problems so that you can perform the algebra more easily.

In this problem, the ratio of:

Soap: Alcohol: Water is
2: 50: 100

Which means that you have multiple two-part ratios:

S:A = 2:50
S:W = 2:100
A:W = 50:100

When they say to double the ratio of soap to alcohol, it's probably best to see that ratio as a fraction (2/50) which makes it easier to "double" (2/50 * 2 = 4/50) to 4/50.

Then they say to halve (divide by 2) the ratio of soap to water: 1/2 * 2/100 = 1/100.

So now you have two ratios:

Soap:Alcohol = 4:50, or you could say 2:25

Soap:Water = 1:100

Now, when dealing with multi-part ratios/inequalities/etc., the key becomes putting it all back together around the common variable. Here, that common term is soap, which we express as 2 (in the 2:25 ratio) and as 1 (in the 1:100 ratio). In order to have a common ratio, we need to have soap in the same terms, and 2 is the Least Common Multiple. To get common terms, we then express the ratios with a base of 2 for soap:

Soap:Alcohol = 2:25
Soap: Water = 2:200

Soap:Alcohol:Water = 2:25:200

Because we now then have 100 cubic centimeters of alcohol, we need to multiply everything by 4 to keep it relative (because our base for alcohol is 25), so we would then have:

8800

And therefore there are 800 ccs of water, and the correct answer is E.


As a big-picture takeaway, when you have a 3-part ratio or inequality, it's often best to break it into two 2-part problems, and then you have to synchronize your results around that common term to finish if off.

shovan85 wrote:S:A:W = 2:50:100

Altered
S:A = 4:50 (ratio doubled)
S:W = 1:100 (halved)
=>S:W = 4:400 (as we have Soap:Alcohol in altered is 4:50, we need 4 for Soap here )
So new ratio S':A':W' = 4:50:400
We can rephrase Soap = 4y, Alcohol = 50y, Water=400y

Now 50y = 100 cubic centimeters
=> 400y = 800 cubic centimeters

IMO E

thank u.. for me a little bit confusing can u explain me clearly.....

pzazz12 wrote:

shovan85 wrote:S:A:W = 2:50:100

Altered
S:A = 4:50 (ratio doubled)
S:W = 1:100 (halved)
=>S:W = 4:400 (as we have Soap:Alcohol in altered is 4:50, we need 4 for Soap here )
So new ratio S':A':W' = 4:50:400
We can rephrase Soap = 4y, Alcohol = 50y, Water=400y

Now 50y = 100 cubic centimeters
=> 400y = 800 cubic centimeters

IMO E

thank u.. for me a little bit confusing can u explain me clearly.....

S:A:W = 2:50:100 (I have abbreviated Soap as S, Alcohol as A and Water as W)

So breaking down S:A = 2:50 (or we can say 2/50)

Now, double this ratio, then S':A' = 4:50 (I have abbreviated Soap in the NEW ratio as S', Alcohol as A')

Again breaking down the old ratio S:W = 2:100 (or we can say 2/100)

Now, half this ratio, then S':W' = 1:100 (I have abbreviated Water in the NEW ratio as W')


Thus we have the total scenario for the new ratio (after the modifications done as per the Question)

S':A' = 4:50 and S':W' = 1:100

What is common to both of the ratio: It is SOAP.

For better understandability let me rephrase the two ratios as below:

4 parts of Soap to 50 parts Alcohol
and
1 part of Soap to 100 parts of Water
=> 4 parts of Soap to 4*100 = 400 parts Water

Now, to the balanced new ratio is 4:50:400

So new ratio S':A':W' = 4:50:400

We can say Soap = 4y, Alcohol = 50y, Water=400y

Now 50y = 100 cubic centimeters
=> 400y = 800 cubic centimeters

Hope this is clear.

S:A:W=2:50:100

Altered ratio
----------------------------------

S':A'=4:50 or S'/A'=4/50-->1

S':W'=1:100 or S'/W'=1/100-->2

2/1 gives,


A'/W'=1/8

The new ratio is S':A':W' 4:50:400

Given alcohol is 100 units

Multiply new ratio by 2

S':A':W'=8800

Water =800 units


Hope this clarify!!!

rtaha2412 and pzazz12:

Let me show you a "different way" of dealing with double ratios, through a tool that I "created" to help students deal with the ratios involved. The reason why "the bridge technique" works is, in fact, explained in the posts above but I guess that the "merit" of this tool is exactly to be able to deal with this arguments "internally" (that is, without our explicit actions).

I hope you like it.

Regards,
Fabio.

fskilnik wrote:rtaha2412 and pzazz12:

Let me show you a "different way" of dealing with double ratios, through a tool that I "created" to help students deal with the ratios involved. The reason why "the bridge technique" works is, in fact, explained in the posts above but I guess that the "merit" of this tool is exactly to be able to deal with this arguments "internally" (that is, without our explicit actions).

I hope you like it.

Regards,
Fabio.

awesome Technique...

goyalsau wrote:awesome Technique...

Thank you, goyalsau, but I must say that all that I did (years ago) was to "combine" the ideas shown above in an easy-to-understand diagram, just that. (Brian´s solution is really the same thing that I did, compare!)

There is another (much harder) double ratio problem that I also answered here (look for "alloy" in the post name)... have a look at it because it is also important, as far as technique is concerned. Rahul´s solution (there) is also important, be sure you understand mine and his.

Regards,
Fabio.