Start with the answer: fit circles, not just area
A school exhibition has a shallow rectangular display tray, 116 mm wide and 92 mm long. It needs to hold round cardboard tokens, each 16 mm in diameter. The design calls for at least 4 mm of clear space between neighbouring circular edges and at least 6 mm between each token and the surrounding tray wall. How many tokens can the students place in a regular rectangular grid? The answer is 20: five columns and four rows, using the full width and height conditions rather than just a rough area division.
This guide teaches a useful G3 Mathematics distinction. An area calculation tells us how much surface exists. It does not automatically tell us whether a collection of shapes fits inside a boundary. For a rectangular grid of round objects, the correct starting point is a line of diameters and gaps. We then test how many complete columns and rows are allowed, locate their centres, check margins, and only afterwards discuss area percentages, scale diagrams and design choices.
The tokens are fictional cardboard markers for a school display, not medicines and not a model for real pharmaceutical packaging. The example follows the Chemistry mole-calculation lesson in Volume 1153, but deliberately changes the examination skill: instead of predicting chemical quantities, we are solving a geometry-and-integer layout problem.
Exam scope and important boundaries
The 2027 Singapore–Cambridge SEC G3 Mathematics subject uses syllabus K310, with 4052 noted as the earlier reference code. The examination has Paper 1 and Paper 2, each lasting 2 hours 15 minutes and carrying 90 marks for 50% of the assessment. Geometry, mensuration, measurement units, circles, proportional reasoning and real-world mathematical modelling are relevant foundations for this exercise. Paper 2 includes a final question focused on the application of Mathematics in a real-world situation.
The tray example is original practice, not an official specimen question. The general K310 syllabus expects relevant working, correct units, appropriate accuracy, clear interpretation and evaluation of a mathematical model. Unless a question specifies otherwise, non-exact numerical answers are generally given to three significant figures. An answer that is a count of objects, however, must be an integer that fits all restrictions. Rounding 5.4 columns up to six would produce a physically impossible layout.
The quick rule for a regular grid
For n equally sized circular tokens of diameter d in one row, with a minimum clear gap g between neighbouring edges and minimum side margin m at both ends, the required length is L = nd + (n − 1)g + 2m. The n diameters contribute nd. There are exactly n − 1 internal gaps, not n. The two outer margins contribute 2m. Each term uses a length unit, so the expression has the correct dimensions.
For the present tray, d = 16 mm, g = 4 mm and m = 6 mm. The total required width is 16n + 4(n − 1) + 12 = 20n + 8 mm. Since width is at most 116 mm, solve 20n + 8 ≤ 116. This gives n ≤ 5.4, so the greatest whole number of columns is five. The same formula applies to rows because the tokens and minimum clearances are unchanged; only the available side length differs.
Check the vertical direction independently
Let r be the whole number of rows. The required height is 16r + 4(r − 1) + 12 = 20r + 8 mm. The available height is 92 mm, so 20r + 8 ≤ 92 and r ≤ 4.2. Therefore the largest number of complete rows is four. If the design is a rectangular grid, its capacity is 5 × 4 = 20 tokens.
Notice the mathematical structure. The width limit rules out six columns because they would need 20(6) + 8 = 128 mm. The height limit rules out five rows because they would need 20(5) + 8 = 108 mm. Both are greater than their available dimensions. A student who states ‘20’ without these feasibility checks may have guessed correctly but has not demonstrated why the layout is valid.
What the word maximum means here
The count of 20 is the maximum for the specified regular rectangular grid with rows and columns aligned. It is not a proof that no staggered arrangement of circles could hold more. Different packing patterns can change how circles share space; proving a global maximum over all imaginable arrangements requires a different geometric argument. A careful answer says ‘20 in the prescribed grid’, not ‘20 is the absolute maximum for every possible arrangement’.
This distinction matters in examinations. A model may ask for a largest whole number under given constraints. We should solve precisely the model asked for and name its limits when the question requires evaluation. Avoid adding an unsupported general claim just because a single layout has been optimised. Clear restrictions protect both the mathematics and the interpretation.
Use coordinates to verify the grid
The five-column arrangement occupies 5(16) + 4(4) = 96 mm from the left edge of its first circle to the right edge of its last. The tray width is 116 mm, leaving 20 mm of unused width. If the arrangement is centred, each side margin is 10 mm, which exceeds the required 6 mm. The centre of the first circle is 10 + 8 = 18 mm from the left wall, and subsequent centres are 20 mm apart.
Measured from the left wall, the x-coordinates of the circle centres are 18, 38, 58, 78 and 98 mm. The row arrangement occupies 4(16) + 3(4) = 76 mm vertically, leaving 92 − 76 = 16 mm. Centred top and bottom margins are 8 mm. The centre y-coordinates are therefore 16, 36, 56 and 76 mm. The final circle reaches 76 + 8 = 84 mm, leaving 8 mm to the bottom wall.
The purpose of the centre-to-centre pitch
When identical circles have diameter d and the required clear gap is g, adjacent centres must be at least d + g apart. Here 16 + 4 = 20 mm. That centre-to-centre distance is called the pitch in this simple geometric model. The formula keeps the geometry coherent because the circle extends a radius of 8 mm on each side of its centre. A 20 mm pitch leaves exactly 4 mm between neighbouring circular boundaries.
Students sometimes subtract the gap from the diameter or use the radius as though it were the full width. Drawing one pair of neighbouring circles to scale reveals the error. Their two radii sum to 16 mm across the centre line. Add a 4 mm edge clearance and the centre separation becomes 20 mm. Every coordinate in the grid should differ by the same pitch along a row or column.
Why dividing by circle area cannot determine capacity
The tray area is 116 × 92 = 10,672 mm². One token has radius 8 mm, so its circular area is π × 8² = 64π mm², approximately 201 mm². A careless calculation might divide 10,672 by 201 and conclude that 53 tokens fit. That would ignore the required gaps, outer margins and geometric impossibility of filling all available surface with circles.
The quotient can be an upper bound based on area when all tokens lie inside the tray and do not overlap, but it is generally too optimistic for a real packing pattern. In this question, the prescribed regular grid is what determines the feasible count. Its 20 circles have a total face area of 1,280π mm², approximately 4,021 mm². Dividing by tray area gives about 37.7% covered by cardboard faces. That percentage describes surface coverage, not a fraction of chemical content or a measure of display quality.
How the margins affect available space
Outer margins and internal gaps serve different jobs. For n circles there are two end margins but only n − 1 internal gaps. The outer margins are measured from the tray boundary to the nearest circle edge. The gap is measured between the edges of neighbouring circles. Mixing these measurements can lead to an off-by-one error that affects the maximum count.
The simplest check is the one-token case. If n = 1, there are no internal gaps, so the formula must reduce to d + 2m. Our expression nd + (n − 1)g + 2m does exactly that. If a proposed formula adds g when n = 1, it is counting a gap between objects that do not exist. Testing a formula on a tiny case is a strong habit in algebra and design problems.
What changes when the model circles are smaller
Suppose a revised exhibit uses cardboard discs of diameter 12 mm but keeps the minimum 4 mm internal gap and 6 mm wall margin. For columns, required length is 12n + 4(n − 1) + 12 = 16n + 8. Within 116 mm, n ≤ 6.75, so six columns fit. For rows, 16r + 8 ≤ 92 gives r ≤ 5.25, so five rows fit. The redesigned regular grid can therefore contain 6 × 5 = 30 model tokens.
The increase from 20 to 30 is an increase of ten tokens, or 50% relative to the original count. But a larger capacity does not mean the discs contain more material, are as legible or represent the same quantity of chemical in a genuine product. The exercise compares geometric models with stated token dimensions. Any conclusion about other properties would require information that the question does not supply.
Construct a scale drawing without altering the rules
A scale of 1:2 means every drawing length is half its actual length. On that drawing the 116 mm width becomes 58 mm, the 92 mm height becomes 46 mm, the 16 mm token diameter becomes 8 mm, the 4 mm minimum gap becomes 2 mm and the 6 mm minimum wall margin becomes 3 mm. Every linear measurement uses the same scale factor, otherwise the picture is distorted.
A scale drawing helps a reader see the arrangement, but the arithmetic should still be based on actual dimensions unless the question explicitly asks you to work in drawing units. If you double the number of tokens because the circles look small on paper, the scale has been misunderstood. A scale changes the representation, not the physical capacity. The area on the drawing is one quarter of the corresponding actual area because both linear dimensions are halved.
Reverse questions are harder than forward questions
Instead of asking how many tokens fit, an examiner may ask for the minimum tray width needed for six columns. Use n = 6 in the length formula: L = 16(6) + 4(5) + 12 = 128 mm. The minimum width is 128 mm at the stated clearances. If the available width is only 116 mm, the design misses by 12 mm. Rounding, a diagram or an optimistic average cannot remove that shortfall.
Likewise, the minimum height for five rows of 16 mm discs is 16(5) + 4(4) + 12 = 108 mm. A tray 92 mm high is short by 16 mm. This reverse view is especially useful because it makes the limiting dimension obvious. The column count is limited by width, while the row count is limited by height; fixing only one dimension may not increase total capacity.
Make a design recommendation that the data support
If the school needs at least 24 round tokens and must retain the 16 mm diameter, the original 116 by 92 mm regular-grid tray fails because it fits 20. With five columns, a fifth row would raise capacity to 25, but the height would have to grow from 92 to at least 108 mm. Alternatively, with four rows, six columns would raise capacity to 24, but the width would have to reach at least 128 mm. Both are mathematical options, not automatic design approvals.
Which option is better depends on constraints not yet specified: shelf width, available height, cost, handling and the exhibit board. A clear answer can compare the two required changes and then state that additional information is needed to choose. An examiner may award credit for identifying that an arithmetic optimum is not always a complete real-world decision.
Turn a word problem into a sequence of checks
Read the entire question, underline diameter, gap, margin and tray dimensions, and label each unit. Define n for columns and r for rows. Build two inequalities rather than trying to guess a count from area. Solve them, take the largest non-negative whole numbers that satisfy each restriction, multiply to obtain grid capacity, and then substitute the chosen numbers back into the original formulas.
Only after establishing a feasible arrangement should you calculate optional quantities such as unused length, percentages of coverage, scaled dimensions or costs. This order protects against an attractive answer that cannot be built. It also keeps methods visible to the marker. A suitable line of working often does more than a page of explanations written around an unsupported final number.
Exam practice 1: The width needed for four circles
Question. A regular row uses 16 mm diameter, 4 mm internal edge gap and 6 mm side margins. What is the minimum width for four circles? Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. There are four diameters, three internal gaps and two outer margins. Width = 4(16) + 3(4) + 2(6) = 88 mm. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. There are not four gaps between four objects. Count the actual intervals before inserting the numbers. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 2: Check five columns
Question. Show that five model circles fit into the 116 mm tray width. Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. Required width = 5(16) + 4(4) + 12 = 108 mm, leaving 8 mm beyond the minimum margins. The arrangement is feasible. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. The inequality is a minimum-width requirement, not an instruction that the tokens must touch the margin. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 3: Reject six columns
Question. Why does the sixth column not fit in the original tray? Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. Six diameters use 96 mm, five gaps use 20 mm and two margins use 12 mm, totalling 128 mm. This exceeds 116 mm by 12 mm. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. A proposed arrangement needs all three parts of the length, not just the six diameters. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 4: Find the row limit
Question. How many rows fit in a tray height of 92 mm with the original circle dimensions? Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. From 20r + 8 ≤ 92, we obtain r ≤ 4.2. The greatest feasible whole number is four. A fifth row requires 108 mm. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. A count of 4.2 rows has no useful physical meaning for an all-or-nothing grid of whole objects. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 5: Count the complete layout
Question. Five columns and four rows are feasible. How many tokens does the grid hold? Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. The rectangular layout contains 5 × 4 = 20 positions. Every row has the same number of columns and the arrangement respects both separate dimension limits. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. Do not add five and four. The question describes a two-dimensional array of positions. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 6: Find unused width
Question. If a centred five-column grid has a footprint of 96 mm, what width is unused? Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. The unused horizontal length is 116 − 96 = 20 mm. If centred, 10 mm remains on each side. That already includes the necessary 6 mm minimum margin. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. Do not subtract the minimum margins twice when the footprint means the outer edges of the first and last circles. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 7: Find unused height
Question. A four-row grid has outer-edge span 76 mm. How much height remains in a 92 mm tray? Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. The leftover is 92 − 76 = 16 mm, or 8 mm above and 8 mm below when the arrangement is centred. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. Use the row footprint 4(16)+3(4), not 4(16)+4(4). A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 8: Locate the first centre
Question. The horizontal edge margin of the centred arrangement is 10 mm. Where is its first centre? Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. A circle radius is 8 mm, so first x = 10 + 8 = 18 mm. The last centre at 98 mm leaves 10 mm from its right edge to the wall. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. The margin ends at the circle edge, not its centre; radius must be added. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 9: Find adjacent centres
Question. How far apart are centres of touching-clearance circles? Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. The minimum centre-to-centre distance is 16 + 4 = 20 mm. The circles have an 8 mm radius each, so 20 − 8 − 8 = 4 mm of edge clearance. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. The pitch is the sum of full diameter and the between-edge gap, not radius plus gap. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 10: Check the final centre
Question. Why does the x-coordinate 98 mm satisfy the right-side margin? Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. The rightmost circle extends to 98 + 8 = 106 mm. Subtracting from the tray width gives 116 − 106 = 10 mm of margin, exceeding the required 6 mm. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. Checking a centre alone is insufficient because the physical circle extends beyond that centre. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 11: Check vertical coordinates
Question. For four rows at y=16,36,56,76 mm, what is the bottom margin? Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. The last radius extends to 76 + 8 = 84 mm; the bottom wall is at 92 mm, leaving 8 mm. The top margin is also 16 − 8 = 8 mm. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. Check the outer edges of the first and last tokens, not only spacing between centres. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 12: Reverse a six-column width
Question. What width is required to increase capacity to six columns without changing token diameter or gaps? Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. The minimum width is 6(16)+5(4)+12 = 128 mm, which is 12 mm wider than the original tray. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. Reverse questions use the same formula. They do not justify changing the required clearance silently. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 13: Reverse a five-row height
Question. What minimum height supports five rows of 16 mm circles? Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. The minimum height is 5(16)+4(4)+12 = 108 mm. Compared with 92 mm, the tray needs an additional 16 mm of height. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. The count of gaps is four, because gaps occur between neighbouring pairs. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 14: Compare 24-token layouts
Question. With 16 mm circles, compare a six-column/four-row design and a five-column/five-row design. Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. Six by four holds 24 tokens and needs at least 128 × 88 mm. Five by five holds 25 positions and needs at least 108 × 108 mm. Both satisfy a target of 24 if the surrounding space permits. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. The plan with one extra position may still be less suitable if available shelf height is limited. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 15: Change token diameter
Question. What is the maximum original-tray grid when the diameter falls to 12 mm? Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. The length formula becomes 16n + 8 mm. Six columns and five rows fit, giving 30 tokens, as both resulting required dimensions fit 116 × 92 mm. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. You must recalculate pitch and the number of rows. Changing diameter alters both directions. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 16: Change only the gap
Question. The diameter is 16 mm, margins are 6 mm and internal gap falls to 2 mm. How many columns fit? Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. Required length = 16n + 2(n − 1) +12 = 18n +10. Since 18n +10 ≤116, n≤5.888…, so five columns fit. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. A smaller gap need not immediately allow another whole object; compare against the next integer threshold. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 17: Increase the margin
Question. The original width remains 116 mm but the minimum side margin rises to 10 mm. Do five columns still fit? Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. Required width = 5(16)+4(4)+2(10)=116 mm. Five columns fit exactly with no extra horizontal space. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. A design can be feasible at equality. ‘At least 10 mm’ includes exactly 10 mm. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 18: Find circle area
Question. Find the face area of one 16 mm diameter disc. Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. Radius = 8 mm. Area = πr² = 64π mm², about 201 mm² to three significant figures. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. The diameter must be halved before squaring; using πd² overestimates area by a factor of four. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 19: Find total circle area
Question. How much of the tray is covered by 20 non-overlapping disc faces? Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. Combined face area = 20 × 64π = 1280π mm² ≈4021 mm². Because circles do not overlap, their areas can be added. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. This number is area, not the capacity in tokens or the volume of the display tray. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 20: Calculate coverage
Question. What percentage of the tray base is covered by the 20 disc faces? Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. Coverage = 1280π/10672 ×100% ≈37.7%. Use the same square-millimetre unit for numerator and denominator. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. Do not describe the remaining 62.3% as uniformly usable space; some lies in narrow gaps or edge strips. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 21: Reject area-only capacity
Question. Dividing tray area by one disc area gives more than 50. Why is that not the feasible grid answer? Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. That quotient ignores geometric placement, internal gaps and wall clearances. It is not proof that 50 circles can occupy the tray in the required regular grid. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. Area is a necessary constraint, not a sufficient packing method. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 22: Convert surface area
Question. Convert 10,672 mm² into cm². Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. Since 1 cm = 10 mm, 1 cm² = 100 mm². Thus 10,672 mm² = 106.72 cm². Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. Do not divide by 10: the area conversion factor is the square of the length conversion factor. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 23: Draw at half size
Question. What should the drawn diameter be for a 16 mm disc at scale 1:2? Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. The drawn diameter is 8 mm; the gap is 2 mm and the outer minimum margin is 3 mm. Every length is halved. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. Changing only the circles and leaving the drawn tray unchanged would not be a consistent scale drawing. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 24: Scale the tray dimensions
Question. Find the dimensions of a 116 × 92 mm tray on a 1:2 diagram. Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. The drawing measures 58 × 46 mm. The physical grid still holds the same number of tokens; only the representation changes. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. A reduction in drawing length does not reduce the stated physical tray capacity. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 25: Area scale in a drawing
Question. How much of the drawing’s area represents the real 10,672 mm² tray at scale 1:2? Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. Area scale is (1/2)² = 1/4, so the drawing area is 10,672/4 = 2668 mm². It also equals 58 × 46. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. Do not halve the area; two independent linear dimensions are scaled. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 26: One-token formula check
Question. What does the minimum-width formula become for n=1? Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. L = d + 2m, because (n−1)g = 0. With d=16 and m=6, the width is 28 mm. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. A one-token test is a practical way to detect counting an extra internal gap. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 27: Compare arrangements without claiming global optimality
Question. The five-by-four grid is verified. Can a student conclude no other arrangement could hold 21 discs? Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. Not from the grid calculation alone. The result establishes a maximum within the specified regular aligned grid. A different staggered arrangement is a different problem requiring a separate proof or analysis. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. A limited mathematical model must not be reported as an unrestricted theorem. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 28: Limit by the shorter dimension
Question. Why does a wider tray not automatically add more rows? Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. Rows are governed by the independent vertical inequality 20r+8≤92. Increasing width leaves that inequality unchanged; a fifth row still requires more height. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. A two-dimensional design has at least two independent constraints. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 29: Decide which enlargement serves 24 tokens
Question. A designer can widen the original tray by 12 mm or increase its height by 16 mm. Which is sufficient? Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. Widening from 116 to 128 mm allows six columns with four rows, so 24 positions. Heightening from 92 to 108 mm allows five columns with five rows, so 25 positions. Either reaches at least 24. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. The preferred change depends on surrounding dimensions and cost, not just the numerical count. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 30: Compare a 12 mm disc’s area
Question. A 12 mm disc has radius 6 mm. What is its face area? Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. Area is 36π mm², approximately 113 mm². Thirty such discs have a combined area of 1080π mm² ≈3393 mm². Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. Smaller tokens can give more grid positions while the summed face area becomes smaller; count and area answer different questions. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 31: Check whether a 24-token claim is supported
Question. A public board says the original tray holds 24 full-size tokens because ‘it has enough total area’. Is that statement valid? Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. No. The fixed 116 by 92 mm tray with the stated margins and regular-grid rule fits five columns and four rows, or 20. Correct the capacity and identify the relevant restrictions. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. A numerical claim must describe the actual model and not conceal a failing dimensional constraint. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 32: Place centres on a coordinate diagram
Question. Write the 20 centre positions without listing every pair in prose. Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. Use x ∈ {18,38,58,78,98} and y ∈ {16,36,56,76} mm, then take every ordered pair (x,y). This produces five times four, or 20 locations. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. The x- and y-values are independent coordinate sets for this rectangular grid, not 9 points obtained by adding their counts. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 33: Decide whether a measured 3 mm gap meets the rule
Question. A drawing shows two tokens with centres 19 mm apart, each radius 8 mm. Does the gap satisfy 4 mm? Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. Edge gap =19−8−8=3 mm. It fails the minimum 4 mm requirement by 1 mm. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. A diagram that looks close enough may still fail a numerical constraint; calculate the actual edge clearance. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 34: Find required pitch from a new clearance
Question. Suppose the minimum clear gap rises from 4 to 5 mm for 16 mm discs. What is the new pitch? Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. The minimum centre-to-centre distance is 16+5=21 mm. Recalculate the grid inequality as 16n+5(n−1)+12≤116, not by changing only the result. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. A changed condition requires new working. Do not reuse a solved inequality with its old parameter. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 35: Distinguish face area from volume
Question. The cardboard tokens represent round tablets, but no thickness is supplied. Can their total volume be calculated? Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. No. Circle area is known, but volume of a cylindrical token additionally requires thickness. A numerical volume cannot be inferred from diameter alone. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. This is the essential test of sufficiency: the question must supply every dimension used by the requested formula. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
Exam practice 36: Estimate rather than assert a manufacturing design
Question. Could this cardboard model prove that real medicinal tablets can be packed this way? Decide first whether the requested result is a length, area, coordinate, whole-number capacity or justified interpretation. Label the original dimensions before calculating; the correct unit is part of the answer.
Worked solution. No. The exercise fixes ideal circles, exact dimensions, clearances and a regular grid. A real package would require other specifications beyond school Mathematics. Check whether the result obeys all original constraints. Where the question asks for a maximum count, test the next whole number and show that it fails. Where the question asks for a geometric length, reinsert the number into the measured layout.
Common error and repair. A well-explained limitation is not a weakness; it shows that the mathematical conclusion has the correct scope. A neat final number alone may hide the wrong model. Use a small diagram, an inequality or a unit check to make the logic visible and correct the first mistaken step rather than adjusting the answer by guesswork.
A useful revision routine for Mathematics Paper 1 and Paper 2
Start without the calculator by drawing one row of circles and counting the diameters and gaps. On the second practice session derive L = nd+(n−1)g+2m and test n=1 and n=2. On the third, solve one width inequality and one height inequality, then explain why the results must be whole numbers. On the fourth, plot circle centres and check clearances using radius and pitch. On the fifth, calculate covered area and convert mm² to cm². On the sixth, practise reverse design decisions and scale drawings.
Finish with a timed mixed problem that removes one detail and asks whether there is enough information. Review your work against three questions: Did I count gaps correctly? Did I keep length, area and capacity distinct? Did I claim only what the regular-grid model proves? These checks apply far beyond exhibition trays. They also improve work on seating layouts, printed labels, storage grids and other real-world settings.
Frequently asked questions
Is circle packing a named compulsory G3 topic? The official syllabus contains circle geometry, mensuration, inequality and modelling skills. This tray is an original combined application, not a promised named examination task.
Why is the number of internal gaps one less than the number of tokens? Each gap lies between adjacent objects, so n objects in a straight row create n−1 internal intervals.
Why does the answer use whole numbers? The problem requires complete, non-overlapping tokens in a regular grid. A fraction of a token does not count as another position.
Does more tray area always mean more capacity? Not necessarily. The width, height and minimum clearances can prevent a new row or column despite unused area.
Is 20 the absolute mathematical packing maximum? It is the maximum for the stated regular aligned grid, not a proof about every possible staggered layout.
Why measure the centre distance? For equal circles it is diameter plus edge gap, allowing direct checks of spacing.
Can I calculate area with the diameter? First divide by two to find the radius, then use A=πr².
Why do scale drawings change area by a square factor? Both the horizontal and vertical dimensions change by the same length scale.
Do I need a calculator for all these steps? No. Much of the structure is counting and inequalities. A calculator is useful for π and final approximations where allowed.
Why link a Chemistry article to a Mathematics geometry problem? The shared exhibition context makes the transition understandable while the assessed mathematical skill is entirely different.
Can the grid design be used to recommend medicine packaging? No. This is a hypothetical cardboard-token geometry exercise, with no claim about pharmaceutical handling or quality.
What should I write if information is missing? Identify the missing dimension or constraint and explain why a numerical conclusion cannot yet be obtained.
Continue learning with official and local owners
The SEAB 2027 G3 school-candidate directory is the authoritative route to examination syllabuses. Read the G3 Mathematics K310 syllabus for current scope and format. The question here is an original teaching application, not an official assessment item.
For earlier Mathematics skills, use the Secondary 3 geometry and mensuration owner and the Secondary 1 Mathematics foundation. For the earlier cross-subject distinction between claims and supporting numbers, revisit Volume 1152. Families coming from PSLE can start at the Sengkang transition into G3.
The lasting examination habit is to model the right thing before calculating. A count of positions is not an area; a scale image is not a larger physical tray; a maximum under grid restrictions is not a theorem about every arrangement. Once the problem’s meaning is secure, the arithmetic becomes both simpler and more trustworthy.
