Ratio, percentage, rate and speed are not separate tricks. They are different ways of describing how quantities compare and change. The difficult part is often not arithmetic. It is deciding what the reference quantity is, whether the relationship is additive or multiplicative, whether a rate is average or instantaneous, and whether a percentage refers to the original value or the final value.
This Secondary 3 Mathematics Learning Guide develops ratio, direct and inverse proportion, percentage comparison, percentage increase and decrease, reverse percentages, average rate, average speed and unit conversion through real contexts. It belongs to the Secondary Mathematics Hub.
The official 2027 SEC G3 Mathematics syllabus K310 includes ratios involving rational numbers, map scales, direct and inverse proportion, percentage comparison and reverse percentages, average rate and average speed, and unit conversion. This guide is G3-oriented; schools may distribute these ideas across different year levels.
Use this route: diagnostic · ratio structure · direct and inverse proportion · percentage control · rate and speed · practice · answers. All examples are original teaching examples.
Comparison Comes Before Calculation
A ratio compares quantities of the same kind or quantities that have been made comparable. A rate compares quantities with different units. A percentage compares a part or change against a reference quantity scaled to 100. A speed is a rate relating distance and time.
The most useful first question is therefore: what is being compared with what? If that relationship is wrong, a correct calculator sequence can still answer the wrong problem.
A Six-Question Diagnostic
Simplify 18:24. Divide $84 in the ratio 2:5. If y is directly proportional to x and y=12 when x=3, find y when x=7. If y is inversely proportional to x and y=15 when x=4, find y when x=10. Increase 250 by 12%. Finally, convert 72 km/h to m/s.
The answers are 3:4; $24 and $60; 28; 6; 280; and 20 m/s. Each error points to a different capability: ratio simplification, part-whole allocation, proportional constant, invariant product, percentage multiplier or unit conversion.
Ratios Describe Relative Size
The ratio 3:5 says that for every 3 equal parts of the first quantity, there are 5 equal parts of the second. It does not say the quantities are 3 and 5. They could be 6 and 10, 15 and 25, or 1.5 and 2.5.
Equivalent ratios are created by multiplying or dividing every term by the same nonzero factor. Thus 18:24 simplifies to 3:4 by dividing both terms by 6.
Part-to-Part and Part-to-Whole Ratios
If red:blue = 2:3, then the ratio compares two parts. The total number of equal parts is 5. Therefore red as a fraction of the total is 2/5, while blue is 3/5.
Confusing part-to-part with part-to-whole is a common source of errors in percentages and probability. Always identify whether the denominator should be one category or the entire total.
Worked Example 1: Divide in a Ratio
$126 is divided between A and B in the ratio 4:5. Find each share. The total number of parts is 9. One part is 126/9 = 14.
A receives 4×14 = $56. B receives 5×14 = $70. Check: 56+70=126 and 56:70 simplifies to 4:5.
Ratios With Rational Numbers
A ratio such as 1.5:2.25 can be simplified by removing decimals. Multiplying both terms by 100 gives 150:225, which simplifies to 2:3.
Fractions can be handled similarly. For 2/3 : 5/6, multiply both terms by the lowest common denominator 6 to obtain 4:5.
Map Scale Is a Ratio Between Drawing and Reality
A scale of 1:25,000 means 1 unit on the map represents 25,000 of the same units in reality. If a map distance is 4.8 cm, the real distance is 4.8×25,000 = 120,000 cm = 1.2 km.
Area scale is different. A length scale factor k produces area factor k². This is why a 1:25,000 length scale does not mean 1 cm² represents 25,000 cm². The area factor is 25,000².
Direct Proportion Preserves a Constant Ratio
If y is directly proportional to x, then y/x is constant. We write y=kx, where k is the constant of proportionality.
If doubling x doubles y, tripling x triples y and zero input gives zero output, a direct-proportion model may be appropriate. The graph of y=kx is a straight line through the origin.
Worked Example 2: Direct Proportion
The cost C dollars is directly proportional to mass m kg. When 3 kg costs $18, find the cost of 7.5 kg.
C=km. From 18=3k, k=6 dollars per kilogram. Therefore C=6(7.5)=$45.
The constant k carries units: dollars per kilogram. This reminds us what the proportionality means physically.
Inverse Proportion Preserves a Constant Product
If y is inversely proportional to x, then xy is constant. We write y=k/x.
If x doubles, y halves. If x is multiplied by 5, y is divided by 5. This is a multiplicative relationship, not a subtraction pattern.
Worked Example 3: Inverse Proportion
For a fixed journey, time t hours is inversely proportional to average speed v km/h. At 60 km/h, the journey takes 3 hours. Find the time at 90 km/h.
tv=k. From 3×60=180, k=180. Hence t=180/90=2 hours.
The invariant product is distance: speed×time = 180 km. This gives the inverse-proportion model a physical interpretation.
Direct or Inverse? Test the Invariant
Do not guess from wording. For direct proportion, test whether y/x is constant. For inverse proportion, test whether xy is constant. A relationship can increase without being directly proportional, and it can decrease without being inversely proportional.
For example, y=2x+5 increases as x increases but is not directly proportional because the graph does not pass through the origin and y/x is not constant.
A Percentage Always Has a Reference Quantity
“20%” alone is incomplete. Twenty percent of what? Percentage calculations are reliable only when the reference quantity is explicit.
To express A as a percentage of B, calculate A/B×100%. The denominator B is the reference quantity. Reversing A and B changes the meaning.
Worked Example 4: Percentage Comparison
A school’s enrolment rises from 800 to 920. Find the percentage increase. The increase is 120. The original value 800 is the reference.
Percentage increase = 120/800×100% = 15%.
Using 920 as the denominator would answer a different question: the increase as a percentage of the final value.
Percentage Multipliers
Increasing a quantity by r% multiplies it by 1+r/100. Decreasing by r% multiplies it by 1−r/100.
A 12% increase uses multiplier 1.12. A 12% decrease uses 0.88. This multiplier view is especially useful for repeated percentage changes and reverse percentages.
Worked Example 5: Successive Percentage Changes
A value of $500 increases by 20% and then decreases by 20%. Find the final value.
After the increase: 500×1.20 = 600. After the decrease: 600×0.80 = 480.
The changes do not cancel because the second 20% uses a different reference quantity. The net multiplier is 1.20×0.80=0.96, a 4% overall decrease.
Reverse Percentage Means Undo the Multiplier
If a final value of 360 is obtained after a 20% increase, then original×1.20=360. Therefore original = 360/1.20 = 300.
Do not subtract 20% of the final value. The original quantity is the base to which the 20% increase was applied.
Worked Example 6: Reverse Percentage
After a 15% discount, an item costs $102. Find the original price. A 15% discount leaves 85% of the original, so final = 0.85×original.
Original = 102/0.85 = $120.
Percentages Greater Than 100%
A value can be more than 100% of another. If A is 150 and B is 60, then A as a percentage of B is 150/60×100% = 250%.
This simply means A is two and a half times B. Percentages above 100% are not errors when the numerator exceeds the reference quantity.
Rate Connects Different Units
A rate such as $4.80 per kilogram, 15 litres per minute or 72 km/h relates two quantities with different units. Units are part of the answer and can reveal whether the calculation was set up correctly.
If 45 litres flow in 3 minutes, average flow rate = 45/3 = 15 L/min. If the same flow rate continues for 8 minutes, volume = 15×8 = 120 L.
Average Speed Is Total Distance Divided by Total Time
Average speed is not generally the arithmetic mean of two speeds. The correct definition is total distance / total time.
Worked Example 7: Two Journey Speeds
A car travels 120 km at 60 km/h and then 120 km at 80 km/h. Find the average speed for the whole journey.
First segment time = 120/60 = 2 h. Second segment time = 120/80 = 1.5 h. Total distance = 240 km and total time = 3.5 h.
Average speed = 240/3.5 ≈ 68.6 km/h. It is not 70 km/h because the car spends more time travelling at the lower speed.
Equal Times Give a Different Average
If the car travels for one hour at 60 km/h and one hour at 80 km/h, the total distance is 140 km in 2 hours, so average speed is exactly 70 km/h.
This contrast shows why the structure of the journey matters. Equal distances and equal times lead to different averaging behaviour.
Unit Conversion: Build a Chain That Cancels Units
To convert km/h to m/s, multiply by 1000 m per km and divide by 3600 s per hour. Therefore 1 km/h = 5/18 m/s.
Thus 72 km/h = 72×5/18 = 20 m/s. Reversing the conversion gives m/s to km/h by multiplying by 18/5.
Worked Example 8: Mixed Units
A runner covers 1500 m in 6 minutes. Find average speed in m/s and km/h.
Six minutes = 360 seconds. Speed = 1500/360 ≈ 4.17 m/s. Multiply by 18/5 to get 15.0 km/h.
The two numerical values differ because the units differ. Comparing 4.17 and 15.0 without units would be meaningless.
Worked Example 9: Rate in a Real Context
A machine packs 360 items in 24 minutes at a constant average rate. How many items would it pack in 1.5 hours at the same rate?
Rate = 360/24 = 15 items/min. One and a half hours = 90 minutes. Expected output = 15×90 = 1350 items.
The model assumes the average rate remains constant. Real machines may pause or vary, so the mathematical answer belongs to the stated model.
Four Common Errors
Using the final value as the percentage base: repair by naming the reference quantity before calculating.
Treating direct and inverse proportion as vocabulary: repair by testing constant ratio y/x or constant product xy.
Averaging speeds arithmetically without checking time: repair by returning to total distance divided by total time.
Converting only the number, not the unit structure: repair by writing conversion factors so unwanted units cancel.
Independent Practice
1. Simplify 2.4:3.6.
2. Divide $195 in the ratio 4:9.
3. A map scale is 1:50,000. Find the real distance represented by 7.2 cm.
4. y is directly proportional to x. If y=18 when x=6, find y when x=11.
5. y is inversely proportional to x. If y=8 when x=15, find y when x=12.
6. Increase 640 by 7.5%.
7. Decrease 480 by 18%.
8. After a 25% increase, a value is 500. Find the original.
9. After a 30% discount, a price is $84. Find the original price.
10. A value increases by 10% then by another 10%. Find the overall percentage increase.
11. Convert 90 km/h to m/s.
12. Convert 12 m/s to km/h.
13. A cyclist travels 30 km in 45 min and then 20 km in 30 min. Find average speed for the whole journey.
14. A tank fills 96 L in 8 min. At the same average rate, how long will 210 L take?
15. Explain why increasing a price by 20% and then decreasing the new price by 20% does not restore the original price.
Explained Answers
1. 2.4:3.6 = 24:36 = 2:3.
2. Total parts=13; one part=$15. Shares are $60 and $135.
3. 7.2×50,000=360,000 cm=3.6 km.
4. y=kx, so k=3 and y=33.
5. xy=120, so y=120/12=10.
6. 640×1.075=688.
7. 480×0.82=393.6.
8. Original=500/1.25=400.
9. Original=84/0.70=120.
10. Net multiplier=1.1×1.1=1.21, so overall increase=21%.
11. 90×5/18=25 m/s.
12. 12×18/5=43.2 km/h.
13. Total distance=50 km. Total time=75 min=1.25 h. Average speed=40 km/h.
14. Rate=12 L/min. Time=210/12=17.5 min.
15. The 20% decrease is taken from the larger new price. Multipliers give 1.2×0.8=0.96, leaving 96% of the original.
A Reliable Checking Routine
For ratio and proportion, check the invariant. For percentage, check the base. For rate, check the units. For speed, reconstruct total distance and total time. For reverse percentage, multiply your recovered original by the stated percentage multiplier and confirm that it returns to the final value.
Teacher and Parent Prompts
Ask “What is the reference quantity?” before a percentage calculation. Ask “What stays constant?” before a proportion problem. Ask “What are the units of this rate?” before accepting a numerical answer. These questions reveal whether the relationship has been understood or merely calculated.
Continue the Secondary 3 Learning Route
Continue with Properties of Circles, Chords and Tangents, Arc Length, Sector Area, Radians and Composite Mensuration, and Congruence, Bisectors and Geometrical Construction.
These topics become secure when the learner can identify the comparison, preserve the correct reference quantity, and keep units and proportional structure intact. Return to the Secondary Mathematics Hub.
