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Question 1196702: Mike wants to make 10 ml of a 69% sugar solution by mixing together a 25% sugar solution and a 80% sugar solution. How much of each solution must he use?
Found 4 solutions by josgarithmetic, greenestamps, ikleyn, math_tutor2020:
Answer by josgarithmetic(39617)   (Show Source):
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
The response from the other tutor shows the setup for solving the problem by the standard method for "mixture" problems. Of course you want to use a method something like that if a formal algebraic solution is required.
But in problems like this where the numbers are "nice" -- as they are in this problem -- a quick and easy solution can be obtained by seeing where the percentage of the mixture lies between the percentages of the two ingredients.
Look at the three percentages (on a number line, if it helps) and observe/calculate that 69 is 44/55 = 4/5 of the way from 25 to 80. That means 4/5 of the mixture needs to be the 80% ingredient.
Since the mixture is to be 10 ml, that means using 8 ml of the 80% sugar solution and 2 ml of the 25% sugar solution.
ANSWER: 8 ml of the 80% solution; 2 ml of the 25% solution
CHECK:
.80(8)+.25(2) = 6.4+.5 = 6.9
.69(10) = 6.9
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Postscript....
If this method of solving the problem looks like magic, observe that the formal solution from the other tutor arrives at that same fraction 44/55 for the fraction of the mixture that must be the 80% sugar solution....
Answer by ikleyn(52786)  (Show Source):
You can put this solution on YOUR website! .
Any chemist, familiar with the subject (and any person familiar with Science) will tell you,
looking into the solution by @josgarithmetic, that this solution is INCORRECT.
It is because at such high concentrations, as given in this problem, the volume of the mixture
increases as you dissolve more sugar in it.
In this respect, the sugar mixtures are DIFFERENT from (behave differently than) salt mixtures.
Therefore, in general, it is not recommended in Science to give such problems with high concentration
to students to solve, because real behavior of mixtures in such cases is NON-LINEAR - - - and significantly NON-LINEAR.
Those who know the subject, also know that it is BAD STYLE and a SIGN of ILLITERACY to give to students
such questions/assignments with high concentrations, where and when real behavior of mixtures is non-linear.
For low concentrations (till 10%), the linear behavior assumption is still admissible
and can be used without a risk to bring shame.
At this forum, there was only one tutor @KMST, who adequately knew this subject and defended the same position,
but she rarely comes to the forum.
What I described in this my post, is the info from any classic textbook/handbook
on mixtures, but I have no a reference at my hands.
Instead, I found a relevant info in the Internet from a knowledgeable amateur, which I place below.
/\/\/\/\/\/\/\/\/\/\/\/
+------------------------------------------------------+
| Does adding sugar to water change its volume? |
+------------------------------------------------------+
https://www.quora.com/Does-adding-sugar-to-water-change-its-volume
- - - - Frank van Wensveen answers - - - -
I have done several tests with various types of sugars
(beet/cane/table sugar, dextrose sugar, dry malt extract) and they are all
identical to each other within a few percent.
The long and the short of it is that
EVERY 1000 GRAMS OF SUGAR, WHEN DISSOLVED IN WATER,
DISPLACES ABOUT 500mL OF WATER.
(I.e. every gram of sugar displaces half its weight in water.)
In other words, if you have 3 litres of water and you dissolve one kg of sugar into it,
you end up with 3,5 litres of total volume of the solution.
- - - - end of the Frank van Wensveen' answer - - - -
The Physical cause, why the behavior of salt solutions and sugar solution is so different
is the fact that their molecules are different:
the salt molecule NaCl is a compact molecule consisting of two atoms, only, while
the sugar molecule  is huge molecule of big volume, consisting of 12+22+11 = 45 atoms.
- - - End of my post. - - -
Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!
I'll use the Alligation method as described on slide 6 from this link here
https://www.slideserve.com/lixue/alligation
The target is 69%
The gap from 25% to the target is 69-25 = 44
The gap from 80% to the target is 80-69 = 11
The ratio 44:11 reduces to 4:1
We'll need 4 times as much of the 80% solution compared to the 25% solution
x = amount of 25% solution
4x = amount of 80% solution
x+4x = 10 mL of solution total
5x = 10
x = 10/5
x = 2
10-x = 10-2 = 8
Answers:
2 mL of the 25% solution
8 mL of the 80% solution
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