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How to Perform in the new G3 SEC Examinations | Learner’s Guide Vol 0111 | Mathematics: Ticket-Revenue Audit — Build and Check Two Simultaneous Equations

G3 SEC Mathematics · Learner’s Guide Vol 0111. Build two equations from counts and receipts, solve, then check the original conditions.

A school sells two kinds of tickets. Adult tickets cost $15 each; student tickets cost $9. Sixty tickets are sold for a total of $720. How many of each kind were sold? You might try numbers until one pair fits. A better method is to use two pieces of information together: the number of tickets and the money collected. The count and the revenue describe the same sales, but in different units.

This G3 SEC Mathematics Learner’s Guide Vol 0111 teaches you to translate a two-price ticket story into simultaneous linear equations, solve them by substitution or elimination, inspect integer and capacity constraints, and check every answer against the original receipts. We will then move to refunds, changed prices, missing data and apparently correct solutions that do not fit the question. All worked examples are original exercises rather than official examination questions.

You do not need clever tricks to solve this kind of problem. You need a dependable order: name what is unknown; build the count equation; build the money equation; solve the pair; verify both original statements. It is the same continuity habit used in Volume 0087 on successive transfers and Volume 0091 on multi-stage journeys: carry the right quantities into the next step without silently changing their meanings.

Chapter index

1. Write down what each number is counting

A ticket problem often gives four kinds of number: price per ticket, number of tickets, total receipts and perhaps a capacity or discount. Before calculating, decide which is which. A price of $15 per adult ticket is not the total paid by all adults. A count of 60 tickets is not $60. A revenue of $720 is money, not the number of people attending.

For the opening example, let a be the number of adult tickets and s the number of student tickets. The quantities a and s are counts; they must be non-negative whole numbers in this setting. Together they account for all 60 tickets, giving a+s=60. One adult ticket contributes $15 and one student ticket $9, so the total amount is 15a+9s=720.

The two equations are about the same event. The first counts tickets; the second counts money. That is why they can be solved together. If you accidentally write 15a+9s=60, you have attached the money expression to a count. The algebra might still produce symbols, but its model is already wrong.

An important check is to look at units. In a+s=60, every term is a count of tickets. In 15a+9s=720, the terms are dollar amounts. Each equation is dimensionally coherent. Never add a number of tickets directly to an amount of money unless a defined conversion, such as price, gives the relationship.

A short translation table can help: adult tickets a, unit price $15, total adult revenue $15a; student tickets s, unit price $9, total student revenue $9s. The totals at the bottom of the count and money columns become the two equations. This is a simple bookkeeping idea, not an additional syllabus formula.

2. What the official 2027 K310 syllabus requires

The SEAB 2027 G3 Mathematics syllabus K310 includes solving simultaneous linear equations in two variables by substitution, elimination and graphical methods, and formulating equations to solve problems. Ticket-count and receipt contexts are original applications of those published skills, not named compulsory ticket question types.

The same K310 syllabus gives Paper 1 and Paper 2 equal weighting of 50%, with each lasting 2 hours 15 minutes and carrying 90 marks. Paper 1 contains about 26 compulsory short-answer questions; Paper 2 has 9 to 10 questions and concludes with a real-world application problem. Essential working matters, and calculators are permitted. Do not infer that this particular ticket scenario will appear simply because real-world situations can be assessed.

The syllabus also specifies default final accuracy for non-exact numerical answers, subject to the question’s wording. Counts of tickets should be whole numbers. Do not round a fractional number of tickets to the nearest integer as though that repairs an incorrect model. A fractional solution may show that the data do not describe an ordinary whole-ticket sale or that you translated the conditions incorrectly.

For the wider algebra learning route, visit the Mathematics Hub and the G3 SEC Learner’s Guide. This article owns the narrow job of constructing and auditing two equations from counts and receipts. General equation-solving technique is a prerequisite, not a substitute for interpreting the story.

Read the examination syllabus for the year you will sit. The definitions and numerical prices in this guide are classroom examples and do not describe real school event fees.

3. Solve the first case by substitution

Begin with a+s=60 and 15a+9s=720. From the first equation, s=60−a. Substitute this expression into the revenue equation: 15a+9(60−a)=720. Expand to obtain 15a+540−9a=720, so 6a=180 and a=30. Then s=60−30=30.

There were 30 adult tickets and 30 student tickets. This is not finished until the original conditions are checked. Count: 30+30=60. Revenue: 30×$15+$30×9 would mix a currency symbol with a count awkwardly; write $450+$270=$720. Both statements hold. The answer also respects the non-negative integer requirement.

Notice what the subtraction 15−9=6 means. Each adult ticket produces $6 more revenue than a student ticket. Replacing one student ticket with an adult ticket while keeping the number of tickets fixed raises receipts by $6. Relative to selling 60 student tickets at $9 each, the baseline revenue would be $540. Actual revenue is $720, an excess of $180. The excess represents 180/6=30 adult replacements.

That second explanation gives a fast independent check. It is not a different answer. It is the same algebra understood as a comparison against a simple baseline. It can be especially useful when a real-world question hides the equations among extra details.

If you obtained a=60, check the receipt: 60 adult tickets at $15 would yield $900, not $720. If you obtained a=20 and s=40, count works but revenue is $300+$360=$660, not $720. A solution must satisfy both equations, not whichever is easier.

The best working shows enough intermediate steps to make the route auditable without using paragraphs where three algebra lines will do.

4. Solve the same case by elimination

Write the pair again: a+s=60 and 15a+9s=720. Multiply the first equation by nine to obtain 9a+9s=540. Subtract this from the revenue equation. The student-ticket terms cancel, leaving 6a=180. Therefore a=30 and s=30.

Elimination works because the equations describe the same variables. Multiplying a complete equation by the same non-zero factor preserves its solutions. Subtracting equal quantities from the two sides also preserves equality. This is why the method is valid, not merely why it looks neat on paper.

You could instead eliminate a by multiplying the count equation by 15: 15a+15s=900. Subtract the actual revenue equation 15a+9s=720 to obtain 6s=180, so s=30. Both routes work. Choose the one that makes the numbers and signs easiest to manage.

A common error is multiplying only one term in an equation. If you multiply a+s=60 by nine, the result is 9a+9s=540, not a+9s=540. Another error is reversing a subtraction in one place but not the other. Keep the entire equation together as one relationship.

Elimination and substitution should agree. If they do not, inspect the first divergence rather than assuming one technique is inherently unreliable. Check the original equations, then each transformation. The goal is to find the wrong move, not rework everything blindly.

You can also use a graphical view: the equations represent two straight lines in the a–s coordinate plane. Their intersection is the common solution (30,30). This is useful conceptually even when arithmetic is more efficient.

5. Make sure the answer is physically possible

A ticket count cannot be negative. Nor can it exceed the number of tickets sold. If an algebraic solution yields a=75 when only 60 tickets were sold, the original data or interpretation must be examined. A negative result could arise because a price was assigned to the wrong category, the revenue total was misread or the scenario cannot hold under the stated assumptions.

The prices also give a quick revenue bound. If 60 tickets cost either $9 or $15, the total revenue must lie between 60×$9=$540 and 60×$15=$900, inclusive. A reported total of $950 would be impossible under this two-price, no-extra-charge model. This is a useful check to perform before solving.

An intermediate total such as $720 is within those bounds. That does not prove it produces whole-number ticket counts, but it establishes that it is not ruled out by the simple minimum and maximum. Once the equations are solved, the counts must still be integers.

Suppose the same event reports total revenue of $721 instead. The equations yield 6a=181, so a is not a whole number. The model cannot represent ordinary whole-ticket sales at the two specified prices with no other adjustments. It would be wrong to round a to 30 and claim the receipts matched exactly. You would need to check for a fee, discount, refund, transcription error or another stated category.

This is why Volume 0071 on constraints and feasibility is useful. Algebra generates candidates; contextual constraints determine which candidates actually answer the question.

Use a final check with three lines: total tickets, total revenue and category plausibility. This is a stronger habit than writing only the pair of numbers.

6. A second worked case with less convenient numbers

A community performance sells adult tickets for $18 and concession tickets for $10. The organiser records 72 tickets and $1,024 in receipts. Let a be adult tickets and c be concession tickets. Then a+c=72 and 18a+10c=1024.

From the first equation, c=72−a. Substitute: 18a+10(72−a)=1024. Expand: 18a+720−10a=1024. Therefore 8a=304 and a=38. Then c=72−38=34.

Verify count: 38+34=72. Verify receipts: 38×18=684 and 34×10=340, giving 684+340=1024. Both match. The result is a non-negative pair of integers within the total capacity.

Now suppose a student misreads “$1,024” as “$1,240”. That amount is still within the maximum possible revenue of 72×18=1296, so a simple bound does not immediately detect the error. But solving yields 8a=520 and a=65, c=7, a different pair. Both solutions are arithmetically coherent for their respective totals. The correct answer depends on reading the original amount accurately.

A useful technique is to write the known data above the equations before working. That reduces the chance of silently changing a number as you copy or rearrange. If the question includes a table, use the table as the authoritative reference, not a number remembered from one sentence.

This case shows why calculation and source reading are connected. A correct algebraic solution to incorrectly copied data is not the right response to the actual question.

7. The difference between gross sales and money after refunds

Refunds complicate a ticket question because they introduce a time boundary. Suppose 80 tickets were originally sold at two prices: adult $16, student $10. The recorded gross receipts were $1,100. Let a and s be the original numbers sold. Then a+s=80 and 16a+10s=1100.

Solve: s=80−a, so 16a+10(80−a)=1100. Then 6a+800=1100 and a=50; s=30. Check: 50×16+30×10=800+300=1100. These are the original sales, before any refund.

Now suppose three adult tickets and two student tickets are refunded in full. The number of valid tickets afterwards is 80−5=75. The net retained receipts are 1100−3×16−2×10=1100−48−20=$1032. The remaining tickets are adult 47 and student 28. Check retained receipts independently: 47×16+28×10=752+280=1032.

This is an important distinction. The original equations describe gross tickets and gross receipts. The after-refund figures describe a later state. If the question asks how many adult tickets were originally sold, the answer is 50, not 47. If it asks how many remain valid after refunds, the answer is 47 adult and 28 student. A learner who mixes the two moments can produce numbers that each look reasonable but belong to different problems.

Do not subtract all refunds at the student price simply because the category is forgotten. The amount refunded depends on the ticket type. Nor should you assume a processing fee unless the question supplies one.

For more general state updates, Volume 0089 on changed conditions teaches how to preserve the valid earlier work and rebuild only the dependent quantities.

8. When discounts change which price belongs in the equation

Imagine an exhibition with a standard ticket at $20 and a discounted ticket at $12. A notice says each standard-ticket buyer can receive a $3 discount with a voucher, but the question gives no number of voucher users. Can you solve a two-category problem using only standard and concession counts? Not necessarily. The standard category now contains two possible actual prices if vouchers were used.

Before making equations, ask whether the price is the listed price or the actual amount paid. Revenue equations use the appropriate actual payment for each transaction under the stated model. A label such as “adult” does not fix a unique contribution if a conditional discount applies.

Suppose the task clarifies that all adult buyers paid $17 after discount and students paid $12. If there are 50 tickets with total receipts $710, then a+s=50 and 17a+12s=710. Substitute s=50−a: 17a+600−12a=710, so 5a=110, a=22 and s=28. Check: 22×17+28×12=374+336=710.

Without the clarification about which adult buyers used a voucher, a model with one adult price may be underdetermined. The correct task is to identify what additional information is needed, not to invent a neat ticket split. This is a practical application of Volume 0101 on evidence sufficiency.

The same discipline applies to service charges and optional add-ons. A “total collected” might include extra fees or merchandise. If so, the revenue equation needs to account for them. Always ask what the stated total includes.

A strong solver first builds an accurate model, then performs accurate algebra. The reverse order makes even perfect calculations unreliable.

9. Recognise when two equations are not independent

Suppose an event report says “Forty adult and student tickets were sold in total” and later says “The number of adult tickets plus the number of student tickets equals forty”. Those are two sentences, but only one independent equation. They both say a+s=40. You still cannot determine a and s individually from that information alone.

A second useful equation must add a different constraint, such as total revenue with distinct prices or a stated difference between the counts. If the prices were both $10 and receipts $400, the revenue equation 10a+10s=400 is simply ten times the count equation. It adds no new information about the split. Many pairs still satisfy it.

Graphically, dependent equations can represent the same line rather than two lines intersecting at one point. The absence of a unique intersection is not a failure of your algebra; it is a property of the supplied conditions. If your work reduces to an identity such as 0=0, the equations may be dependent.

If an inconsistent total leads to 0=5, the conditions cannot both hold under the model. Do not try to invent a pair after reaching a contradiction. Check transcription and assumptions; if the question deliberately presents inconsistent information, explain the impossibility.

This distinction matters when reading long word problems. Two numbers are not automatically two independent facts. The model needs constraints that actually narrow the unknowns.

A useful starting question is, “Does the second statement tell me anything that the first could not?” If it merely restates the same total, search for another independent condition before attempting to solve.

10. A third worked case: reconstruct counts from a comparison

Sometimes the second equation comes from the difference between groups rather than revenue. Suppose a workshop issues 56 tickets and sells eight more student tickets than adult tickets. Let a and s be the counts. Then a+s=56 and s=a+8. Substitute into the total equation: a+(a+8)=56, so 2a=48, a=24 and s=32.

Now add prices: adult $14, student $8. The revenue is 24×14+32×8=336+256=$592. You have first solved the counts and then computed the receipts. That differs from the earlier problems where receipts formed the second equation.

A question might ask you to check a reported gross amount of $592 against the ticket split. You can do so without setting up a new pair of unknown equations, because the counts are already determined. Use the information in the order that requires the least unnecessary work.

Suppose an organiser says the student category brought in more revenue because it sold more tickets. Here it sold 32 versus 24, but student revenue was $256 versus adult revenue $336. The claim is false: greater count does not necessarily mean greater money. Price per ticket matters.

This is an example of comparison basis. “More tickets” refers to count; “more revenue” refers to money. A learner who treats them as synonyms can give an attractive but incorrect explanation even after solving the equations.

A short table with count, price and subtotal is often the most reliable presentation for multi-part questions. It makes each unit and comparison visible.

11. Unit checks, rounding and final interpretation

A ticket count must be an integer in a standard physical-ticket context. A dollar amount may include cents, depending on the prices given. Do not round counts without justification, and do not attach “tickets” to a money value. Units and the model’s domain are part of the answer.

When a price is quoted in dollars and cents, it can be helpful to express both price and revenue consistently in cents to avoid floating-point or transcription confusion. For example, $12.50 becomes 1250 cents and $7.50 becomes 750 cents. The equations remain exact under the stated figures.

If a question gives estimated revenue rather than an exact amount, the counts may not be uniquely determined. Treat the word “approximately” seriously. The acceptable integer pairs may need to be compared with the tolerance or range supplied. Do not assert a unique exact count from an approximate total unless the other conditions justify it.

A useful plausibility check is that the average revenue per ticket, total revenue divided by ticket count, lies between the two ticket prices if every ticket was sold at one of those prices and there are no other adjustments. In the first case it is $720/60=$12, between $9 and $15. In the 72-ticket case it is $1024/72≈$14.22, between $10 and $18.

However, an average price between the two listed prices does not alone guarantee whole-number solutions. The integral-count check still matters. The average is an early warning, not a replacement for the equations.

A final written interpretation might say: “Thirty adult and thirty student tickets were sold. These totals give sixty tickets and receipts of $720, so both recorded constraints are met.” It communicates the result and its verification calmly.

12. Original practice clinic: twelve independent questions

A. Adult tickets cost $15 and student tickets $9. Sixty tickets bring in $720. Build the equations and solve the two counts without looking at the worked opening.

B. Standard tickets cost $18 and concessions $10. Seventy-two tickets bring in $1,024. Find both counts and check the receipts.

C. An event sells 90 tickets: regular $14 and concession $8, collecting $960. How many of each were sold?

D. Sixty tickets are sold at $12 and $7, with total receipts $570. Find the two counts.

E. A workshop sells 56 tickets, eight more student than adult. Find both counts, then calculate receipts if prices are $14 and $8.

F. A sale records 80 tickets and $1,100 in gross receipts at adult $16 and student $10. Three adult tickets and two student tickets are fully refunded. Find the number of valid tickets and net receipts.

G. A student models 30 adults at $15 and 30 students at $9 using the equation 15a+9s=60. Explain the units error and correct the right-hand side.

H. A source gives 40 total tickets, with both ticket categories costing $10 and total receipts $400. Can the counts be found uniquely? Why?

I. A sale of 50 tickets at $17 and $12 yields $710. Solve the adult and student counts under a two-price model.

J. Sixty tickets at $9 or $15 are reported to bring in $950. Is that possible without extra charges or adjustments? Explain without solving a pair of equations.

K. A source gives 60 total tickets and $721 at adult $15 and student $9, with no extra payments. Explain the issue that appears when solving for whole-ticket counts.

L. A service notice says some adult buyers received a $3 voucher discount, but it gives no count of voucher users. Explain why a two-price count-and-revenue equation may no longer determine a unique adult/student split.

M. A venue allows no more than 100 attendees. The algebra gives 65 adults and 42 students. What context check must be applied?

N. A pair of algebraic equations reduces to 0=0 after elimination. What does that suggest about uniqueness, and what should be checked in the problem information?

O. A pupil obtains a=38 and c=34 for an event with prices $18 and $10. Write two independent checks that test the solution against a reported 72 tickets and $1,024.

Work through A–E with algebra. Use F–L to practise model interpretation and changes of state. Use M–O to check feasibility and the meaning of a solved pair.

13. Worked answers and diagnostic corrections

A. a+s=60 and 15a+9s=720. Substitute s=60−a: 6a=180, so a=30 and s=30. Revenue check $450+$270=$720.

B. a+c=72 and 18a+10c=1024. Substitution gives 8a=304, hence a=38 and c=34. Check 38×18+34×10=1024.

C. r+c=90; 14r+8c=960. Substitution gives 6r=240, so r=40 and c=50. Check 560+400=960.

D. Let x be $12 tickets and y be $7 tickets. x+y=60; 12x+7y=570. Substitute y=60−x: 5x+420=570, so x=30 and y=30.

E. a+s=56; s=a+8. Thus a=24, s=32. Revenue=24×14+32×8=$592.

F. Original counts a=50 and s=30. After five refunds, 75 tickets remain valid. Net receipts=1100−3×16−2×10=$1032.

G. The left side 15a+9s is money in dollars, not a count of tickets. The correct revenue equation is 15a+9s=720; the count equation is a+s=60.

H. No. The money equation is 10(a+s)=400, equivalent to the count equation a+s=40. A unique split requires an independent constraint.

I. a+s=50; 17a+12s=710. Then 5a=110, so a=22 and s=28. Check $374+$336=$710.

J. Maximum revenue is 60×$15=$900. A total of $950 exceeds that bound under the stated assumptions, so the report cannot hold.

K. The equations give 6a=181, so adult tickets would not be an integer. A normal whole-ticket model cannot produce exactly $721 from these counts and prices with no adjustments.

L. Adult tickets may now have two actual prices; without the number of discounted buyers, another unknown enters the revenue equation. More relevant information is required.

M. The total is 65+42=107, which exceeds the venue’s capacity of 100. An algebraic pair must satisfy the real-world constraint.

N. The two equations may describe the same constraint and thus fail to identify a unique pair. Check whether the second statement adds genuinely independent information.

O. Count: 38+34=72. Money: 38×18+34×10=684+340=1024. The units and both totals match.

For every error, name the first faulty decision: wrong price association, changed time basis, ignored discount, non-independent equation, integer violation, capacity violation or arithmetic mistake. That is a more useful correction than rewriting the entire worked solution.

14. From step-by-step working to examination speed

On the first day of learning, use a full table. On the second, build equations without a table but label both unknowns. On the third, solve equivalent cases by substitution and elimination and choose the shorter route. Speed should come after each step has meaning.

Under timed conditions, start by writing the two equations rather than estimating ticket counts in your head. The equations compress the story while preserving the conditions. The calculation can then be performed reliably with ordinary algebra.

Do not calculate a long answer and only afterwards read the phrase “after refunds”. That phrase changes which state is being asked for. Underline time markers such as originally, after, remaining and net before defining variables. If your unknown describes original tickets, say so.

Use a one-line unit check: ticket equation counts people or tickets; revenue equation totals money. Add a simple lower-and-upper revenue bound when the numbers are unfamiliar. That quick check can reveal an impossible statement earlier than algebra.

When you finish, reverse the route. Insert the values into the original problem rather than into a rearranged equation alone. Check totals and non-negative integers. If there is a refund or changed condition, verify the final state separately.

This is the algebra version of a strong reading habit: a conclusion must satisfy every relevant part of the source.

15. A seven-day route from foundation to independent work

Day 1: language. Work with five short ticket stories. Identify prices, counts and revenue, and name the unknowns without solving. This separates the model from the arithmetic.

Day 2: count and revenue. Build simultaneous equations for two ticket types. Use a table to check the units of each term.

Day 3: solving. Solve the same pair by substitution and elimination. Explain why both should give the same solution.

Day 4: contextual checks. Practise integer counts, maximum and minimum revenue, and venue capacity constraints. Reject impossible results with specific reasons.

Day 5: changed conditions. Add refunds or clearly stated discounts. Keep gross and net tickets and receipts separate.

Day 6: insufficient information. Work on cases where prices are the same or voucher use is unknown. Learn to distinguish missing evidence from weak algebra.

Day 7: independent clinic. Solve Questions A–O without notes, then record errors by type. Redo the weakest three with new numbers after several days.

Parents can ask, “Does this equation count tickets or money?” and later, “Does your pair satisfy both totals?” Those questions expose the underlying model rather than simply checking a numerical result.

16. The answer should balance two ledgers

A two-price ticket problem is not primarily a puzzle about letters x and y. It is an accounting story with two consistent records: how many items were sold and how much was collected. Simultaneous equations make the records agree. A good final answer proves the agreement.

Continue to the Mathematics Hub, the G3 SEC Learner’s Guide and Volume 0105 on command words. When a question changes after a refund, Volume 0089 on revised conditions offers a useful parallel.

Official examination scope is verified in the 2027 SEAB G3 syllabus index and the K310 Mathematics syllabus. The prices, counts and situations are original practice data and do not represent actual school ticket sales.