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class 9 ganita manjari chapter 10
📖 Textbook Reference: 10.1.2 Mixtures
When two different concentrations are mixed, the resulting concentration will always lie strictly between the concentrations of the two ingredients.
It is mathematically impossible to mix two substances and end up with a concentration that is lower than the lowest ingredient or higher than the highest ingredient. This defines the absolute "limits" of any mixture.
8. Dorjee has collected 1 litre of water from the Dead Sea! Using the information in the table, answer the following questions.
(i) What is the salinity of the mixture if he mixes 1 litre of water from the Dead Sea with 2 litres of purified drinking water?
(ii) Is it possible to mix water from the Dead Sea and purified drinking water to get a mixture with the salinity of groundwater? What quantity of purified drinking water should Dorjee mix with 1 litre of water from the Dead Sea to get a mixture having the salinity of groundwater?
(iii) Is it possible to mix water from the Dead Sea and groundwater to get a mixture with the salinity of purified drinking water? Why/Why not?
(i) What is the salinity of the mixture if he mixes 1 litre of water from the Dead Sea with 2 litres of purified drinking water?
(ii) Is it possible to mix water from the Dead Sea and purified drinking water to get a mixture with the salinity of groundwater? What quantity of purified drinking water should Dorjee mix with 1 litre of water from the Dead Sea to get a mixture having the salinity of groundwater?
(iii) Is it possible to mix water from the Dead Sea and groundwater to get a mixture with the salinity of purified drinking water? Why/Why not?
| Water Source | Salinity |
|---|---|
| Dead Sea | ≈ 34% |
| Other seas and oceans | ≈ 3.5% |
| Ground water | ≈ 0.01% |
| Purified drinking water | ≈ 0.001% |
1
Part (i) & (ii): Use the weighted average formula for mixtures: $\frac{w_1 x_1 + w_2 x_2}{w_1 + w_2} = \text{Target \%}$.
2
Part (iii): A mixture's concentration can never fall below the concentration of its most diluted ingredient. You cannot mix two salty things to get something purer than both!
1
Hint for (i): $w_1 = 1\text{L}, x_1 = 34\%$. $w_2 = 2\text{L}, x_2 = 0.001\%$. Total volume = $3\text{L}$. Calculate the average.
2
Hint for (ii): Set the equation equal to $0.01\%$. Let $v$ be the unknown volume of purified water. $\frac{1(34) + v(0.001)}{1 + v} = 0.01$. Solve for $v$.
3
Hint for (iii): The lowest ingredient is Groundwater ($0.01\%$). The target is Purified Water ($0.001\%$). Is it possible to go lower than the lowest ingredient?
!
The Decimal Trap: When doing algebra with $0.001\%$ and $0.01\%$, be very careful with decimal places. $0.01 \times v$ minus $0.001 \times v$ is $0.009 \times v$. Do not confuse the zeroes!
Solution (i): Finding the Mixture
Total Salinity = $(1 \times 34) + (2 \times 0.001) = 34 + 0.002 = 34.002$
Total Volume = $1 + 2 = 3\text{ Litres}$
$\text{New Salinity} = \frac{34.002}{3} = 11.334\%$
Solution (ii): Diluting to Groundwater
Yes, it is possible because $0.01\%$ is between $34\%$ and $0.001\%$.
Let $v$ be the volume of purified water needed.
$\frac{1(34) + v(0.001)}{1 + v} = 0.01$
$34 + 0.001v = 0.01(1 + v)$
$34 + 0.001v = 0.01 + 0.01v$
$34 - 0.01 = 0.01v - 0.001v$
$33.99 = 0.009v$
$v = \frac{33.99}{0.009} = 3776.67\text{ Litres}$
Solution (iii): The Impossible Mix
No, it is mathematically and physically impossible.
The target salinity ($0.001\%$) is strictly lower than both ingredients available ($34\%$ and $0.01\%$).
A mixture's concentration can never fall below the lowest concentration of its ingredients.
A dairy farmer has two tanks of milk. Tank A has 8% fat content, and Tank B has 2% fat content. Can the farmer mix milk from these two tanks to create a product with 1% fat content? Why or why not?
1
This tests the exact same logic as Part (iii) of the main question. The concept of Absolute Limits of a mixture.
Logical Explanation
No, it is not possible.
The lowest fat content the farmer possesses is $2\%$.
When mixed with $8\%$ milk, the resulting mixture will always fall somewhere strictly between $2\%$ and $8\%$. It can never be driven down to $1\%$ without introducing a third liquid (like water or skim milk) that has a fat content lower than $1\%$.
In part (ii) of the main problem, why does it require such a massive amount of purified water (over 3700 Litres) just to dilute 1 Litre of Dead Sea water to the level of groundwater?
A. Because purified water has 0% salinity.
B. Because the gap between 34% (Dead Sea) and 0.01% (Target) is massive, requiring an enormous "weight" of clean water to pull the average down.
C. Because groundwater is heavier than ocean water.
D. Because 34 divided by 0.01 is exactly 3400.
1
A weighted average acts like a seesaw. The further you want to drag the average away from an extremely potent ingredient (34%), the more volume (weight) you need on the other side.
Logical Explanation
The Dead Sea water ($34\%$) is incredibly salty. The target ($0.01\%$) is incredibly fresh.
To force the final average all the way down from $34\%$ to $0.01\%$, the volume of the fresh water must overwhelmingly dwarf the volume of the salty water.
This is mathematically proven by the ratio $\frac{33.99}{0.009} \approx 3776$. You need nearly 3777 parts fresh water for every 1 part Dead Sea water!
Ans: Option (B)
In part (i) of the main problem, what is the exact amount of pure salt inside the final 3 Litre mixture?
A. 11.334 L
B. 0.34 L
C. 0.34002 L
D. 34.002 L
1
The total amount of pure salt in the final mixture is simply the sum of the pure salt from the two starting ingredients.
Numerator Calculation
Salt from Dead Sea water = $34\%$ of $1\text{L} = 0.34\text{ Litres}$.
Salt from Purified water = $0.001\%$ of $2\text{L} = \frac{0.001}{100} \times 2 = 0.00002\text{ Litres}$.
Total Pure Salt = $0.34 + 0.00002 = 0.34002\text{ Litres}$.
Note: In the formula, we usually just add $(34 \times 1) + (0.001 \times 2)$ to get $34.002$, but since we are looking for the absolute litres of salt, we must divide that by $100$. So it is $0.34002\text{ L}$.
Ans: Option (C)
