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aiMOOC

Ratio sharing and scale - Mathematics Year 10

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Introduction

A ratio 3: 5 does not mean that the first share is 3/5 of the total. A map scale compares lengths in the same units, and a length scale factor does not also serve as an area scale factor.

Solve five ratio problems and three scale problems. Use unit-parts, whole-fractions and inverse checks to explain your choices, including non-integer shares and conversions between centimetres, metres and kilometres.

Solve five ratio problems from different given quantities

A ratio compares quantities in a specified order. If A: B=3: 5, then A=3 u and B=5 u for a common unit-part u. The total is 8 u, the difference is 2 u, and the first part is 3/8 of the total. The fraction A/B=3/5 instead compares the first share directly with the second share. Name the denominator’s meaning before calculating.

For total£168 in ratio 3: 5, add 3+5=8 parts. One part is 168/8=£21; the shares are 3×21=£63 and 5×21=£105. A second route uses fractions of the total: 3/8×168=63 and 5/8×168=105. The checks 63+105=168 and 63/105=3/5 test both conservation of total and the required relationship.

Five complete ratio problems. The supplied datum may be a total, one share or a difference. Those are different quantities and cannot all be divided by the total number of parts without interpretation.

Task	Given information	Total ratio parts	Value of one part	Completed result
1	Total£168; ratio 3: 5	8	£21	£63 and£105
2	Total 54 items; ratio 2: 3: 4	9	6 items	12, 18 and 24 items
3	First share 36; ratio 3: 5	8	12	First 36, second 60, total 96
4	Total 2.5 L; ratio 2: 3	5	0.5 L	1 L and 1.5 L
5	Difference 18; ratio 5: 2	7	6	Larger 30, smaller 12, total 42

In row 2, 2+3+4=9 parts and 54/9=6 items per part. The shares 12, 18, 24 add back to 54 and each share divided by its ratio component gives 6. The three-way check is stronger than checking only that the final numbers add correctly.

In row 3, the known 36 is the first 3 parts, so u=36/3=12. The other 5 parts make 60 and the total is 96. Dividing 36 by 8 would incorrectly treat the supplied first share as the complete total. In row 4, a whole 2.5 L divided into 5 parts gives.5 L per part; ratio parts are abstract counts, while their values can be non-integers with units.

In row 5, the difference between 5 parts and 2 parts is 3 parts. Thus u=18/3=6, and the shares 30 and 12 differ by 18. Their total 42 corresponds to 7 parts. Dividing the difference 18 by 7 confuses difference-parts with total-parts even though 7 is the correct total ratio count.

Equivalent ratios preserve a common multiplier. For 3: 5 and 6: 10, the multiplier is 2. Adding 2 to get 5: 7 does not preserve equivalence because 3×7=21 but 5×5=25. Cross-products provide a second check when comparing ratios. Convert units first: 50 cm: 2 m is 50: 200=1: 4, not 25: 1.

Solve three scale problems and distinguish dimensions

A length scale 1: N means one drawing unit represents N of the same real-world length unit. A scale 1: 25000 therefore makes 1 cm on a map correspond to 25000 cm in the modelled world. Convert only after that same-unit relationship is clear. Since 100 cm=1 m and 100000 cm=1 km, 4 cm represents 100000 cm=1 km.

Three complete scale problems. Separate a map or plan conversion from enlargement of a geometric shape. The last row asks for area as well as new side lengths.

Task	Given information	Same-unit or dimensional working	Requested result
1	Map 1: 25000; map distance 4 cm	4×25000=100000 cm	1 km
2	Plan 1: 50; plan length 7.5 cm	7.5×50=375 cm	3.75 m
3	Similar rectangles 4 cm×6 cm; length factor 2.5	10 cm×15 cm; 24 cm²×6.25	Area 150 cm²

For row 1, the unit-route 4 cm×25000=100000 cm followed by centimetres-to-kilometres conversion gives 1 km. An alternative first converts the scale: 1 cm represents.25 km, so 4 cm represents 4×.25=1 km. Reverse-check by 1 km=100000 cm and 100000/25000=4 cm on the map.

For row 2, 7.5 cm×50=375 cm=3.75 m. A second route notes that 1 drawing centimetre represents.5 m, so 7.5×.5=3.75 m. Dividing 375 by 50 reconstructs 7.5 cm. Mixing the unconverted 375 cm with a label 375 m would introduce a factor-of 100 error.

For similar rectangles in row 3, multiply both perpendicular sides by 2.5. New dimensions are 10 cm and 15 cm. Area changes from 4×6=24 cm² to 10×15=150 cm². The area factor is $k^{2} = (2.5)^{2} = 6.25$ , not 2.5. The independent ratio 150/24=6.25 agrees with squaring the length factor.

The squared factor follows from multiplying two scaled lengths: (ka)(kb)=k² ab. For a similar solid, three independent length factors give volume factor k³. This dimensional reasoning requires geometrically similar shapes; sharing one length or one area does not by itself prove similarity. A non-uniform stretch changes different directions by different factors.

The original Similar Shapes tutorial identifies corresponding sides and finds a missing length. Follow how correspondence is established, then distinguish its length calculations from our area-factor task. Its displayed dimensions are not measurements of our rectangle. Original: corbettmaths, Similar Shapes - Missing Sides. The written explanations and native tasks work without sound or the external player.

A drawing is not automatically a valid measuring instrument: screen zoom and printing can alter displayed size. Use its written scale and stated data. When a task supplies dimensions directly, those stated dimensions control the calculation; measuring pixels is not an independent exact check of the mathematical answer.

Transfer methods and repair first errors

Choose a method by identifying what the supplied quantity represents. Unit-parts are convenient for an unknown total or a known difference; fractions of the whole can be quick when the total is given. Either method needs a conservation check and a ratio check. A result can satisfy one condition while failing the other.

For unfamiliar transfer, an invented 64 L mixture has component ratio 5: 3. There are 8 parts, each 8 L, so the quantities are 40 L and 24 L. Add 8 L only to the second component: the new amounts 40 L and 32 L give ratio 5: 4. The total is 72 L. The original 5: 3 ratio cannot be reused after changing only one component.

A second way checks the new first-to-total fraction: 40/72=5/9, so the second-to-total fraction is 4/9. These fractions sum to 1 and correspond to part-part ratio 5: 4. Reporting 5/4 as the first fraction of the total would exceed 1 and expose the denominator error immediately.

For a map transfer with scale 1: 50000, a route is drawn as three segments.8 cm, 1.2 cm and 1.6 cm. Add the drawn lengths to 3.6 cm, then multiply by 50000 to get 180000 cm=1.8 km. Alternatively each map centimetre represents.5 km, and the real segments are.4, .6, .8 km, which sum to 1.8 km. This is length along the modelled route, not a claim about straight-line separation or actual terrain travel time.

For reverse area reasoning, similar shapes with area ratio 9: 4 have positive length ratio 3: 2. Taking the square root of the area factor 9/4 gives 3/2, not 9/4. If an original side is 8 cm, the corresponding enlarged side is 12 cm. Check by comparing length ratio 12/8=3/2 and its square 9/4. A negative root is inadmissible for a scale between positive lengths.

First-error repair. A learner shares£168 using 168×3/5=100.8 and calls that the first share. The first error is interpreting the second component 5 as the total ratio parts. Replace 3/5 by 3/8 to obtain 63. Another learner enlarges a 24 cm² area by length factor 2.5 and gets 60 cm². The first error is using a one-dimensional scale factor for a two-dimensional quantity; the corrected area is 24×6.25=150 cm².

Retain all five ratio solutions and all three scale solutions, including the datum’s meaning, units, independent checks and a corrected misconception. On a later attempt, cover the results and use the new mixture or route values. Record the actual response and uncertainty rather than treating repeated reading as evidence of independent problem solving.

Interactive Tasks

Quiz: Test Your Knowledge

Share£168 in ratio 3: 5. First share? (£63) (!£100.80) (!£21) (!£105)

In ratio 3: 5, the first fraction of the total is? (3/8) (!3/5) (!5/8) (!8/3)

54 items in ratio 2: 3: 4. One ratio part? (6 items) (!9 items) (!12 items) (!18 items)

First share 36 in ratio 3: 5. Total? (96) (!60) (!36) (!57.6)

2.5 L in ratio 2: 3. Second share? (1.5 L) (!1 L) (!0.5 L) (!2 L)

Difference 18 in ratio 5: 2. Larger share? (30) (!12) (!42) (!6)

Map 1: 25000. Four centimetres represents? (1 km) (!100 km) (!0.1 km) (!4 km)

Plan 1: 50. Length 7.5 cm represents? (3.75 m) (!375 m) (!0.15 m) (!37.5 m)

Length factor 2.5. Area factor? (6.25) (!2.5) (!5) (!15.625)

64 L mixture 5: 3; add 8 L to second component. New ratio? (5: 4) (!5: 3) (!5: 11) (!13: 11)

Memory Game

Ratio part	One common abstract unit in a ratio allocation
Part-whole	Comparison of one share with the combined total
Part-part	Comparison of one share with another share
Scale	Correspondence between drawing and actual lengths
Similarity	Same shape with common corresponding-length factor
Area factor	Square of a similar length scale factor

Drag and Drop

£21	One ratio part in the£168 source allocation
£63	First share in that allocation
1 km	Distance represented by 4 cm at 1: 25000
3.75 m	Length represented by 7.5 cm at 1: 50
150 cm²	Area after enlarging 24 cm² by length factor 2.5

Crossword Puzzle

RATIO	Ordered comparison between quantities
TOTAL	Combined value of all shares
DIFFERENCE	Subtraction of the smaller from the larger
SIMILARITY	Same-shape correspondence under a common length factor
SCALE	Drawing-to-actual length relationship
DIMENSION	Independent length direction contributing to area or volume

LearningApps

Optional extra practice. Choose an appropriate activity with your teacher. If the external collection is unavailable, use the native activities and independent tasks in this course.

Cloze Text

Open Tasks

Easy

Identify the datum: Classify each ratio table row as total, share or difference.
Find the first fraction: Distinguish 3/8 of total from 3/5 of the second share.
Read a map scale: Explain 1 cm to 25000 cm before converting.
Explain area dimensions: State why both side lengths scale.

Standard

Solve all five ratio tasks: State unit-part, shares and independent checks.
Solve the map and plan tasks: Use two unit routes and reverse conversion.
Complete the similar-rectangle task: Find new sides, area and area factor.
Repair two first errors: Correct the denominator and dimensional-scale mistakes.

Advanced

Change one mixture component: Find 5: 4 after adding 8 L to the second component.
Check a route in two ways: Find 1.8 km from three map segments.
Reverse an area factor: Derive length ratio 3: 2 from area ratio 9: 4.
Evaluate representations: Explain why screen-pixel measurement is not an exact independent check.

Learning Check

Five ratio results: Identify the supplied quantity and find a common unit-part.
Two drawing scales: Convert same-unit lengths and reverse-check.
Similar rectangle: Distinguish length and area factors.
Mixture transfer: Change only the specified component.
Route and reverse-area transfer: Use dimensions and positive lengths.
Checks and retained evidence: Explain errors and what the work demonstrates.

Worked checks and assessment guidance

£63/£105; 12/18/24 items; 36/60 with total 96; 1/1.5 L; 30/12 with total 42. Unit-parts are 21, 6, 12, .5 and 6. In the last row, the difference 18 corresponds to 3 parts, not the total 7 parts.
4 cm×25000=100000 cm=1 km; 7.5 cm×50=375 cm=3.75 m. Alternative per-centimetre rates.25 km and.5 m agree. Dividing actual centimetre values by scale denominators restores 4 cm and 7.5 cm.
4×6 cm becomes 10×15 cm. Area 24 cm² becomes 150 cm². Factor 150/24=6.25 equals 2.5²; using 2.5 directly for area gives the wrong 60 cm².
64 L at 5: 3 gives 40 L and 24 L. Add 8 L to the second to get 40: 32=5: 4, total 72 L. First-to-total fraction is 40/72=5/9, not 5/4.
Three segments total 3.6 cm at 1: 50000 represent 1.8 km. Similar area ratio 9: 4 gives positive length ratio 3: 2, so 8 cm maps to 12 cm. Square 12/8 to recover 9/4.
Keep five ratio and three scale tasks with units, conservation and relationship checks.168×3/5 uses the wrong whole; 168×3/8=63. Similarity assumptions matter. The modelled map route is not actual journey time or certified measurement skill. Record only observed later retrieval.

Learning evidence

Keep five complete ratio allocations or reconstructions, three scale solutions, independent conservation/relationship/unit checks, mixture and route transfer, and a repaired first error.

OER and Wikipedia

Department for Education:GCSE Mathematics subject content and assessment objectives. Source area:GCSE Mathematics subject content > Scope of study > Ratio proportion and rates of change - Items 2 to 8. Year10 allocation is editorial within the GCSE pathway;Higher-only extensions are explicitly labelled. Original examples,teaching and feedback;Commons images retain their actual attribution,and videos identify original publishers. Reviewed12November2026.

Cuisenaire-Rods.jpg, Celcom at en.wikipedia, CC BY 3.0. Unchanged original; caption and learning activities are original.

Illustration for the area of a rectangle.svg, Stephan Kulla, CC0. Unchanged original; caption and learning activities are original.