Why does the coordinate read y divided by x sometimes give the correct rate?

Only when the line passes through the origin (0, 0). On such a line the starting y and x are both zero, so rise equals y2 and run equals x2, making gradient equal y2 divided by x2. This coincidence breaks down on any line that does not start at the origin.

How do you find the ratio of two pay rates from a two-section pay graph?

Find the gradient of each section separately, then form the ratio of those gradients. For a basic section from (0,0) to (8,120) the gradient is 120 divided by 8 = 15 pounds per hour. For an overtime section from (8,120) to (12,200) the gradient is 80 divided by 4 = 20 pounds per hour. The ratio of rates is 15:20 = 3:4. Forming the ratio of the end y-values (120:200 = 3:5) is wrong because it reflects both the rate and the duration, not the rate alone.

GCSE Maths Higher · AQA · Compound measures and rates

Rate from a graph is the gradient

Q: Why is the rate from a height-time graph from (0,5) to (10,35) equal to 3 cm/s and not 3.5 cm/s?

Because the rate is the gradient, which is rise divided by run: (35 minus 5) divided by (10 minus 0) = 30 divided by 10 = 3 cm/s. The container started at height 5 cm, so only 30 cm of height was added over 10 seconds. The coordinate read 35 divided by 10 = 3.5 treats the end point as if the line started at the origin, which it does not.

When a graph question asks for a rate (speed, filling rate, pay rate), the instinct is to read a coordinate: take the y-value at the end point and divide by the x-value. On a height-time graph from (0, 5) to (10, 35) that gives $35 / 10 = 3.5$ cm/s. It is wrong. The container started at height 5 cm, so only 30 cm of height was added: gradient = $(35 - 5) / (10 - 0) = 3$ cm/s.

The coordinate read coincidentally gives the right answer on a line through the origin, which masks the error. On any other line it fails: for (2, 30) to (6, 70) the coordinate read gives 70 / 6 = 11.67, but the gradient is (70 − 30) / (6 − 2) = 10. On a two-section pay graph the ratio of pay rates is the ratio of the two gradients (3:4), not the ratio of the two end y-values (3:5). These are reliable mark-losers at grades 7 to 9.

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How to spot it in your own work

You wrote rate = 35 / 10 = 3.5 by reading the end coordinate, ignoring that the line starts at height 5, not zero.
You found a gradient on an origin line and assumed the coordinate read always works, then applied it incorrectly to a non-origin line.
You formed the ratio of pay rates as 120 : 200 = 3:5 instead of finding the gradient of each section and using 15 : 20 = 3:4.
You estimated the speed at a point on a curved distance-time graph by reading the height at that point instead of computing the gradient of the tangent drawn there.

An exam question that triggers it

Here is the structure of a typical AQA Higher rate-from-graph question:

A container is being filled with water. A height-time graph shows a straight line through the points (0, 5) and (10, 35), where the horizontal axis is time in seconds and the vertical axis is height in centimetres.

Work out the rate at which the height increases.

The misconception writes $\text{rate} = 35 / 10 = 3.5$ cm/s. The fix: gradient $= (35 - 5) / (10 - 0) = 30 / 10 = 3$ cm/s.

Why students fall for this

Students learn speed = distance / time and carry a shortcut into graph questions: find the biggest y-value, divide by the corresponding x-value. This works on distance-time graphs that start at the origin (a very common classroom example), so the error goes unnoticed until a graph with a non-zero intercept appears.

The origin-line coincidence is the core trap. On a line through (0, 0), rise equals y and run equals x, so gradient = y / x. Students memorise this as a universal rule. Exam questions at grades 7 to 9 deliberately use non-origin lines, pay graphs with two sections, and curves with tangents to expose the error.

The two-section pay-graph variant is particularly deceptive: forming the ratio of the end y-values (120:200 = 3:5) feels natural because both values appear labelled on the graph, while the gradient calculation requires an extra subtraction step that students skip under exam pressure.

The fix: Always compute gradient as rise over run between two points

Step 1: identify two points on the line. Read the coordinates of two clearly labelled points. For a straight line, any two points give the same gradient.

Step 2: compute rise and run. Rise = y2 − y1. Run = x2 − x1. For (0, 5) and (10, 35): rise = 35 − 5 = 30; run = 10 − 0 = 10.

Step 3: divide. Gradient = rise / run = 30 / 10 = 3 cm/s. Do not divide the end y-value by the end x-value; that only works on a line through the origin.

Step 4 (ratio questions): find each gradient separately. For a two-section graph, compute the gradient of each section independently. Form the ratio of those gradients, not the ratio of the end y-values.

Step 5 (curves): use the tangent gradient. Draw the tangent to the curve at the required point. Read two points on the tangent line and compute its gradient. That gradient is the instantaneous rate at that point.

Worked example

Height-time line: rate from a non-origin straight-line graph.

Two points: (0, 5) and (10, 35).
Rise and run.
$\text{rise} = 35 - 5 = 30 \text{ cm}, \quad \text{run} = 10 - 0 = 10 \text{ s}$
Gradient.
$\text{gradient} = \frac{30}{10} = 3 \text{ cm/s}$
Trap: $35 / 10 = 3.5$ (coordinate read, wrong because the line starts at height 5, not zero).

Pay graph: ratio of rates from a two-section graph (JUN22 style).

Basic rate. $(0, 0) \to (8, 120)$ : $120 / 8 = 15$ pounds/h.
Overtime rate. $(8, 120) \to (12, 200)$ : $(200 - 120) / (12 - 8) = 80 / 4 = 20$ pounds/h.
Ratio of rates. $15 : 20 = 3 : 4$ . Trap: $120 : 200 = 3 : 5$ (ratio of y-values, not rates).

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Common questions

Why is the rate 3 cm/s and not 3.5 cm/s on the height-time graph from (0, 5) to (10, 35)?: Because the line does not start at the origin. At time 0 the height is already 5 cm. Only 30 cm of height is added over 10 seconds: $\text{gradient} = (35 - 5) / (10 - 0) = 30 / 10 = 3$ cm/s. The coordinate read 35 / 10 = 3.5 assumes the line starts at zero, which it does not.
When does the coordinate read y / x give the correct rate?: Only when the line passes through the origin (0, 0). On such a line y1 = 0 and x1 = 0, so rise = y2 and run = x2, making gradient = y2 / x2. This is a coincidence, not a general rule. It fails on every line with a non-zero intercept.
How do you find the ratio of pay rates on a two-section pay graph?: Find the gradient of each section separately. Basic section (0, 0) to (8, 120): $120 / 8 = 15$ pounds/h. Overtime section (8, 120) to (12, 200): $80 / 4 = 20$ pounds/h. Ratio = $15 : 20 = 3 : 4$ . The trap 120:200 = 3:5 uses total pay amounts, not hourly rates.

Related misconceptions

Average speed is not the mean of the speedsThe companion compound-measures misconception: for a multi-stage journey use total distance / total time, not the arithmetic mean of the speeds.
Density formula direction and unitsThe rearrangement error in density = mass / volume: multiplying when the formula requires dividing, and comparing rates in different units without converting first.

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