GCSE Rates of change on graphs Practice Questions

Free GCSE Rates of change on graphs practice questions with full step-by-step worked solutions. Covers gradient as rate, units of a rate, direct proportion, intercept as fixed charge. Practise exam-style problems and check your method.

gradient as rateunits of a ratedirect proportionintercept as fixed chargerecognising graph shapesinverse proportion
GCSE Foundation70 questionsStep-by-step solutions
Question 1
1 markeasy
Water runs into an empty tank at a constant rate. The graph of yy against xx is a straight line through the origin and through the point (4,10)(4, 10), where xx is the time, in minutes, and yy is the volume of water in the tank, in litres. Work out the rate at which the tank is filling, in litres per minute.
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Worked solution

  1. Work out the rise (the change in y)

    rise=10−0=10\text{rise} = 10 - 0 = 10

    The volume of water in the tank changes by 1010 litres\text{litres} between the two points.

  2. Work out the run (the change in x)

    run=4−0=4\text{run} = 4 - 0 = 4

    The time changes by 44 minutes\text{minutes} over the same section.

  3. State the rate with its units

    2.5 litres per minute2.5\text{ litres per minute}

    The rate is 2.52.5 litres per minute\text{litres per minute} — a gradient is only a rate once its units are attached.

Answer
2.5 litres per minute2.5\text{ litres per minute}
Question 2
2 markseasy
A candle burns down at a constant rate. The graph of yy against xx is a straight line through the points (0,24)(0, 24) and (8,8)(8, 8), where xx is the time, in hours, and yy is the height of the candle, in cm. Work out the rate at which the height of the candle is decreasing, in cm per hour.
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Worked solution

  1. Work out the rise (the change in y)

    rise=8−24=−16\text{rise} = 8 - 24 = -16

    The height of the candle changes by −16-16 cm\text{cm} between the two points.

  2. Work out the run (the change in x)

    run=8−0=8\text{run} = 8 - 0 = 8

    The time changes by 88 hours\text{hours} over the same section.

  3. State the rate with its units

    2 cm per hour2\text{ cm per hour}

    The rate is 22 cm per hour\text{cm per hour} — a gradient is only a rate once its units are attached.

Answer
2 cm per hour2\text{ cm per hour}
Question 3
2 marksintermediate
The graph of yy against xx is a smooth curve showing that yy is inversely proportional to xx. The curve passes through the point (2,18)(2, 18). Use the graph to work out the value of yy when x=6x = 6.
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Worked solution

  1. Recognise the shape of the graph

    xy=kxy = k

    An inverse-proportion graph is a hyperbola: it falls steeply at first, then levels off, getting closer and closer to both axes without ever touching them. It never passes through the origin.

  2. Write down the rule for inverse proportion

    y=kx⇔xy=ky = \frac{k}{x} \quad\Leftrightarrow\quad xy = k

    For an inverse-proportion curve the product xyxy is the same at every point.

  3. Use the given point to find k

    k=2×18=36k = 2 \times 18 = 36

    Substituting the point (2,18)(2, 18) gives k=36k = 36.

  4. Write down the equation of the curve

    y=36xy = \frac{36}{x}

    Every point on this curve satisfies xy=36xy = 36.

  5. Substitute the required x-value

    y=366y = \frac{36}{6}

    Put x=6x = 6 into the equation.

  6. State the value of y

    y=6y = 6

    When x=6x = 6, the curve gives y=6y = 6.

Answer
y=6y = 6
Question 4
3 markshard
Water runs into an empty tank at a constant rate. The graph of yy against xx is a straight line through the origin and through the point (8,20)(8, 20), where xx is the time, in minutes, and yy is the volume of water in the tank, in litres. Use the graph to work out the value of xx when y=35y = 35.
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Worked solution

  1. Read what each axis shows

    x: time (minutes)y: volume of water in the tank (litres)x:\ \text{time (minutes)}\qquad y:\ \text{volume of water in the tank (litres)}

    The gradient of the line is a rate: it is measured in the yy-unit per the xx-unit, so its units here are litres per minute\text{litres per minute}.

  2. Choose two points on the line

    (0,0)and(8,20)(0, 0) \quad \text{and} \quad (8, 20)

    Any two points on a straight line give the same gradient, so use the two points that are stated exactly.

  3. Work out the rise (the change in y)

    rise=20−0=20\text{rise} = 20 - 0 = 20

    The volume of water in the tank changes by 2020 litres\text{litres} between the two points.

  4. Work out the run (the change in x)

    run=8−0=8\text{run} = 8 - 0 = 8

    The time changes by 88 minutes\text{minutes} over the same section.

  5. Divide the rise by the run

    gradient=riserun=208=2.5\text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{20}{8} = 2.5

    The gradient of the line is 2.52.5.

  6. Attach the units to the gradient

    2.5 litres per minute2.5\text{ litres per minute}

    A gradient without units is only half an answer: this rate is 2.52.5 litres per minute\text{litres per minute}.

  7. Substitute the given y-value into the equation

    35=2.5x35 = 2.5x

    Put y=35y = 35 into y=2.5xy = 2.5x and solve for xx.

  8. Solve for x

    2.5x=35⇒x=352.5=142.5x = 35 \quad\Rightarrow\quad x = \frac{35}{2.5} = 14

    Divide by the rate 2.52.5 to undo the multiplication.

  9. Check by working forwards again

    2.5×14=352.5 \times 14 = 35

    Putting x=14x = 14 back into the equation returns y=35y = 35, so the answer checks out.

  10. State the value of x

    x=14x = 14

    The volume of water in the tank reaches 3535 litres\text{litres} when the time is 1414 minutes\text{minutes}.

Answer
x=14x = 14
Question 5
5 markschallenging
A van hire firm charges a fixed booking fee plus a cost for each day of hire. The graph of yy against xx is a straight line through the points (3,95)(3, 95) and (7,155)(7, 155), where xx is the time, in days, and yy is the total hire cost, in pounds. Work out the fixed charge, in pounds.
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Worked solution

  1. Read what each axis shows

    x: time (days)y: total hire cost (pounds)x:\ \text{time (days)}\qquad y:\ \text{total hire cost (pounds)}

    The gradient of the line is a rate: it is measured in the yy-unit per the xx-unit, so its units here are pounds per day\text{pounds per day}.

  2. Choose two points on the line

    (3,95)and(7,155)(3, 95) \quad \text{and} \quad (7, 155)

    Any two points on a straight line give the same gradient, so use the two points that are stated exactly.

  3. Work out the rise (the change in y)

    rise=155−95=60\text{rise} = 155 - 95 = 60

    The total hire cost changes by 6060 pounds\text{pounds} between the two points.

  4. Work out the run (the change in x)

    run=7−3=4\text{run} = 7 - 3 = 4

    The time changes by 44 days\text{days} over the same section.

  5. Divide the rise by the run

    gradient=riserun=604=15\text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{60}{4} = 15

    The gradient of the line is 1515.

  6. Attach the units to the gradient

    15 pounds per day15\text{ pounds per day}

    A gradient without units is only half an answer: this rate is 1515 pounds per day\text{pounds per day}.

  7. Find where the line crosses the y-axis

    c=50c = 50

    The line cuts the yy-axis at 5050, so there is a fixed amount of 5050 pounds\text{pounds} before the rate is applied.

  8. Write the equation of the line

    y=15x+50y = 15x + 50

    This is y=mx+cy = mx + c with m=15m = 15 and c=50c = 50.

  9. Check the equation with the other point

    15×7+50=15515 \times 7 + 50 = 155

    Substituting x=7x = 7 gives y=155y = 155, which matches the point (7,155)(7, 155), so the gradient and intercept are right.

  10. Say what the gradient means here

    15 pounds per day\text{15 pounds per day}

    For each extra day\text{day}, the total hire cost goes up by 1515 pounds\text{pounds}.

  11. Avoid the classic error of turning the fraction upside down

    runrise=460=115\frac{\text{run}}{\text{rise}} = \frac{4}{60} = \frac{1}{15}

    Run divided by rise gives 115\frac{1}{15} day per pound\text{day per pound} — the rate the other way round, not what was asked for.

  12. Check whether the graph shows direct proportion

    yx=1557=1557\frac{y}{x} = \frac{155}{7} = \frac{155}{7}

    The line is straight but it does NOT pass through the origin, so yy is NOT directly proportional to xx — the ratio yx\frac{y}{x} changes from point to point.

  13. Use the intercept to find the fixed charge

    y=15x+c⇒95=15×3+c⇒c=50y = 15x + c \quad\Rightarrow\quad 95 = 15 \times 3 + c \quad\Rightarrow\quad c = 50

    Substituting the point (3,95)(3, 95) into y=mx+cy = mx + c gives c=50c = 50. The intercept is the FIXED charge; the gradient is the RATE.

  14. Keep the gradient and the intercept apart

    m=15 (pounds per day)c=50 (pounds)m = 15\text{ (pounds per day)}\qquad c = 50\text{ (pounds)}

    The gradient 1515 is charged for every extra day\text{day}; the intercept 5050 is paid once, whatever the time.

  15. State the fixed charge

    50 pounds50\text{ pounds}

    The fixed charge is 5050 pounds\text{pounds} — the value of yy when x=0x = 0.

Answer
50 pounds50\text{ pounds}

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