A conversion graph is a straight line that turns one unit into another, such as kilograms into pounds or Singapore dollars into ringgit. To read it, find the given value on one axis, trace across or up to the line, then trace across or down to the other axis. To build your own linear model from real-life data, find the gradient m from two known points and the intercept c, then write y = mx + c.
Conversion graphs look easy because there is no algebra to solve, only a line to read. That is exactly why careless mistakes cost marks: misreading a scale that jumps in twos instead of ones, forgetting the axis does not start at zero, or rounding a reading when the question wants an exact value from the equation instead.
Real-life linear model questions are harder because they ask you to build the equation first, often from a word problem such as a taxi fare or a phone plan, before you can answer the question at all. Students who skip straight to guessing numbers lose marks that a clean y = mx + c would have secured.
Step 1
Check what each axis represents and its scale
Step 2
Locate the given value on one axis
Step 3
Trace to the line, then to the other axis
Step 4
Read off the converted value
A straight-line graph converts Singapore dollars (x-axis) into Malaysian ringgit (y-axis). The x-axis runs from 0 to 40, and the line passes through the labelled points (0, 0) and (20, 68).
Solution: use the graph to convert S$35 to ringgit
m = 68 ÷ 20 = 3.4 ringgit per dollar.x = 35 on the S$ axis, trace up to the line, then trace across to the RM axis.35 × 3.4 = 119.Quick check: if the graph passes through the origin, the rate is simply y ÷ x from any labelled point. Use that rate to work out any reading precisely, instead of squinting at gridlines.
A taxi charges a flat booking fee plus a fixed rate per kilometre. A 5 km trip costs $9.50 and a 12 km trip costs $17.90. Find the equation connecting the fare $C and the distance d km, then find the fare for a 20 km trip.
Solution
(5, 9.50) and (12, 17.90).m = (17.90 − 9.50) ÷ (12 − 5) = 8.40 ÷ 7 = 1.20, so the taxi charges $1.20 per kilometre.C = md + c to find c, the flat booking fee: 9.50 = 1.20 × 5 + c, so 9.50 = 6.00 + c, giving c = 3.50.C = 1.2d + 3.5. Check it against the other point: 1.2 × 12 + 3.5 = 14.40 + 3.50 = 17.90 ✓.d = 20: C = 1.2 × 20 + 3.5 = 24.00 + 3.50 = 27.50. The fare is $27.50.The step students get wrong
Students often plug a data point straight into y = mx + c and forget that c here is a real-world quantity, such as a flat booking fee or a starting charge, not just leftover algebra. Always name what m and c mean in the context of the question ("cost per km" and "flat fee") before substituting numbers back in. This also lets you sanity-check the answer: a negative flat fee or a negative rate per km usually means an arithmetic slip.
Many exam graphs use a broken or non-zero starting axis to save space, and the scale on each axis is often different, for example every gridline on the y-axis worth 5 units while every gridline on the x-axis is worth 2 units. Always read the axis labels and count gridlines carefully before taking any reading, rather than assuming each small square is worth 1 unit.
Exam tip: when a question gives you a conversion graph and asks for a value beyond the drawn range (extrapolation), use the gradient and intercept from the graph to calculate it with the equation, since you cannot read a point that is not on the page.
What is the difference between a conversion graph and a distance-time graph?
A conversion graph turns one unit into another using a fixed rate, such as dollars into ringgit, and usually passes through the origin. A distance-time graph plots how a quantity changes over time, and its gradient represents speed rather than a conversion rate.
How do I find the equation of the line shown on a conversion graph?
Pick any two clearly labelled points on the line, find the gradient m between them, then substitute one point into y = mx + c to find c. This is the same process used for any equation of a straight line question, just applied to a real-life context.
Do I always read values straight off the graph, or can I calculate them?
If the question says "use the graph," you should trace and read a value directly, and small rounding is accepted. If it asks you to "calculate" or gives you an equation to find first, use the equation, since it gives an exact answer that a hand-drawn reading cannot match.
Why does my flat fee (the value of c) come out negative?
A negative c is possible in pure algebra but rarely makes sense in a real-life model like a taxi fare or phone bill. If your calculated flat fee is negative, recheck your gradient and your substitution: it is more likely an arithmetic error than a genuine feature of the situation.
— Mr Gan Math Tuition
Mr. Gan helps students turn word problems into a clean linear equation, not a guess from the gridlines.
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