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Welcome to our Math lesson on Finding the New Ratio When One Quantity Changes, this is the fifth lesson of our suite of math lessons covering the topic of Ratios, you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.
Finding the New Ratio When One Quantity Changes
Knowing how to find the new ratio when one of the quantities involved in the original ratio changes is particularly useful in chemistry, as this phenomenon is very common when dealing with chemical solutions, when the amount of one of the quantities of an element used in a mixture changes to provide a new mixture. Let's consider an example to explain this situation.
Example 9
The ratio X : Y for two substances contained in a mixture is 3 : 5 and the amount of mixture is 240 g.
- What will the new ratio between X and Y be if we add in the mixture other 20 g from the substance X?
- What are the original and the new ratio of the substance X to the total?
Solution 9
- First, we have to find how many grams from each substance were originally in the mixture. For this, we use the known procedure of expressing both quantities in terms of their GCF. Thus, since X = 3k and Y = 5k, we have
X + Y = 240
3k + 5k = 240
8k = 240
k = 240/8
k = 30 g
Thus, X = 3 × 30 g = 90g
and Y = 5 × 30 g = 150 g
With the new quantity of 50 g added, the substance X becomes 90 g + 20 g = 110 g, while the quantity Y doesn't change (it remains 150 g). Hence, the ratio between X and Y in the new mixture is Rnew = Xnew/Y = 110 g/150 g = 11/15
- The original ratio R1 of the substance X to the total can be calculated in two ways: (1) through the greatest common factors k, and (2) through the masses in grams. Thus, we have
R1 = 3k/3k + 5k = 3k/8k = 3/8
or R1 = 90 g/240 g = 90 g ÷ 30/240 g ÷ 30 = 3/8
As for the new mixture, we no longer have the same GCF, so we use only the second method to find the new ratio R2 of the substance X to the total. Thus, R2 = 110 g/110 g + 150 g = 110 g/260 g = 11/26
We can use the same procedure when removing a quantity from one of the components involved in a ratio as well. Let's consider another example.
Example 10
The ratio of salt to the total in a new alimentary product was 3:200. However, consumers left negative feedback because it was too salty. Therefore, the producers decided to remove 5 g salt from each kilogram of the actual product without adding other things in it. What is the new ratio of salt to the new alimentary product?
Solution 10
The ratio 3 : 200 means that in one kilogram of the original product there were 5 × 3 g = 15 g of salt, as 1 kg = 1000 g = 5 × 200 g.
When removing 5 g salt from each kilogram of the product, the total of food in each package becomes 1000 g - 5 g = 995 g and the salt in each package becomes 15 g - 5 g = 10 g. Therefore, the new ratio of salt to the food is
R = 10 g/995 g = 10 g ÷ 5/995 g ÷ 5 = 2/249
In this way, the food became less salty than it was before.
More Ratios Lessons and Learning Resources
Ratio and Proportion Learning Material| Tutorial ID | Math Tutorial Title | Tutorial | Video
Tutorial | Revision
Notes | Revision
Questions |
|---|
| 4.1 | Ratios | | | | |
| Lesson ID | Math Lesson Title | Lesson | Video
Lesson |
|---|
| 4.1.1 | Definition of a Ratio | | |
| 4.1.2 | Ratio in a Number Line | | |
| 4.1.3 | Finding the Fraction of One Quantity Out of the Total | | |
| 4.1.4 | Part : Whole vs Part : Part Ratios | | |
| 4.1.5 | Finding the New Ratio When One Quantity Changes | | |
| 4.1.6 | Dividing a Given Ratio | | |
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- Continuing learning ratio and proportion - read our next math tutorial: Rates. Applications of Ratios and Rates in Practice
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