Secondary 1 Mathematics Classroom | Chapter 3: Ratio and Proportion | G2/G3

SECONDARY 1 MATHEMATICS CLASSROOM · CHAPTER 3 · RATIO AND PROPORTION · G2/G3

Ratio and Proportion: Find What Stays the Same Before You Scale Anything

In this classroom, you will not begin by cross-multiplying two fractions. You will begin by naming the quantities, fixing their order, finding what one equal part means and identifying which relationship must stay unchanged.

A ratio is a multiplicative comparison. A proportion says that the same comparison is preserved as quantities change. The same ratio can describe many different-sized situations: 2:3 can mean 2 and 3, 20 and 30, or 1.4 and 2.1. Another condition—a total, a difference, one known quantity or a scale factor—is what fixes the actual size.

Classroom rule: labels first, equal parts second, constraint third, scaling fourth, verification last.

The current Secondary One G2 and G3 Mathematics syllabuses include ratio and proportional reasoning, with exact sequencing and depth varying by subject level and school. This classroom teaches the shared ratio core first, then labels direct and inverse proportional modelling carefully where greater depth may depend on the learner’s school sequence.

Official reference: MOE G2 and G3 Mathematics Syllabuses.

Navigate: ratio meaning · equivalent ratios · equal parts · totals and differences · linked ratios · proportion · scale · changing ratios · misconception clinic · guided practice · examination transfer · exit ticket.

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Featured Answer: What Is a Ratio?

A ratio compares quantities multiplicatively. If red:blue = 2:3, the red quantity corresponds to two equal parts while the blue quantity corresponds to three equal parts. The ratio describes relative size, not the actual amount of either quantity.

Therefore all of these satisfy red:blue = 2:3:

• 2 red and 3 blue;
• 10 red and 15 blue;
• 40 red and 60 blue;
• 1.2 kg red material and 1.8 kg blue material.

The scale changes. The comparison remains.

The Simple Classroom Answer

A ratio tells you the shape of the comparison. Another condition tells you its size.

• Order tells you which quantity each term belongs to.
• Equivalent ratios are created by multiplying or dividing every term by the same factor.
• One part is the common scale connecting ratio terms to actual quantities.
• A total corresponds to the sum of the ratio parts.
• A difference corresponds to the difference of the ratio parts.
• A known component corresponds to that component’s number of parts.
• Linked ratios require the shared quantity to be placed on a common scale.
• Proportion preserves a ratio or multiplier under stated conditions.

How to Use This Classroom

1. Write the quantity labels before the ratio numbers.
2. Convert comparable measurements to the same unit before simplifying.
3. Ask what the known amount represents: total, difference, one component or a change.
4. Find one equal part only after the correct constraint is identified.
5. Attempt every Your Turn question before opening its solution.
6. Check both the ratio and every stated condition.
7. Return after a delay and solve a changed version without notes.

1. Start With Labels, Not Bare Numbers

Teacher: Write:

red : blue = 2 : 3.

Ask what the 2 represents and what the 3 represents. The answer should name the quantities: two equal parts of red for every three equal parts of blue.

If the order is reversed, blue:red = 3:2. Ratio order is part of the mathematical statement.

Never write the ratio before you know what each term is labelling.

2. Part-to-Part Is Not Part-to-Whole

A box contains 12 red and 18 blue counters, with no other colours.

• red:blue = 12:18 = 2:3;
• red as a fraction of blue = 12/18 = 2/3;
• red as a fraction of the whole box = 12/30 = 2/5.

All three statements are correct because their reference quantities are different.

3. The Denominator Tells You What You Are Comparing Against

If A:B = 3:7, then A/B = 3/7. But if A and B are the only two categories, A as a fraction of the whole is 3/(3+7) = 3/10.

This distinction becomes extremely important in the next chapter on percentages because “percentage of what?” is the same reference-quantity question.

4. A Ratio Does Not Fix the Actual Amounts

If A:B = 3:5, then:

A = 3k and B = 5k

for some common scale k.

The ratio tells you how many equal parts each quantity receives. The value of k tells you the size of one part.

5. Recognise When There Is Not Enough Information

“Two ribbons have lengths in the ratio 3:5. Find their lengths.”

This is not enough information. Possible pairs include 3 cm and 5 cm, 6 cm and 10 cm, 30 cm and 50 cm and infinitely many other positive pairs.

A total, a difference or one known length would fix the scale.

Your Turn 1

A:B = 4:9. Which of these statements must be true?

1. A = 4 and B = 9.
2. A/B = 4/9.
3. A is 4/13 of the total, if A and B are the only categories.
4. B:A = 9:4.

Answer

Statements 2 and 4 must be true. Statement 3 is also true if A and B are the only categories. Statement 1 need not be true because the scale is unknown.

6. Comparable Quantities Need Common Units

Find the ratio of 1.2 m to 80 cm.

Do not simplify 1.2:80. Convert first:

1.2 m = 120 cm.

Then:

120:80 = 3:2.

The unit conversion comes before ratio simplification.

7. Time Conversion Must Respect Base 60

Find the ratio of 45 minutes to 1.5 hours.

1.5 hours = 90 minutes, not 150 minutes.

Therefore:

45:90 = 1:2.

8. Equivalent Ratios Come From Common Scaling

Multiplying or dividing every term of a ratio by the same positive factor preserves the comparison.

4:6 = 8:12 = 20:30 = 2:3.

Each pair has the same first-to-second multiplier relationship.

9. Adding the Same Amount Does Not Preserve a Ratio

Start with 2:3.

Doubling both terms gives 4:6, still 2:3.

Adding 2 to both terms gives 4:5, which is different.

Multiplicative comparison and additive difference are different structures.

10. Simplest Whole-Number Form Removes Common Factors

Simplify 42:63.

The HCF of 42 and 63 is 21, so divide both terms by 21:

42:63 = 2:3.

This reconnects to Chapter 1: HCF is a natural tool for ratio simplification.

11. Three-Term Ratios Use the Same Rule

Simplify:

18:27:45.

Divide every term by 9:

2:3:5.

Every term belongs to the same linked comparison. Do not simplify only two terms and leave the third unchanged.

12. Ratios Can Contain Decimals

Simplify 0.75:1.2.

Multiply both terms by 100:

75:120 = 5:8.

The common multiplier removes decimal notation without changing the ratio.

13. Ratios Can Contain Fractions

Simplify:

3/4 : 5/8.

Multiply both terms by 8:

6:5.

Do not compare numerators alone. The fractions must first be placed on a common scale.

14. Teacher Model 1: Simplify a Mixed-Number Ratio

Simplify:

1 1/2 : 2 1/4.

Convert to improper fractions:

3/2 : 9/4.

Multiply both by 4:

6:9 = 2:3.

Check: 1.5 is two-thirds of 2.25.

Your Turn 2

1. Simplify 54:81.
2. Simplify 14:21:35.
3. Simplify 0.6:1.5.
4. Simplify 2/3 : 5/6.
5. Find the ratio of 2.4 m to 80 cm.

Answers

2:3. 2:3:5. 2:5. 4:5. 2.4 m=240 cm, so 240:80=3:1.

15. One Ratio Part Is a Scale Unit

If A:B = 5:7, write A = 5k and B = 7k. The value k is the size of one equal ratio part.

Every ordinary ratio word problem asks, directly or indirectly, how k is determined from the information supplied.

16. A Known Total Uses the Sum of the Parts

A and B share $156 in the ratio 5:7.

Total parts = 5 + 7 = 12.

One part = 156 ÷ 12 = 13.

• A = 5×13 = $65;
• B = 7×13 = $91.

Check: 65+91=156 and 65:91=5:7.

17. Teacher Model 2: Three-Way Sharing

Three groups share 240 cards in the ratio 2:3:5.

Total parts = 10.

One part = 240 ÷ 10 = 24.

The shares are:

• 48;
• 72;
• 120.

Check: 48+72+120=240 and the ratio simplifies to 2:3:5.

18. A Known Difference Uses the Difference of the Parts

A:B = 4:7 and B exceeds A by 54.

The difference in ratio parts is 7−4=3.

Three parts = 54, so one part = 18.

• A = 4×18 = 72;
• B = 7×18 = 126.

Check: 126−72=54 and 72:126=4:7.

19. The Wrong Divisor Reveals the Wrong Interpretation

If the known amount is a difference, dividing by the sum of the ratio parts answers the wrong question.

Ask: which collection of equal parts does the known number represent?

20. One Known Component Uses That Component’s Part Count

A:B = 6:5 and B = 45.

Five parts = 45, so one part = 9.

A = 6×9 = 54.

The total is 99.

Do not add 6+5 unless the known amount is the total.

21. Ratios Convert Naturally Into Fractions of the Whole

If red:blue = 3:7 and no other category exists, total parts = 10.

• red is 3/10 of the whole;
• blue is 7/10 of the whole;
• red is 3/7 of blue.

Reference quantity determines the denominator.

22. Convert a Fraction of the Whole Back Into a Ratio

If 3/8 of a collection is red and all remaining objects are blue, then 5/8 are blue.

red:blue = 3:5.

The ratio 3:8 would compare red with the whole, not red with blue.

Your Turn 3

1. Share 180 in the ratio 4:5.
2. A:B = 3:8 and the difference is 65. Find A and B.
3. A:B = 7:4 and B = 36. Find A.
4. Red:blue = 2:3. What fraction of the whole is blue?
5. Two-fifths of a group are girls and the rest are boys. Find girls:boys.

Worked answers

80 and 100. Difference parts=5, so one part=13; A=39 and B=104. Four parts=36, so A=63. Blue is 3/5 of the whole. Girls:boys=2:3.

23. Linked Ratios Share a Quantity That Must Match

Suppose:

A:B = 2:3

and

B:C = 4:5.

The B in both statements represents the same quantity, but it appears as 3 parts in the first ratio and 4 parts in the second. These part scales must be aligned.

24. Match the Shared Quantity Using a Common Multiple

The LCM of 3 and 4 is 12.

Scale A:B by 4:

8:12.

Scale B:C by 3:

12:15.

Therefore:

A:B:C = 8:12:15.

Chapter 1 reappears again: LCM provides the common part count.

25. Check Both Original Ratios After Combining

From 8:12:15:

• A:B = 8:12 = 2:3;
• B:C = 12:15 = 4:5.

A combined ratio is correct only if every original relationship survives.

26. Do Not Join Ratios Merely Because the Same Letter Appears

If one B refers to boys in Class 1A and another B refers to boys in Class 1B, they are not the same quantity. Context determines whether the shared label genuinely represents the same object.

Mathematical convenience cannot create an identity the problem did not provide.

27. Teacher Model 3: Three Linked Quantities

A:B = 3:5 and B:C = 2:7.

Match B. The LCM of 5 and 2 is 10.

• A:B = 6:10;
• B:C = 10:35.

Therefore:

A:B:C = 6:10:35.

Your Turn 4

1. A:B = 4:7 and B:C = 3:5. Find A:B:C.
2. P:Q = 5:6 and Q:R = 8:9. Find P:Q:R.

Answers

Match B at 21: A:B=12:21 and B:C=21:35, so 12:21:35. Match Q at 24: P:Q=20:24 and Q:R=24:27, so 20:24:27.

28. Proportion Means a Comparison Is Preserved

Suppose 3 identical notebooks cost $7.50 under a constant unit-price model. If the same price per notebook applies, the ratio of cost to number of notebooks remains constant.

One notebook costs $2.50.

Twelve notebooks cost:

12 × 2.50 = $30.

The proportional assumption is the constant unit price.

29. The Unitary Method Finds One Unit First

If 7 identical notebooks cost $17.50, one costs:

17.50 ÷ 7 = $2.50.

Then 12 cost $30.

This route exposes what the constant multiplier means.

30. A Scale-Factor Method Can Be Faster

If 4 kg of material costs $18 under a constant price-per-kilogram model, 10 kg is 10/4 = 2.5 times as much material.

Cost = 18 × 2.5 = $45.

The same scale factor must act on both corresponding quantities.

31. A Proportion Equation Is Another Representation

If 6 folders cost $15 and C is the cost of 10 folders under the same constant-price model:

15/6 = C/10.

Both fractions mean dollars per folder. Solving gives C = 25.

The equation is valid because corresponding quantities are compared in the same order.

32. Cross-Multiplication Is a Consequence of Equality

If a/b = c/d and b,d are non-zero, multiplying both sides by bd gives:

ad = bc.

This is why cross-products agree. The method is not created by drawing diagonal arrows. It comes from preserving equality.

33. Direct Proportion Has a Constant Multiplier

In a directly proportional relationship:

y = kx.

The constant k is the multiplier from x to y.

Doubling x doubles y. Tripling x triples y.

34. A Direct-Proportion Table Has a Constant Quotient

x	y	y/x
2	5	2.5
4	10	2.5
7	17.5	2.5

The stable quotient shows y = 2.5x.

35. A Fixed Charge Breaks Direct Proportion

Suppose a service costs $4 plus $2 per unit:

C = 4 + 2n.

Doubling n does not double the total cost because the fixed $4 does not scale.

There is a constant variable rate of $2 per unit, but the total cost is not directly proportional to n.

36. Use the Zero-Input Check for Direct Proportion

For y = kx, if x = 0 then y = 0.

Therefore a non-zero fixed starting amount is evidence that the total relationship is not direct proportion.

Use this as a diagnostic, together with the constant quotient test.

37. Teacher Model 4: Direct Proportion

y is directly proportional to x. When x = 6, y = 15.

Since y = kx:

15 = 6k, so k = 2.5.

When x = 14:

y = 2.5×14 = 35.

38. Extension: Inverse Proportion Preserves a Product

Where your school sequence includes inverse proportion, do not define it merely as “one goes up while the other goes down”. The defining structure is:

xy = k

or equivalently y = k/x.

For a fixed 120 km journey at constant speed, speed × time = 120. If speed doubles, time halves under that model.

Keep this as extension if it has not yet been taught in your course.

Your Turn 5

1. Five identical pens cost $12.50 under a constant price model. Find the cost of 14 pens.
2. y is directly proportional to x. When x=4, y=18. Find y when x=10.
3. Is C=5+3n directly proportional to n? Explain.

Worked answers

One pen costs $2.50, so 14 cost $35. For direct proportion, k=18/4=4.5, so y=45. C=5+3n is not directly proportional to n because of the fixed 5; C/n is not constant and C is not zero when n=0.

39. Scale Is a Ratio Between Representation and Reality

A map scale of 1:25,000 means one unit of length on the map represents 25,000 of the same units on the ground.

If a route measures 3.6 cm on the map:

3.6×25,000 = 90,000 cm = 900 m.

Apply the scale with matching units, then convert to a convenient final unit.

40. Reverse Scale Questions Divide by the Scale Factor

A ground distance of 1.5 km is shown on a 1:25,000 map.

Convert:

1.5 km = 150,000 cm.

Map distance:

150,000 ÷ 25,000 = 6 cm.

Predict direction: the map length should be much smaller than the ground distance.

41. Scale Drawings Require Corresponding Lengths

When a figure is enlarged or reduced by a scale factor, all corresponding lengths must scale by the same factor.

This prepares for later similarity. Do not assume area scales by the same linear factor; area and volume relationships belong to later geometry development.

42. Teacher Model 5: Scale Diagram

A classroom floor plan uses scale 1:100. A wall is 7.4 cm on the plan.

Actual length:

7.4×100 = 740 cm = 7.4 m.

The numerical value happens to return to 7.4 after changing centimetres to metres, but the unit conversion is essential to the meaning.

43. A Changing-Ratio Problem Needs a Bridge Between Two States

A before-and-after ratio cannot be solved unless something links the states.

• one quantity may remain unchanged;
• the total may remain unchanged;
• the difference may remain unchanged;
• a stated amount may be added, removed or transferred.

Identify the bridge before comparing the ratio parts.

44. Do Not Assume One Part Has the Same Size Before and After

Ratio parts are relative units inside each state. If the state changes, the actual size of one ratio part may also change.

Parts can be matched only when an unchanged quantity or another valid relationship justifies the match.

45. Teacher Model 6: One Quantity Stays Unchanged

Initially red:blue = 2:3. Ten red counters are added, blue is unchanged, and the new ratio is 4:3.

Blue is 3 parts both before and after and is unchanged, so the part size matches in this case.

Red rises from 2 parts to 4 parts, an increase of 2 parts. Those 2 parts equal 10 counters.

One part = 5.

Original counts:

• red = 10;
• blue = 15.

After adding 10 red, the ratio is 20:15 = 4:3.

46. Equal Additions Preserve the Difference

If the same amount is added to both quantities, their difference remains unchanged, but their ratio usually changes.

For example, 2:5 becomes 4:7 after adding 2 to both displayed terms. The difference remains 3; the ratio changes.

47. Teacher Model 7: Equal Additions

Red:blue = 2:5. Twelve counters are added to each colour. The new ratio is 4:7. Find the original counts.

Let the original counts be 2k and 5k.

After addition:

(2k+12):(5k+12) = 4:7.

So:

7(2k+12) = 4(5k+12).

14k+84 = 20k+48.

k = 6.

Original counts are 12 and 30. After adding 12, they become 24 and 42, which simplifies to 4:7.

48. Transfers Preserve the Total Inside the Group

A:B = 3:7 and the total is 80. A receives 12 from B.

Initially:

• A = 24;
• B = 56.

After transfer:

• A = 36;
• B = 44.

New ratio = 36:44 = 9:11.

The total stays 80 because nothing crosses the boundary of the two-person system.

49. Addition, Removal and Transfer Have Different Conservation Rules

Change	What may stay unchanged?
same amount added to both	difference
amount transferred from A to B	combined total
amount removed from the system	total decreases
one category unchanged	that actual quantity

State the invariant before writing equations.

50. Resource-Limit Problems Add a Feasibility Constraint

A mixture uses powder:water = 3:5. If only 180 g of powder is available, three parts equal 180 g, so one part is 60 g.

Water required = 5×60 = 300 g.

Total mixture = 480 g.

You cannot make 800 g at the same ratio using only 180 g of powder. Hitting the total by adding extra water would change the composition.

51. Whole Objects Can Restrict the Scale Factor

If one pack requires 3 red pieces and 5 blue pieces, and you have 20 red and 37 blue pieces:

• red permits 20÷3 = 6 complete packs with remainder;
• blue permits 37÷5 = 7 complete packs with remainder.

Only six complete packs can be made. The limiting resource controls the whole-number answer.

52. The Method-Selection Table

Information supplied	What it controls	First move
ratio + total	sum of parts	divide total by sum of ratio terms
ratio + difference	difference of parts	divide difference by difference of ratio terms
ratio + one component	that component’s part count	divide known component by its ratio term
two linked ratios	shared quantity	match shared term using a common multiple
constant unit model	constant quotient or multiplier	find one unit or the scale factor
before/after ratio	stated invariant or change	identify what remains fixed or write an equation

53. Misconception Clinic: Reverse the Ratio Order

If red:blue = 2:3, blue:red = 3:2. Label the quantities before simplifying so a correct calculation is not attached to the wrong order.

54. Misconception Clinic: Treat 2:3 as Two-Thirds of the Whole

Two-thirds compares the first quantity with the second. If there are only two categories, the first quantity is 2/5 of the whole.

55. Misconception Clinic: Simplify Different Units Directly

Convert 1.2 m and 80 cm into the same unit before writing 120:80. Ratio terms cannot be meaningfully simplified while the measurement units disagree.

56. Misconception Clinic: Add the Ratio Parts for Every Problem

Add the parts only when the known amount is the total. Use the difference of the parts for a known difference and the relevant term for one known component.

57. Misconception Clinic: Equivalent Ratios Come From Equal Addition

Equivalent ratios require common multiplication or division. Equal addition preserves a difference, not generally a ratio.

58. Misconception Clinic: Join Linked Ratios Without Matching the Shared Quantity

A:B=2:3 and B:C=4:5 do not become 2:3:5. The B part counts 3 and 4 belong to different ratio scales until they are matched.

59. Misconception Clinic: Cross-Multiply Before Defining the Fractions

First make sure each fraction compares corresponding quantities in the same order. Cross-products only preserve a proportion that was formed correctly.

60. Misconception Clinic: Both Quantities Increase, So They Are Directly Proportional

Many non-proportional relationships increase. Direct proportion requires a constant multiplier y/x and passes through zero in the model.

61. Misconception Clinic: Equal Additions Preserve the Ratio

Adding the same amount to both quantities preserves their difference. It generally changes their ratio because the relative sizes change.

62. Misconception Clinic: A Transfer Changes the Combined Total

If an amount moves from A to B and stays inside the two-category system, the combined total is unchanged. Add to one side and subtract from the other.

63. Misconception Clinic: A Fractional Number of Objects Is Automatically Acceptable

Continuous quantities such as mass can take fractional values. Counts of indivisible objects usually cannot. Return the mathematical scale factor to the physical context before accepting the result.

64. Guided Practice Set A: Meaning and Simplification

1. Simplify 48:72.
2. Simplify 18:30:42.
3. Simplify 0.8:1.4.
4. Simplify 5/6 : 3/4.
5. Find the ratio of 1.8 m to 60 cm.

Solutions

2:3. 3:5:7. 4:7. Multiply both fractions by 12 to get 10:9. 1.8 m=180 cm, so 180:60=3:1.

65. Guided Practice Set B: Total

A and B share 198 tokens in the ratio 4:5.

Worked solution

Nine parts=198, so one part=22. A=88 and B=110. Check total 198 and ratio 4:5.

66. Guided Practice Set C: Difference

A:B = 5:9 and B−A = 64. Find A and B.

Worked solution

Difference parts=4. One part=16. A=80 and B=144.

67. Guided Practice Set D: Known Component

Flour:sugar = 7:3. Sugar mass is 150 g. Find the flour mass and total mixture mass.

Worked solution

Three parts=150 g, so one part=50 g. Flour=350 g. Total=500 g.

68. Guided Practice Set E: Fractions of a Whole

1. Red:blue=4:7. What fraction of the whole is red?
2. Three-eighths of a group are in A and the rest are in B. Find A:B.
3. A:B=2:5. What is A as a fraction of B?

Solutions

4/11. A:B=3:5. A/B=2/5.

69. Guided Practice Set F: Linked Ratios

1. A:B=3:4 and B:C=6:5. Find A:B:C.
2. P:Q=2:7 and Q:R=3:4. Find P:Q:R.

Solutions

Match B at 12: A:B=9:12 and B:C=12:10, so 9:12:10. Match Q at 21: P:Q=6:21 and Q:R=21:28, so 6:21:28.

70. Guided Practice Set G: Direct Proportion

1. Eight identical items cost $28 under a constant-price model. Find the cost of 15.
2. y is directly proportional to x. When x=5, y=12. Find y when x=18.
3. Explain why y=3x+4 is not directly proportional to x.

Solutions

One item=$3.50, so 15 cost $52.50. k=12/5=2.4, so y=43.2. y=3x+4 has a fixed 4; y/x is not constant and y is not zero when x=0.

71. Guided Practice Set H: Scale

A map uses scale 1:50,000.

1. A route is 4.2 cm on the map. Find its actual distance in kilometres.
2. An actual distance is 3 km. Find the map distance in centimetres.

Worked solution

4.2×50,000=210,000 cm=2.1 km. 3 km=300,000 cm, and 300,000÷50,000=6 cm.

72. Guided Practice Set I: Changing Ratio

Initially A:B=3:5. Ten is added to A only, B stays unchanged, and the new ratio is 5:5. Find the original values.

Worked solution

B remains 5 parts in both states, so the part size matches. A rises from 3 parts to 5 parts, an increase of 2 parts equal to 10. One part=5. Original A=15 and B=25.

73. Guided Practice Set J: Transfer

A:B=2:3 and the total is 100. A transfers 10 to B. Find the new ratio.

Worked solution

Initially A=40 and B=60. After transfer A=30, B=70. New ratio=3:7.

74. Challenge Practice: Missing Information

A:B=4:9. A student claims A=20 and B=45. Is the claim possible? Is it uniquely determined by the ratio?

Answer

It is possible because 20:45 simplifies to 4:9. It is not uniquely determined; many other pairs have the same ratio. Another condition is needed to fix the scale.

75. Challenge Practice: Same Ratio, Different Totals

Group X has red:blue=2:3 with 40 objects. Group Y has red:blue=2:3 with 100 objects. Find the counts and explain what is the same and what is different.

Worked solution

Group X has 16 red and 24 blue. Group Y has 40 red and 60 blue. The multiplicative comparison is the same; the scale and total quantities are different.

76. Challenge Practice: Ratio and Algebra

A:B=2:5. After 18 is added to each, the ratio becomes 4:7. Find the original values.

Worked solution

Let values be 2k and 5k. Then (2k+18)/(5k+18)=4/7. So 14k+126=20k+72, giving 54=6k and k=9. Original values are 18 and 45. After addition they are 36 and 63, ratio 4:7.

77. Challenge Practice: Limited Resources

A kit requires red:blue pieces=4:7. There are 38 red and 70 blue pieces. What is the greatest number of complete kits that can be made, and what remains?

Worked solution

Red allows floor(38/4)=9 kits. Blue allows floor(70/7)=10 kits. Therefore 9 complete kits can be made. They use 36 red and 63 blue, leaving 2 red and 7 blue.

78. Challenge Practice: Scale and Unit Conversion

A model is built at scale 1:75. A real wall is 9 m long. Find the model length in centimetres.

Worked solution

9 m=900 cm. Model length=900÷75=12 cm.

79. Examination Method: Write the Labels Beside the Ratio

Use:

boys:girls = 3:5

instead of a bare 3:5 while the question is being interpreted. This prevents silent reversal.

80. Examination Method: Write What the Known Amount Represents

• “9 parts = total 180”;
• “4 parts = difference 64”;
• “3 parts = sugar mass 150 g”.

This one line exposes the model before arithmetic begins.

81. Examination Method: Convert Units Before Simplifying

Write the conversion line explicitly. A correct ratio written from incompatible numerical units is still wrong.

82. Examination Method: Check Two Conditions, Not One

For ratio with total, check both total and simplified ratio.

For ratio with difference, check both difference and ratio.

For changing ratios, check original ratio, stated change and final ratio.

83. Examination Method: Predict Direction Before Scaling

A map distance should be much smaller than the ground distance when the scale is 1:50,000. A larger order quantity should cost more under a positive constant unit-price model.

Use the context to catch a reversed multiplier.

84. Examination Method: Do Not Claim Proportion Without a Constant Relationship

If a fixed fee, discount threshold, changing rate or other condition exists, the whole relationship may not be directly proportional. State the model being used.

85. Examination Method: Use Algebra When Parts Cannot Be Matched Safely

For changing-ratio questions, let original quantities be ak and bk when no obvious unchanged part allows a direct comparison. Write the new condition as an equation.

86. Oral Classroom Check

1. What does A:B=2:3 actually say?
2. Why is A not necessarily 2?
3. What is the difference between A/B and A as a fraction of the whole?
4. Why must units match before simplifying a measurement ratio?
5. Why does equal multiplication preserve a ratio but equal addition usually not?
6. When do you add the ratio parts?
7. When do you subtract the ratio parts?
8. How do you combine linked ratios?
9. What proves direct proportion?
10. What stays unchanged in a transfer between two categories?

The student should answer using a small numerical example. If every explanation turns into “cross multiply”, return to the equal-part model.

87. Exit Ticket

1. Simplify 45:75.
2. Find the ratio of 1.5 m to 60 cm.
3. Share 144 in the ratio 5:7.
4. A:B=3:8 and B−A=50. Find A and B.
5. A:B=2:3 and B:C=4:7. Find A:B:C.
6. Six identical items cost $21 under a constant-price model. Find the cost of 10.
7. A map uses scale 1:25,000. A route is 8 cm on the map. Find the ground distance in kilometres.
8. A:B=3:5 with total 64. A transfers 8 to B. Find the new ratio.

Exit-ticket solution

45:75=3:5. 1.5 m=150 cm, so 150:60=5:2. Total parts=12, one part=12, so shares 60 and 84. Difference parts=5, one part=10, so A=30 and B=80. Match B at 12: A:B=8:12 and B:C=12:21, so 8:12:21. One item=$3.50, so 10 cost $35. 8×25,000=200,000 cm=2 km. Initially A=24, B=40; after transfer A=16, B=48, so 1:3.

88. Homework: Retrieval, Variation and Transfer

Layer 1 — Retrieval

• Define ratio in one sentence.
• Explain part-to-part versus part-to-whole.
• Write the common-unit rule.
• Write the total, difference and known-component routines.
• Explain how linked ratios are matched.
• State two tests for direct proportion.

Layer 2 — Variation

• Four simplification questions, including decimals or fractions.
• Two ratio-and-total questions.
• Two ratio-and-difference questions.
• One known-component question.
• Two linked-ratio questions.
• Two direct-proportion questions.
• One scale question.
• One changing-ratio question.

Layer 3 — Transfer

Create three problems using the same ratio 3:5: one with a total, one with a difference and one with one known component. Solve all three and explain why the first division step is different in each case.

89. The Seven-Day Return Cycle

1. Day 0: complete models and guided practice.
2. Day 1: solve one total, one difference and one component problem.
3. Day 3: combine two linked ratios and solve one scale problem without notes.
4. Day 7: complete a changing-ratio problem and repeat the exit ticket with changed values.

90. A 60-Minute Teaching Lesson

1. 5 minutes: ratio-language retrieval.
2. 10 minutes: units and equivalent ratios.
3. 15 minutes: equal parts from totals, differences and known components.
4. 10 minutes: linked ratios.
5. 10 minutes: direct proportion and scale.
6. 5 minutes: one changed-ratio question.
7. 5 minutes: exit ticket.

91. A 90-Minute Teaching Lesson

1. 10 minutes: ratio diagnostic and labels.
2. 15 minutes: equivalent ratios, units and fractions.
3. 20 minutes: equal-part modelling with three constraints.
4. 10 minutes: linked ratios.
5. 15 minutes: proportion and scale.
6. 10 minutes: changing-ratio modelling.
7. 5 minutes: oral explanation.
8. 5 minutes: exit ticket and return date.

92. The Full Chapter Routine

For ordinary ratio questions:

labels → common units → ratio parts → identify constraint → find one part → scale → verify.

For linked ratios:

identify shared quantity → match its part count → combine → recheck both source ratios.

For direct proportion:

state constant relationship → find unit value or multiplier → scale both quantities consistently → verify.

For changing ratios:

before state → stated change → invariant → after state → equation or matched parts → check all conditions.

93. Why This Chapter Matters Beyond Chapter 3

Ratio is one of the main bridges from arithmetic into algebra. It teaches students to preserve a relationship while scale changes. Percentages are ratios per hundred. Rates compare different kinds of quantities. Similarity uses equal corresponding ratios. Probability uses favourable outcomes relative to a sample space. Trigonometry uses side ratios. Direct proportion becomes a function and a graph.

The deeper habit is this: identify the invariant before calculating. When the invariant is clear, the scale can change without losing the mathematical relationship.

94. Connect Back to Chapters 1 and 2

Chapter 1 supplies HCF and LCM for simplifying and linking ratios. Chapter 2 supplies fractions, decimals, signed arithmetic, units and estimation. Chapter 3 combines them into multiplicative comparison.

• Secondary 1 Mathematics Classroom | Chapter 1: Number Structure
• Secondary 1 Mathematics Classroom | Chapter 2: Directed Numbers, Number Lines, Approximation and Estimation

95. Ready for Chapter 4?

You are ready to move on when you can do all of the following without prompts:

• read ratio order correctly;
• distinguish part-to-part from part-to-whole;
• convert measurements to common units before simplifying;
• simplify ratios containing whole numbers, decimals and fractions;
• use the sum of parts for a known total;
• use the difference of parts for a known difference;
• use a component’s own part count when that quantity is known;
• convert ratios into fractions of the whole;
• combine linked ratios by matching the shared quantity;
• solve simple proportional models by unitary or scale-factor methods;
• recognise when a fixed amount breaks direct proportion;
• apply and reverse a map or drawing scale;
• identify what stays unchanged in a changing-ratio problem;
• check both the ratio and every stated condition.

If one item is weak, return to the smallest section that owns it and complete a changed example. If all are stable, continue to Percentages, where ratio reasoning is expressed on a common scale of one hundred and the reference quantity becomes the central control.

Continue the Secondary 1 Mathematics Learning Route

• Secondary 1 Mathematics Learning Guide | Ratio and Proportion
• Secondary 1 Mathematics Learning Guide | Percentages and Reverse Percentages
• Secondary 1 Mathematics Learning Guide | Rate, Speed and Unit Conversion
• Secondary 1 Mathematics Learning Guide | Word Problems and Mathematical Representation
• Secondary Mathematics Sengkang | S1–S4 Capability Map