Rates of change on graphs Worked Solutions — GCSE Maths

Fully worked, step-by-step solutions to GCSE Rates of change on graphs questions. See exactly how to solve problems on gradient as rate, units of a rate, direct proportion, intercept as fixed charge.

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.

Worked solution

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

    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 xx)

    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
1 markeasy
A garage charges for fuel at a constant price per litre. The graph of yy against xx is a straight line through the origin and through the point (20,30)(20, 30), where xx is the fuel bought, in litres, and yy is the cost of the fuel, in pounds. Work out the cost of the fuel per litre, in pounds per litre.

Worked solution

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

    rise=30−0=30\text{rise} = 30 - 0 = 30

    The cost of the fuel changes by 3030 pounds\text{pounds} between the two points.

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

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

    The fuel bought changes by 2020 litres\text{litres} over the same section.

  3. State the rate with its units

    1.5 pounds per litre1.5\text{ pounds per litre}

    The rate is 1.51.5 pounds per litre\text{pounds per litre} — a gradient is only a rate once its units are attached.

Answer
1.5 pounds per litre1.5\text{ pounds per litre}
Question 3
1 markeasy
A pool is filled from a hose at a constant rate. The graph of yy against xx is a straight line through the origin and through the point (3,45)(3, 45), where xx is the time, in hours, and yy is the volume of water in the pool, in litres. Work out the rate at which the pool is filling, in litres per hour.

Worked solution

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

    rise=45−0=45\text{rise} = 45 - 0 = 45

    The volume of water in the pool changes by 4545 litres\text{litres} between the two points.

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

    run=3−0=3\text{run} = 3 - 0 = 3

    The time changes by 33 hours\text{hours} over the same section.

  3. State the rate with its units

    15 litres per hour15\text{ litres per hour}

    The rate is 1515 litres per hour\text{litres per hour} — a gradient is only a rate once its units are attached.

Answer
15 litres per hour15\text{ litres per hour}
Question 4
1 markeasy
Sand falls from a hopper into a sack at a constant rate. The graph of yy against xx is a straight line through the origin and through the point (5,20)(5, 20), where xx is the time, in seconds, and yy is the mass of sand in the sack, in kilograms. Work out the rate at which sand is falling into the sack, in kilograms per second.

Worked solution

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

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

    The mass of sand in the sack changes by 2020 kilograms\text{kilograms} between the two points.

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

    run=5−0=5\text{run} = 5 - 0 = 5

    The time changes by 55 seconds\text{seconds} over the same section.

  3. State the rate with its units

    4 kilograms per second4\text{ kilograms per second}

    The rate is 44 kilograms per second\text{kilograms per second} — a gradient is only a rate once its units are attached.

Answer
4 kilograms per second4\text{ kilograms per second}
Question 5
1 markeasy
A cleaner is paid at a constant rate for the time worked. The graph of yy against xx is a straight line through the origin and through the point (6,54)(6, 54), where xx is the time worked, in hours, and yy is the pay, in pounds. Work out the hourly rate of pay, in pounds per hour.

Worked solution

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

    rise=54−0=54\text{rise} = 54 - 0 = 54

    The pay changes by 5454 pounds\text{pounds} between the two points.

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

    run=6−0=6\text{run} = 6 - 0 = 6

    The time worked changes by 66 hours\text{hours} over the same section.

  3. State the rate with its units

    9 pounds per hour9\text{ pounds per hour}

    The rate is 99 pounds per hour\text{pounds per hour} — a gradient is only a rate once its units are attached.

Answer
9 pounds per hour9\text{ pounds per hour}

Unlock 65 more Rates of change on graphs questions

Create a free account to work through every GCSE Rates of change on graphs question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Rates of change on graphs practice

Hard QuestionsPractice Questions

Related Ratio & Proportion topics

Algebraic proportionCompound measuresDirect and inverse proportionDividing in a ratioRatios, fractions and linear functionsInstantaneous rate of change