Two labelled solutions create a new kind of problem
A container marked 20 g/dm³ holds a less concentrated sucrose solution than one marked 80 g/dm³. A student needs 300 cm³ of a solution at 50 g/dm³ for an imaginary classroom display. Which amounts of the two stocks should be combined? The tempting answer is to take equal parts because 50 lies midway between 20 and 80. In this particular case that answer works. But the important skill is to know why it works, how to prove it and what changes when the target is not midway.
This guide continues Volume 1025 on chemistry concentration and final volume. The chemistry guide established what the labels mean; this one builds a mathematical model of two-stock mixing. It also draws on Volume 1022 on interpreting numerical evidence and Volume 1024 on responsible claims. The learning goal is not to memorise a particular recipe. It is to translate quantities into equations, choose a feasible solution and explain where the model’s assumptions begin and end.
Use the correct 2027 G3 Mathematics examination
The official SEAB 2027 G3 syllabus directory, updated on 26 October 2026, lists Mathematics as K310, replacing the older GCE reference code 4052 for the new SEC examination. The K310 Mathematics syllabus specifies two compulsory papers, each lasting two hours fifteen minutes, with 90 marks and a 50% weight. Paper 1 includes approximately twenty-six short-answer questions; Paper 2 includes nine or ten questions, ending with an extended real-world application.
The syllabus expects students to show essential working, communicate mathematical reasoning and solve problems where context matters. Non-exact answers are generally given to three significant figures, and angles in degrees to one decimal place, unless another accuracy is specified. Mixture equations here are original practice problems showing how algebra, ratio, constraints and interpretation can work together. They are not a prediction that the actual SEC examination will use sucrose, stock solutions or any particular laboratory scenario.
Start with an explicit model, not an unspoken laboratory assumption
In the mathematical model used in this chapter, each stock solution has a uniform stated concentration. We assume the solute remains dissolved and none is lost in transfer. We also assume the final solution volume equals the sum of the two stock volumes. This additive-volume assumption gives an accessible, checkable school problem; it is not a universal physical law for every mixture of real liquids.
A real preparation would need to check the final volume and composition using appropriate procedure and equipment. If the problem instead supplies measured final volume, use that value rather than silently substituting the sum. Good mathematics begins by identifying the model the question authorises. It must not present a convenient simplification as a fact about all chemical solutions.
Know what each letter means
Let c₁ and c₂ be the concentrations of the two stock solutions in g/dm³. Let v₁ and v₂ be the volumes taken from those stocks in dm³. If the final volume is V and the target concentration is C, the additive-volume model gives v₁ + v₂ = V. Conservation of the dissolved solute mass gives c₁v₁ + c₂v₂ = CV.
These equations answer different questions. The first accounts for the total solution volume. The second accounts for the solute carried by each portion. Writing only the first leaves infinitely many possible splits. Writing only the second may identify combinations with the correct solute mass but the wrong final volume. The two together express what makes a candidate mixture acceptable.
Derive the weighted-concentration formula
The first stock contributes c₁v₁ grams of solute because grams per dm³ multiplied by dm³ gives grams. The second contributes c₂v₂ grams. The total solute mass is their sum. Under the stated model, the completed volume is v₁ + v₂. Therefore the concentration of the resulting mixture is C = (c₁v₁ + c₂v₂)/(v₁ + v₂).
This formula is a weighted average of the stock concentrations. The weights are their volumes relative to the total: v₁/(v₁+v₂) and v₂/(v₁+v₂). Those weights add to one. A mixture that uses more of the high-concentration stock gives that concentration more influence. If the volumes are equal, the weights are equal; if the volumes differ, the ordinary unweighted mean is generally wrong.
Solve the first complete example
Assume the stocks are 20 and 80 g/dm³, and the task is 300 cm³ at 50 g/dm³. Convert 300 cm³ to 0.300 dm³. Let x dm³ be taken from the lower-concentration stock. The higher-concentration portion is then 0.300 − x. The solute equation is 20x + 80(0.300 − x) = 50 × 0.300.
Expand: 20x + 24 − 80x = 15, so −60x = −9 and x = 0.150 dm³. The other portion is also 0.150 dm³. Thus the answer is 150 cm³ from each stock under the model. The solution is not just ‘half and half’; it is a justified split that satisfies both constraints.
Check the answer in two independent ways
First check the total volume: 0.150 + 0.150 = 0.300 dm³. Next check the solute masses: the weaker stock contributes 20 × 0.150 = 3.0 g, and the stronger contributes 80 × 0.150 = 12.0 g. Together they contribute 15.0 g. Dividing by 0.300 dm³ gives the required 50 g/dm³.
The numerical checks are not optional ornament in a real-world problem. They establish that a solved equation fits its original conditions. A student can solve an equation correctly after accidentally defining the variable backwards. Rechecking volume and solute mass independently catches errors that a single substitution may miss.
See the equal-volume shortcut and its limit
When two stocks contribute equal volumes, their concentrations receive equal weights. Therefore the resulting target is the arithmetic mean of the two concentrations under this model. For 20 and 80, the mean is 50 g/dm³. This is why 150 cm³ of each gives the target in our first example.
But that shortcut only applies to equal volumes and to the stated mixing assumptions. Suppose the target is 65 g/dm³. Equal portions would still yield 50, not 65. The mathematics must give more volume to the 80 g/dm³ stock. A shortcut is useful only when its conditions are checked before using it.
Solve an unequal split
Use the same 20 and 80 g/dm³ stocks to make 400 cm³ at 65 g/dm³. Let x dm³ be the volume of the stronger stock and 0.400 − x the weaker. Then 80x + 20(0.400 − x) = 65 × 0.400. Simplifying gives 80x + 8 − 20x = 26, so 60x = 18 and x = 0.300 dm³.
The answer is 300 cm³ of the 80 g/dm³ stock and 100 cm³ of the 20 g/dm³ stock. This makes intuitive sense because the target 65 is much closer to 80 than to 20. A reasonableness check should ask which stock must dominate even before performing the arithmetic.
Use the direct fraction method
Let f be the fraction of the final volume taken from the higher-concentration stock. The other fraction is 1 − f. The target is then C = c_low(1 − f) + c_high f. Rearranging gives f = (C − c_low)/(c_high − c_low). This expression directly tells us what proportion of the final volume comes from the stronger stock.
For the 20, 80 and 65 problem, f = (65 − 20)/(80 − 20) = 45/60 = 0.75. Multiply 0.75 by 400 cm³ to obtain 300 cm³ of the stronger stock. The fraction method is quicker when the algebra is familiar. The simultaneous-equations method remains valuable because it explains exactly what is being conserved.
The proportion can be read as distance between concentrations
The target 65 lies 45 units above the lower concentration 20, across a complete stock-concentration gap of 60 units. That is why 45/60 of the mixture’s volume comes from the upper stock. The nearer the target lies to the stronger stock’s concentration, the greater the fraction that must come from that stock.
This intuitive reading is a safeguard against swapped volumes. If a student obtains only 25% from the stronger stock for a target much nearer the stronger concentration, they should stop and inspect the equation. The model’s direction gives a fast qualitative error check even before any arithmetic is redone.
A target outside the stock range is impossible
Consider a request to make 200 cm³ at 90 g/dm³ using only 20 and 80 g/dm³ stocks. Under positive-volume mixing with no additional solute, every weighted average must lie between 20 and 80. Therefore 90 cannot be produced from these two stocks under the stated model. An algebraic solution would require a negative volume for one stock, which is not physically feasible.
This is an example of a bound. Mathematics can prove impossibility without searching through every proposed volume combination. In examination problems, feasibility is part of the answer. Reporting an algebraic value with a negative volume as though it were a workable recipe would mean the student solved the symbols but misunderstood the context.
The other boundary case matters too
If the requested concentration is exactly 20 g/dm³, the mixture must use only the 20 g/dm³ stock unless both stocks have that same concentration. Any positive portion of the 80 g/dm³ stock would raise the weighted average above 20. Likewise, exactly 80 g/dm³ requires only the stronger stock.
The boundary clarifies why a target between the two concentrations can use positive volumes of both stocks, while a target at an endpoint may use zero of one. If a task additionally insists on positive quantities of each stock, an exact endpoint target would be infeasible. The answer depends on every constraint, not solely on the target number.
Build the model with two simultaneous equations
Some questions give the final volume but hide the target concentration within a different condition. Let x and y represent the volumes of the weak and strong stocks. Write x + y = V for total volume. Then write c_low x + c_high y = required solute mass. Two unknown volumes need two independent conditions. This is a familiar simultaneous-equations structure dressed in a practical setting.
Suppose 500 cm³ is required and the final solution must contain 35 g of solute. The stock concentrations are 40 and 100 g/dm³. Convert the total volume to 0.500 dm³. The equations are x + y = 0.500 and 40x + 100y = 35. Solving gives y = 0.250 and x = 0.250 dm³. The implied final concentration is 35/0.500 = 70 g/dm³.
Why two equations are better than a guessed split
A guess may work quickly when the target is the midpoint, but it is unreliable with arbitrary targets or hidden conditions. The simultaneous equations encode conservation exactly. A student who writes them properly can solve many variants without inventing a special trick for each surface story.
Good equation writing starts with a description of the unknowns and their units. Do not write x = sugar and y = water if both unknowns are stock-solution volumes. Incorrect definitions create incorrect equations. A line such as ‘Let x be the volume in dm³ of the 40 g/dm³ stock’ prevents later ambiguity and helps the examiner follow the reasoning.
Beware of the false average
A classmate suggests that 50 cm³ at 20 g/dm³ mixed with 150 cm³ at 80 g/dm³ must give 50 g/dm³ because the two stock concentrations average to 50. Calculate the correct solute amount. The weaker portion contributes 20 × 0.050 = 1 g; the stronger contributes 80 × 0.150 = 12 g. Total solute is 13 g in 0.200 dm³.
The resulting concentration is 13/0.200 = 65 g/dm³ under the model. The high-concentration stock occupies three quarters of the volume, so it has three quarters of the influence in the weighted mean. An ordinary mean is correct only when the two contributing volumes are equal.
A visual ratio method
Suppose the target is 35 g/dm³ using 20 and 80 g/dm³ stocks. The target lies 15 units above 20 and 45 units below 80. Under the mixing model, the high-concentration fraction is 15/(15+45) = 1/4, so one quarter of the final volume comes from the stronger stock and three quarters from the weaker.
This can be expressed as a weak-to-strong volume ratio of 3:1. The ratio is the reverse of the distances: 80 − 35 = 45 and 35 − 20 = 15. Students sometimes reverse it and use 1:3, producing a much stronger mixture. The weighted-average equation is the final check, regardless of whether you used a ratio diagram.
Find the total volume from a fixed stock allocation
A problem may say that 120 cm³ of the stronger 80 g/dm³ stock is available and the target is 40 g/dm³ using a 20 g/dm³ weaker stock. Let x cm³ be the required weaker stock. Use common volume units consistently in the weighted-average equation: (20x + 80×120)/(x+120)=40, provided both numerator terms use the same proportional unit basis.
Multiply through: 20x + 9600 = 40x + 4800, so 20x = 4800 and x = 240 cm³. The total is 360 cm³. Check with actual solute masses: the 240 cm³ weaker portion has 4.8 g, the 120 cm³ stronger portion has 9.6 g, and 14.4 g in 0.360 dm³ is 40 g/dm³. This example shows why multiplying concentration by cm³ without a conversion can still cancel correctly in a ratio, but it must not be reported as grams without the proper factor.
Keep algebraic units consistent
Writing 20x + 80y is acceptable in a ratio equation when x and y are both in the same volume unit, because the common conversion factor cancels. But if a student claims 20×120 equals 2400 grams when 120 represents cm³, the unit is wrong. For a solute-mass equation that reports grams, convert volumes to dm³ first.
A safe method in examinations is to define all volumes in dm³ before setting up mass balance. Alternatively, state explicitly that x and y are in cm³, with every concentration conversion applied consistently. Either route can work. Mixing litres, cm³ and dm³ in the same equation without conversion is never a valid shortcut.
Use a table to prevent quantity confusion
Create four headings: stock, concentration, volume, solute mass. For a 100 cm³ portion of a 20 g/dm³ stock, write 20 g/dm³, 0.100 dm³, 2.0 g. For a 300 cm³ portion of an 80 g/dm³ stock, write 80 g/dm³, 0.300 dm³, 24.0 g. The total row contains 0.400 dm³ and 26.0 g.
The final concentration is 26.0/0.400 = 65 g/dm³. The table is more than tidy presentation. It keeps the quantities in distinct columns, and each solute-mass entry can be independently checked. The layout works well in extended real-world applications where a paragraph contains many pieces of data.
Mixing does not create solute
The total dissolved solute mass in our ideal model is simply the sum of the masses supplied by the two stock portions. If a worked answer produces more sugar than the sum of the two inputs, the calculation contradicts its own assumptions. Similarly, an answer that silently discards some solute would require a stated loss or removal process.
Conservation is an efficient diagnostic. Rather than checking only whether the final concentration seems reasonable, compute the solute mass implied by the answer and compare it with the input masses. The stock labels tell us a mass for each chosen volume. Their sum is the balance that the calculation must respect.
What happens when one stock is plain water?
If the weaker stock has zero dissolved sucrose and the stronger stock has concentration 100 g/dm³, the same weighted-mean model becomes a familiar dilution problem. For example, a target of 25 g/dm³ uses one quarter of the final volume from the strong stock under the additive-volume and solute-retention assumptions.
That case is a boundary of the general two-stock model. The series already covers serial dilution in Volume 0532 and reverse dilution in an earlier dedicated Mathematics volume. The distinct focus here is blending two nonzero stocks and testing algebraic feasibility under a final-volume requirement.
An unequal stock pair changes the required ratio
Imagine concentrations of 30 and 90 g/dm³ with a target of 70 g/dm³. The strong-stock fraction is (70 − 30)/(90 − 30)=40/60=2/3. For a final volume of 450 cm³, the stronger portion is 300 cm³ and the weaker portion is 150 cm³.
Verify by mass: 90 × 0.300 = 27 g and 30 × 0.150 = 4.5 g. Total is 31.5 g in 0.450 dm³, which is 70 g/dm³. The target is closer to the strong concentration than to the weak one, so the stronger volume should exceed half. The numerical result matches the intuitive direction.
A target that is not an integer does not make a problem impossible
Suppose 25 and 75 g/dm³ stocks must form 320 cm³ at 58 g/dm³. The fraction from the higher stock is (58 − 25)/(75 − 25)=33/50=0.66. Thus 211.2 cm³ comes from the stronger stock and 108.8 cm³ from the weaker stock under the model. Both values are positive and add to 320 cm³.
Real measuring equipment may impose practical resolution limits, but the mathematical problem has a feasible solution. Do not round each volume to the nearest ten before verifying the total and concentration. If a question gives a specific accuracy requirement, report accordingly; otherwise keep adequate precision in intermediate work and a sensible final form.
Check a proposed plan rather than solving from scratch
A student offers a plan: combine 80 cm³ of 20 g/dm³ solution with 120 cm³ of 80 g/dm³ solution to obtain 60 g/dm³. Check the volumes: the total is 200 cm³. The solute amounts are 1.6 g and 9.6 g respectively, giving 11.2 g in 0.200 dm³ and a resulting concentration of 56 g/dm³.
The proposal misses its target by 4 g/dm³. You can reject it with a calculation rather than an opinion. If the final volume must stay 200 cm³, solve for the strong portion: (60−20)/(80−20)=2/3 of 200, or approximately 133.3 cm³; the weak portion is approximately 66.7 cm³. Keep the model assumption and the measuring precision explicit.
Understand the difference between an absolute and a relative error
If the target is 60 g/dm³ and the achieved mathematical result is 56, the absolute concentration difference is 4 g/dm³. The relative shortfall compared with the target is 4/60 × 100%, or about 6.67%. These describe the same deviation in different units and reference frames.
Do not say the result is ‘four percent lower’ merely because the numbers differ by four. That sentence would treat concentration units as percentages without calculating a ratio. A clear answer names the baseline and whether the quantity is an absolute difference or a percentage of the target.
Determine which stock has more influence
For any weighted mixture, the stock contributing the larger fraction of final volume has greater influence on the resulting concentration. That does not always mean it contributes the greater solute mass. A small volume of an extremely concentrated stock may contribute more solute mass than a larger volume of a weak stock. These are two separate questions.
For example, 100 cm³ of 100 g/dm³ stock contributes 10 g, while 200 cm³ of 20 g/dm³ stock contributes 4 g. The weaker stock occupies more volume, but the stronger stock contributes more solute mass. The combined volume is 300 cm³ and combined solute mass 14 g, giving about 46.7 g/dm³. Precise quantity names prevent mistaken conclusions.
Mathematical bounds before detailed calculation
If two concentrations are both positive and different, any nonnegative weighted average of them lies between the lower and higher concentration. This simple principle can be proved: C = c_low + f(c_high − c_low), where f is between zero and one. Therefore f times the positive gap ranges from zero to the entire gap.
The bound gives an immediate answer to impossible-target questions. It can also reveal faulty work. If an otherwise routine mixing calculation produces a final concentration above both stock concentrations, either the volumes, solute amount, assumptions or arithmetic are wrong. The point is not that algebra is unnecessary; it is that reasoning should supervise the algebra.
Include a maximum stock-volume constraint
A school exercise may allow at most 100 cm³ of the strong 80 g/dm³ stock, while the total required final volume is 400 cm³ and the weak stock is 20 g/dm³. The strongest achievable mixture uses the maximum strong-stock volume, with 300 cm³ weak stock. Its modelled concentration is (80 × 0.100 + 20 × 0.300)/0.400 = (8 + 6)/0.400 = 35 g/dm³.
A request for 40 g/dm³ is therefore infeasible under the stock limit, even though forty lies between twenty and eighty. The stock range condition is necessary but not sufficient once quantity constraints appear. This is a stronger form of mathematical modelling: feasibility depends on both the available concentrations and the amounts available.
Solve a minimum-stock requirement
Suppose a target of at least 55 g/dm³ must be achieved using 20 and 80 g/dm³ stocks in a fixed total volume of 500 cm³. Let f represent the proportion from the strong stock. The model gives 20 + 60f ≥ 55, so f ≥ 35/60 = 7/12, approximately 0.5833.
Thus at least about 291.7 cm³ of the final 500 cm³ must come from the strong stock in the idealised model. The inequality states a minimum, not an exact solution. Volumes above that level also satisfy the concentration threshold if all other constraints are unchanged. Inequalities express a broader set of feasible plans than equations do.
Recognise that rounded quantities affect the target
If the ideal answer gives 291.666… cm³ and a practical plan rounds it to 292 cm³, the concentration changes slightly. For a 500 cm³ final volume, 292 cm³ strong and 208 cm³ weak yield (80 × 0.292 + 20 × 0.208)/0.500 = (23.36 + 4.16)/0.500 = 55.04 g/dm³. This still meets an at-least-55 requirement.
Rounding down to 291 cm³ would produce 54.92 g/dm³ and fail that strict threshold. It is useful to recompute after rounding when a problem includes minimum or maximum requirements. A final answer that looks numerically close can be unacceptable if it crosses a hard constraint.
An overdetermined request may have no solution
Suppose a student must prepare exactly 300 cm³ at exactly 50 g/dm³ using 20 and 80 g/dm³ stocks, but is also told to use exactly 200 cm³ of the stronger stock. The required solution from the first two conditions uses 150 cm³ of each. The extra condition fixes the strong portion at a different value, so all three cannot be satisfied together.
To prove the conflict, substitute the 200 cm³ strong portion and 100 cm³ weak portion into the concentration formula. The result is (80 × 0.200 + 20 × 0.100)/0.300 = 60 g/dm³, not 50. Do not invent a fourth, unlisted ingredient to fix the problem. State that the requirements are inconsistent under the given two-stock model.
Set up a cost-constrained variation
A mathematical word problem may assign illustrative prices per litre to the two stocks. Suppose the weak stock costs $2 per litre and the strong stock $5 per litre, with a requirement for 300 cm³ at 50 g/dm³ from 20 and 80 g/dm³ stocks. The concentration and volume constraints fix the split at 0.150 dm³ of each. The cost is 2 × 0.150 + 5 × 0.150 = $1.05.
If a question asks which of several plans is cheapest, calculate only among feasible plans and preserve units. Under the fixed two-stock, exact-volume and exact-target model with distinct concentrations, there is only one split, so there is no freedom to optimise cost. Recognising when optimisation is unnecessary is itself a useful modelling skill.
A mixture model is not a chemical reaction equation
Combining two solutions of the same dissolved substance at different concentrations is different from balancing a chemical equation for a reaction. The mass balance here tracks dissolved sucrose already present in each stock. It does not introduce reactants disappearing into products or require stoichiometric coefficients between different substances.
If a task includes reacting solutes, precipitation, gas loss or another chemical process, the simple mixture relation may no longer describe the final dissolved amount. Students should read the chemistry before applying a familiar formula. This chapter’s model is deliberately narrow: two portions of a common solute, no loss or reaction, and additive volumes for calculation.
A significant-figures check
Imagine an equation yields a strong-stock volume of 0.1372549 dm³, with all data stated to three significant figures. A final answer such as 0.137 dm³ may be appropriate, subject to the problem’s exact accuracy instruction. The calculator’s additional digits can remain in intermediate working. Be careful not to round a volume too early when another calculation depends on it.
An examination question may ask for a whole number of containers, exact fractions or a tolerance interval. Follow that instruction rather than applying three significant figures mechanically to every result. Accuracy is part of meaning: a count of bottles is not the same kind of quantity as a measured concentration.
Read the word ‘at least’ literally
A requirement to make at least 60 g/dm³ does not mean exactly 60. With 20 and 80 g/dm³ stocks, the high-stock fraction must satisfy 20 + 60f ≥ 60, giving f ≥ 2/3. Any fraction from two thirds up to one is mathematically feasible, subject to volume and stock constraints.
Students sometimes solve the equality and stop. That gives the boundary but not the permitted range. A well-formed answer states both the threshold and the direction of the inequality. Words like at least, at most, greater than and no more than can change the task more than a complicated formula does.
Read ‘no more than’ just as carefully
If the final concentration must be no more than 35 g/dm³, using 20 and 80 g/dm³ stocks, the high-stock fraction obeys 20 + 60f ≤ 35, so f ≤ 1/4. For a 400 cm³ final volume, the stronger stock may contribute at most 100 cm³ in the model.
The solution includes every fraction from zero to one quarter, not just the boundary one quarter. If the problem also requires at least 50 cm³ of the stronger stock, the feasible range becomes 50 to 100 cm³. Mathematical modelling often means combining restrictions correctly rather than searching for one spectacular calculation.
Practice one: reconstruct the equation
Question: Two stocks contain 10 and 70 g/dm³ of the same dissolved solute. The target is 240 cm³ at 40 g/dm³. Let x dm³ represent the strong stock. The remaining weak stock has volume 0.240 − x. Write 70x + 10(0.240 − x) = 40 × 0.240. Expanding gives 70x + 2.4 − 10x = 9.6, so 60x = 7.2 and x = 0.120 dm³.
Therefore the mixture uses 120 cm³ of each stock. Explain why an equal split works here: 40 is exactly midway between 10 and 70. The goal of the exercise is not the easy answer but the ability to construct both conservation statements independently.
Practice two: a target near the weak stock
Question: Stocks are 15 and 95 g/dm³. Make 500 cm³ at 31 g/dm³. The fraction from the strong stock is (31 − 15)/(95 − 15) = 16/80 = 0.20. The required split is 100 cm³ strong and 400 cm³ weak. Check: 95 × 0.100 + 15 × 0.400 = 9.5 + 6.0 = 15.5 g, and 15.5/0.500 = 31 g/dm³.
The target is close to 15, so the weaker stock must dominate. If your answer gives 400 cm³ strong, the concentration would instead be much nearer 95. This intuitive check can catch a reversed ratio before the final line.
Practice three: an impossible target
Question: Can 20 and 80 g/dm³ stocks yield 95 g/dm³ by mixing positive volumes of those two stocks only? No. A positive weighted average cannot exceed its highest input concentration under this model. Algebra confirms the problem: f = (95 − 20)/(80 − 20) = 75/60 = 1.25, which would require 125% of the final volume to come from the strong stock.
Since the total of the volume fractions must be exactly 100%, the implied weak fraction is negative. A negative solution volume is a mathematical warning about an invalid physical plan. The correct answer is infeasibility under the stated constraints, not an invented negative-volume recipe.
Practice four: a limited strong stock
Question: You have at most 75 cm³ of an 80 g/dm³ stock and unlimited 20 g/dm³ stock. The final volume must be 300 cm³. What is the maximum feasible concentration? Use the full 75 cm³ of the strong stock and 225 cm³ of the weak. The combined solute mass is 80 × 0.075 + 20 × 0.225 = 6 + 4.5 = 10.5 g.
The maximum modelled concentration is 10.5/0.300 = 35 g/dm³. It is not 80 merely because that is the higher stock. The final volume constraint forces most of the mixture to come from the weaker solution. The strongest answer states the limiting stock and verifies the resulting concentration.
Practice five: check an advertisement
Question: A fictional school poster promises that equal amounts of 30 and 90 g/dm³ solution will yield 75 g/dm³. Is it correct? No. Equal volumes give the arithmetic mean, (30 + 90)/2 = 60 g/dm³, not 75. The claim fails even without knowing the total volume.
To attain 75 under the model, the strong-stock fraction must be (75 − 30)/(90 − 30) = 45/60 = 0.75. Three quarters must come from the strong stock. This demonstrates how an apparently minor phrase—equal amounts—can decide the entire calculation.
Practice six: missing final volume information
Question: A record states that two liquids were mixed, with each stock’s concentration and measured initial volume, but warns that the resulting volume was not equal to the sum of their measured volumes. Can we calculate the exact final concentration with the basic weighted-average denominator? Not without knowing the actual final solution volume or a justified alternative relationship.
The initial stock volumes can still provide their solute masses if their concentrations are known and no solute is lost. But the final concentration is total dissolved solute mass divided by actual final volume. The additive-volume formula cannot simply be applied after the assumption has been explicitly withdrawn.
Practice seven: diagnose a common sign error
Question: In a mixture where the target lies between 20 and 80 g/dm³, a student’s solution gives x = −0.10 dm³ for one stock. What should they do? First verify the variable definition and equation. A feasible two-stock mixture with a target strictly inside the range requires positive volumes of both stocks; a negative value indicates mis-specified algebra, mismatched units or an inconsistent extra constraint.
Do not fix the result by dropping the minus sign. A sign is part of the mathematical evidence. Return to the volume and solute balance equations, reconstruct the substitution and check whether the original constraints admit a nonnegative solution.
Practice eight: express the answer as a ratio
Question: Stock concentrations are 25 and 85 g/dm³; target is 55 g/dm³. What is the weak-to-strong volume ratio? The target is 30 above the weak and 30 below the strong, so the split is equal: 1:1. The formula gives a strong fraction of (55−25)/(85−25)=30/60=1/2.
If the target changes to 65, the strong fraction becomes 40/60=2/3 and the weak-to-strong volume ratio is 1:2. Ratios describe the relative allocation; absolute volumes require an additional total-volume condition.
Practice nine: a parameter instead of a number
Let a weaker stock have concentration a and a stronger one concentration b, where b exceeds a. Let the target t lie between them. The strong-stock fraction is (t−a)/(b−a), and the weak-stock fraction is (b−t)/(b−a). Both fractions are nonnegative and sum to one if a ≤ t ≤ b.
This general result helps students see the structure behind all the examples. Numbers change, but conservation and the bounds remain. Substituting actual concentrations into the symbolic result is faster once the derivation has been understood. Memorising the formula without its conditions invites misuse.
Practice ten: a reverse question
Question: A mixture contains 100 cm³ of 20 g/dm³ stock and 300 cm³ of unknown concentration x. The final 400 cm³ solution is 65 g/dm³ under the model. The total required solute mass is 65×0.400=26 g. The weak stock contributes 20×0.100=2 g, so the unknown stock contributes 24 g in 0.300 dm³.
Therefore x = 24/0.300=80 g/dm³. Check that 100 and 300 cm³ combine to 400 cm³, and the resulting weighted average is 65. This is an inverse problem: the formula is the same, but a different quantity is unknown. Strong students recognise that structure instead of hunting for a new special rule.
Practice eleven: compare two proposed plans
Plan A uses 150 cm³ at 20 g/dm³ and 150 cm³ at 80 g/dm³. Its total solute mass is 3+12=15 g in 0.300 dm³, giving 50 g/dm³. Plan B uses 200 cm³ at 20 and 100 cm³ at 80. Its total solute mass is 4+8=12 g in 0.300 dm³, giving 40 g/dm³.
If a school wants 45 g/dm³, neither plan meets the exact target. The required high-stock fraction is (45−20)/60=5/12, corresponding to 125 cm³ high and 175 cm³ low for a total of 300 cm³. A good answer evaluates available options and, if asked, constructs a corrected feasible plan.
Practice twelve: solve a constrained range
Suppose a 500 cm³ mixture from 20 and 80 g/dm³ stocks must have concentration between 44 and 56 g/dm³ inclusive. Let f be the high-stock fraction. The condition is 44 ≤ 20+60f ≤ 56. Subtract twenty throughout to obtain 24 ≤ 60f ≤ 36, and divide by sixty to get 0.4 ≤ f ≤ 0.6.
The stronger stock can therefore contribute between 200 and 300 cm³ inclusive. Every value in that interval satisfies the stated concentration range under the model. A range problem requires a range answer, not the midpoint 250 cm³ alone.
A diagnostic page: four errors to catch
The first error is dividing by the volume of one stock rather than the final mixture. The second is averaging stock concentrations without volume weights. The third is using cm³ in a solute-mass equation as if those volumes were dm³. The fourth is accepting negative volumes or a target outside the stock range. Each error has a different cause, and the repair should name it precisely.
Corrective practice should be targeted. If the student gets the equations right but forgets to check feasibility, add bound questions. If they make unit errors, practise converting before substitution. If they swap stocks, write the meaning of each variable above the algebra. General advice to ‘be more careful’ is less effective than fixing a specific habit.
A fast verification checklist
Before accepting the final answer, ask whether the two input volumes are nonnegative, whether they add to the required final volume and whether the implied solute mass matches the target concentration. Check the units at every conversion. Finally ask whether the target concentration lies in the feasible range once all stock limits are considered.
A student who does these checks may spend a few extra seconds but avoids losing marks on an implausible final line. The checks are especially valuable when a longer question has several stages: an early unit error can infect everything that follows unless a reasonableness test interrupts it.
Use an equation to explain an effect instead of merely describing it
Imagine the stronger-stock fraction rises from 0.25 to 0.50 with the concentrations fixed at 20 and 80 g/dm³. The target changes from 20+60×0.25=35 to 20+60×0.50=50 g/dm³. The 0.25 increase in fraction raises the concentration by 15 g/dm³. That is the gradient of the linear concentration-versus-fraction relationship.
The formula C=20+60f displays the relationship directly. Its intercept is the weak-stock concentration and its gradient is the gap between the stock concentrations. This connects mixtures to linear functions, graphs and interpretation. You can reason about changes without repeatedly recomputing the full solute balance.
Graph the concentration against the strong-stock fraction
Place f, from zero to one, on the horizontal axis and final concentration C on the vertical axis. For the 20 and 80 g/dm³ stocks, the graph is a straight line from (0,20) to (1,80) under the stated model. The point f=0.5 gives 50; f=0.75 gives 65. The graph makes both the feasible range and the effect of the mixing ratio visible.
Do not label the horizontal axis ‘volume’ if it actually represents a fraction of total volume. The units and definitions matter: f is dimensionless, while C is in g/dm³. A tidy graph with the wrong quantity label can communicate a false mathematical relationship.
What the graph cannot tell you
The straight-line model connects concentration to the proportion of stock solutions under assumed additive volumes and unchanged solute amount. It does not predict how a living tissue sample will respond to the resulting concentration. The biological percentage mass change in Volume 1022 is a different dependent variable and may follow a different relationship.
Do not take the concentration mixing line and call it a graph of osmosis. Two contexts can use similar units or an identical horizontal variable without sharing the same law. Mathematics helps us separate the equations from the phenomena they model.
A higher-level word problem with two decisions
A coordinator wants 600 cm³ of a solution at least 50 g/dm³ but no more than 65 g/dm³ using 20 and 80 g/dm³ stocks. Let f be the strong-stock fraction. The inequality is 50 ≤ 20+60f ≤ 65, yielding 0.5 ≤ f ≤ 0.75. Thus the strong stock must occupy between 300 and 450 cm³ of the total 600 cm³.
If the coordinator also limits the stronger stock to 400 cm³, the feasible interval narrows to 300–400 cm³. The additional constraint changes the answer but does not invalidate the initial concentration calculation. This is what multi-stage modelling often requires: solve the mathematical condition, then intersect it with the real-world restriction.
How to write a conclusion that respects constraints
A useful concluding sentence for that example is: ‘Under the stated additive-volume mixing model, between 300 and 400 cm³ of the stronger stock meets the required concentration range and the stock limit; the remaining volume comes from the weaker stock.’ It states the model, the feasible decision and how the total is completed.
A weaker response says ‘use 350’ without showing whether other solutions are allowed. A single candidate may be valid, but it does not answer a request for all feasible volumes. Read the demanded output—one example, an exact split, a minimum or a range—before choosing the final sentence.
Avoid unearned claims about practical safety
The numerical exercises concern hypothetical solutions and mathematical constraints. They do not provide instructions for unsupervised laboratory work. The school or teacher should determine any actual preparation procedure, apparatus and safety requirements. No calculation alone confirms that a physical preparation has reached its target concentration.
The scientific boundary is simple: the algebra describes a model under stated assumptions. If volumes are not additive, transfer loses solute, a substance reacts or some solid remains undissolved, the assumptions need re-examination. A confident calculation should include a clear statement of its scope.
Link PSLE ratio to SEC algebra
At Primary level, pupils learn fractions, ratios, percentage, units and logical checking. Those foundations are present in the mixture problem: the strong stock occupies a fraction of the final volume, the concentration is a weighted average, and the ratio must respect the total. At Secondary level, students can use simultaneous equations, inequalities, algebraic manipulation and graphs to express more complex constraints.
The Sengkang PSLE Mathematics problem-solving route supports the transition from diagram-based thinking to algebra. For subject-specific secondary practice use the Secondary Mathematics owner. These learning routes are related without claiming PSLE and SEC have identical content or assessment.
A short oral explanation improves mathematical clarity
Ask a student to explain the 20-and-80-to-65 example aloud without reading the equations. A complete explanation might be: ‘The target is closer to the stronger stock, so more than half must come from it. The weighted-average equation gives three quarters of the final volume from the strong stock. The solute-mass check confirms the target.’
If the student cannot explain the sign, the ratio or the bound in everyday words, they may have memorised an algebraic pattern without understanding it. Verbal clarity is a useful diagnostic, although this Mathematics subject does not thereby become an English oral examination.
Frequently asked question: why not just average the two labels?
Because the portions may have different volumes. Equal-volume portions give equal weights, but unequal volumes do not. In the general model the target is the sum of each stock concentration multiplied by its share of the total volume. If one stock occupies three quarters of the mixture, its concentration should have three quarters of the weighting.
An ordinary mean is a special case, not the general rule. The proof is in the solute mass: calculate how many grams each portion contributes and divide by the total volume. The correct concentration emerges without guessing.
Frequently asked question: why does a negative volume appear?
A negative solution volume can arise when an algebraic target lies outside the feasible weighted-average range, or when an equation has been set up incorrectly. It is not a special kind of physical ingredient. The solver should inspect the model and constraints instead of treating the negative value as a recipe.
With positive-volume inputs and no solute loss, the final concentration is bounded by the two input concentrations. That fact can often establish impossibility before any lengthy calculation. Use it as a preliminary feasibility test.
Frequently asked question: can I mix volumes in centimetres and concentrations in g/dm³?
You can write a correctly scaled ratio equation using volumes expressed consistently in cm³, because a common factor cancels in numerator and denominator. But when calculating an actual solute mass in grams, volume in dm³ is required for a g/dm³ concentration unless you explicitly convert the factor. Problems become error-prone when unit conventions are mixed without explanation.
The most reliable route is to convert the given volumes to dm³ before multiplying by g/dm³. Always inspect whether the final requested answer is a concentration, a volume or a mass, and report the corresponding unit.
Frequently asked question: is the weighted-average equation always physically exact?
No. The version used here assumes uniform stocks, conservation of dissolved solute, no reaction or loss, and additive solution volumes. Real mixtures can depart from some of those assumptions, so a task involving actual final volume may require measuring it or using supplied information rather than adding inputs automatically.
Mathematical models are valuable because their assumptions are explicit and their consequences can be tested. Accuracy is not achieved by pretending every simplification is universally true; it is achieved by choosing the right model for the evidence available.
A last comprehensive test
A fictional school wants 480 cm³ at 62 g/dm³ using 20 and 90 g/dm³ stocks. Let f be the strong-stock fraction. Calculate f=(62−20)/(90−20)=42/70=0.6. The strong portion is 0.6×480=288 cm³; the weak portion is 192 cm³. Check: the strong contribution is 90×0.288=25.92 g, and the weak contribution is 20×0.192=3.84 g. Together they give 29.76 g in 0.480 dm³, exactly 62 g/dm³.
Now imagine the stronger stock is limited to 250 cm³. The mathematically required 288 cm³ violates the limit, so the exact target becomes infeasible without changing the available materials or total-volume requirement. A complete answer names both the algebraic solution and the constraint failure. That distinction is the heart of successful real-world Mathematics.
Final learning path
Begin with the meaning of g/dm³ in Volume 1025, then derive the two-stock equation from solute-mass conservation. Practise exact targets before inequalities and capacity limits. When you can explain why a target is impossible or why an unweighted mean is wrong, you have achieved more than the ability to copy a recipe. You can reason about constraints.
For official assessment details consult the 2027 K310 Mathematics syllabus. For the original osmosis setting revisit Volume 1022. The wider skill is transferable: define quantities, write consistent equations, test bounds and communicate a conclusion that the model can actually support. That is how straightforward arithmetic becomes dependable upper-secondary problem solving.
