The same amount for each one
Pens cost $3 each. 1 pen costs $3, 2 pens cost $6 and 3 pens cost $9. Every extra pen adds the same $3 to the cost, because every pen costs the same.
The cost of 1, 2 and 3 pens. Each bar is $3 taller than the one before.
Double one and the other doubles
2 pens cost $6. Double the number of pens to 4, and the cost doubles too: 4 pens cost $12. Multiply the number of pens by 3 and the cost is multiplied by 3: 6 pens cost 3 × $6 = $18. Halve the number of pens and the cost halves.
When two quantities change like this, so that multiplying one by any number multiplies the other by the same number, the two quantities are in direct proportion. Here the cost is in direct proportion to the number of pens.
Another way to see it: divide each cost by its number of pens, and you always get the same answer. $6 ÷ 2 = $3, $12 ÷ 4 = $3, $18 ÷ 6 = $3. The ratio of cost to pens stays fixed at 3 : 1.
2 pens cost $6 and 4 pens cost $12: twice as many pens, twice the cost.
A straight line that starts at zero
Plot the number of pens against the cost and the points lie on a straight line. The line starts at 0, because 0 pens cost $0.
The cost of 0 to 4 pens: a straight line from $0, rising $3 for every pen.
Growing steadily is not enough
A taxi charges $5 to start the ride and then $2 for every kilometer. 1 km costs $7, 2 km cost $9 and 4 km cost $13. Each extra kilometer adds the same $2, but doubling the distance from 2 km to 4 km does not double the cost: $9 doubled is $18, not $13. The cost is not in direct proportion to the distance, because of the $5 you pay before the taxi has moved.
On a graph the line is still straight, but it starts at $5, not at 0. Two quantities in direct proportion always start together at 0.
The taxi fare for 0 to 4 km: a straight line that starts at $5, so the fare is not in direct proportion to the distance.
The usual mistakes
Adding the extra number of items instead of scaling. If 2 pens cost $6, then 4 pens do not cost $6 + 2 = $8. Each pen costs $3, so 4 pens cost 4 × $3 = $12.
Adding the number of items to the price of one. 4 pens at $3 each is 4 × 3 = $12, not 4 + 3 = $7.
Pricing one item too few. 4 pens cost 4 × $3 = $12. $9 is the cost of only 3 of them.
Distance at a steady speed
When two cyclists ride for the same length of time, the distance each one covers is in direct proportion to their speed: twice the speed covers twice the distance. The next problem uses this to find how far apart two cyclists are after they set off together from the same place.
Worked example: Constant Time (Direct Proportion of Distance to Speed)
Question Keith and Leonard started cycling at the same time from the same entrance along a park connector in the same direction. Keith cycled at 24 km/h and Leonard cycled at 16 km/h. (a) How far apart were they after 45 minutes? (b) How long after they set off were they 10 km apart?
1.Ratio of Speed (Keith : Leonard) = 24 : 16 = 3 : 2.
Same start, same time on the road, so distances are in the ratio of the speeds: 24:16 = 3:2. 2.For any given duration, Keith travels 3 units while Leonard travels 2 units (Difference = 1 unit).
For any duration, Keith covers 3 units to Leonard's 2. The gap is always 1 unit. 3.In 1 hour, Keith travels 24 km (3u) and Leonard travels 16 km (2u) ⟹ u = 8 km.
In one hour the units are 24 km and 16 km, so u = 8 km. Slide the time: the gap grows in step. 4.(a) For 45 min (34 of an hour): Gap = 34 × 8 km = 6 km.
(a) 45 min is 34 h: gap = 34 × 8 = 6 km. 5.(b) A 10 km gap is 10 ÷ 8 = 114 hours' worth of gap: 114 hours is 1 h 15 min.
(b) The gap grows 8 km an hour: 10 km takes 10 ÷ 8 = 114 h, which is 1 h 15 min.
Answer: (a) 6 km; (b) 1 hour 15 minutes
Common mistakes
- Calculating the distance Keith traveled and forgetting to subtract Leonard's distance.
- Reading 1.25 hours as 1 h 25 min instead of 1 h 15 min.