Hard GCSE Gradient and intercepts Questions

Challenging, exam-style GCSE Gradient and intercepts questions with worked solutions. Stretch yourself on the hardest equation of a line, evaluating, gradient and intercept, comparing linear models problems.

equation of a lineevaluatinggradient and interceptcomparing linear modelssolving equationsconversion graph
GCSE Foundation34 questionsStep-by-step solutions
Question 1
6 markschallenging
A hot drink cools so that its temperature is T=80−4tT = 80 - 4t degrees Celsius after tt minutes. Work out how many minutes it takes for the drink to cool to 2020 degrees Celsius.
Show worked solution

Worked solution

  1. Read the formula

    T=80−4tT = 80 - 4t

    The drink starts at 80°C and cools as time passes.

  2. Interpret the intercept

    80=start temperature80 = \text{start temperature}

    At t=0t = 0 the drink is 80∘80^\circC.

  3. Interpret the gradient

    −4=°C per minute-4 = \text{°C per minute}

    The temperature falls by 4°C each minute.

  4. Note the sign

    negative⇒cooling\text{negative} \Rightarrow \text{cooling}

    A negative gradient means the temperature is dropping.

  5. Set the temperature to 2020

    20=80−4t20 = 80 - 4t

    We want the time when T=20∘T = 20^\circC.

  6. Subtract 8080 from both sides

    20−80=−4t20 - 80 = -4t

    Move the starting temperature across.

  7. Simplify the left side

    −60=−4t-60 = -4t

    Twenty minus eighty is minus sixty.

  8. Divide by −4-4

    t=−60−4t = \frac{-60}{-4}

    Dividing two negatives gives a positive.

  9. Work it out

    t=15t = 15

    Sixty divided by four is fifteen.

  10. Add units

    15 minutes15 \text{ minutes}

    It takes 15 minutes to cool to 20°C.

  11. Check the start

    80−4(0)=8080 - 4(0) = 80

    At the start the drink is 80°C.

  12. Check the answer

    80−4(15)=2080 - 4(15) = 20

    After 15 minutes the temperature is 20°C.

  13. How much it has cooled

    80−20=6080 - 20 = 60

    The drink has cooled by 60°C in total.

  14. Relate to the rate

    604=15\frac{60}{4} = 15

    Cooling 60°C at 4°C per minute takes 15 minutes.

  15. State the answer

    15 minutes15 \text{ minutes}

    The drink reaches 20°C after 15 minutes.

Answer
1515 minutes
Question 2
6 markschallenging
A phone is charging. Its battery level is B=20+8tB = 20 + 8t percent after tt minutes. Work out how many minutes it takes to reach 100%100\%.
Show worked solution

Worked solution

  1. Read the formula

    B=20+8tB = 20 + 8t

    The battery starts at 2020% and rises with time.

  2. Interpret the intercept

    20=start level20 = \text{start level}

    At t=0t = 0 the battery is at 2020%.

  3. Interpret the gradient

    8=per minute8 = \text{per minute}

    The battery gains 88% each minute.

  4. Full charge means B=100B = 100

    100=20+8t100 = 20 + 8t

    Set the battery level to 100100%.

  5. Subtract 2020 from both sides

    100−20=8t100 - 20 = 8t

    Move the starting level across.

  6. Simplify the left side

    80=8t80 = 8t

    One hundred minus twenty is eighty.

  7. Divide by 88

    t=808t = \frac{80}{8}

    Isolate t.

  8. Work it out

    t=10t = 10

    Eighty divided by eight is ten.

  9. Add units

    10 minutes10 \text{ minutes}

    It takes 1010 minutes to reach full charge.

  10. Check the start

    20+8(0)=2020 + 8(0) = 20

    At the start the battery is 2020%.

  11. Check the end

    20+8(10)=10020 + 8(10) = 100

    After 1010 minutes the battery is 100100%.

  12. Read it from the graph

    line reaches 100\text{line reaches } 100

    The charging line hits 100100% at t=10t = 10.

  13. What if it started emptier?

    lower intercept\text{lower intercept}

    A lower start would need more time to reach 100100%.

  14. What if it charged faster?

    steeper line\text{steeper line}

    A bigger gradient would reach 100100% sooner.

  15. State the answer

    10 minutes10 \text{ minutes}

    The phone reaches 100100% after 1010 minutes.

Answer
1010 minutes
Question 3
6 markschallenging
Work out the equation of the straight line that is parallel to y=2x+5y = 2x + 5 and passes through the point (1,4)(1, 4).
Show worked solution

Worked solution

  1. Read the gradient of the given line

    y=2x+5⇒m=2y = 2x + 5 \Rightarrow m = 2

    The coefficient of x is 2.

  2. Parallel lines have equal gradients

    m=2m = 2

    A parallel line has the same gradient.

  3. Start y=mx+cy = mx + c for the new line

    y=2x+cy = 2x + c

    Only the intercept can differ.

  4. Substitute the point (11, 44)

    4=2(1)+c4 = 2(1) + c

    The new line passes through (11, 44).

  5. Work out 22 ×\times 11

    4=2+c4 = 2 + c

    Two times one is two.

  6. Solve for c

    c=2c = 2

    Subtract 22 from both sides.

  7. Write the equation

    y=2x+2y = 2x + 2

    Put the gradient and new intercept together.

  8. Compare the intercepts

    5 vs 25 \text{ vs } 2

    The lines have different y-intercepts, so they never meet.

  9. Check the point

    2(1)+2=42(1) + 2 = 4

    At x=1x = 1 the equation gives 44, matching (11, 44).

  10. Confirm they are parallel

    both gradient 2\text{both gradient } 2

    Equal gradients mean the lines are parallel.

  11. Find the new line's y-intercept

    (0,2)(0, 2)

    The new line crosses the y-axis at 22.

  12. Find the given line's y-intercept

    (0,5)(0, 5)

    The original crosses the y-axis at 55.

  13. Note the vertical gap

    5−2=35 - 2 = 3

    The lines are always 33 units apart vertically.

  14. Interpret the gradient

    up 2 per 1\text{up } 2 \text{ per } 1

    Both lines rise 22 for every 11 across.

  15. State the equation

    y=2x+2y = 2x + 2

    This is the parallel line through (11, 44).

Answer
y=2x+2y = 2x + 2
Question 4
6 markschallenging
A gym charges a flat £3030 per month for unlimited visits. A pay-as-you-go option costs £55 per visit. Work out the number of visits per month for which the two options cost the same.
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Worked solution

  1. Write the membership cost

    C=30C = 30

    The membership is a flat £3030, whatever the number of visits.

  2. This line is horizontal

    gradient 0\text{gradient } 0

    The cost does not change with visits, so the gradient is 00.

  3. Write the pay-as-you-go cost

    C=5vC = 5v

    Each visit costs £5, so v visits cost 5v pounds.

  4. Interpret the pay-as-you-go gradient

    5=per visit5 = \text{per visit}

    The gradient is the cost of one visit.

  5. Interpret its intercept

    00

    With no visits it costs nothing.

  6. Set the costs equal

    5v=305v = 30

    The options cost the same at the crossover.

  7. Divide by 55

    v=305v = \frac{30}{5}

    Isolate v.

  8. Work it out

    v=6v = 6

    Thirty divided by five is six.

  9. State the crossover

    6 visits6 \text{ visits}

    At 66 visits both options cost £3030.

  10. Check the membership cost

    C=30C = 30

    The membership is £3030 for 66 visits.

  11. Check the pay-as-you-go cost

    5×6=305 \times 6 = 30

    Six visits pay-as-you-go is also £3030.

  12. Compare few visits

    at 4,pay=20\text{at } 4, \text{pay} = 20

    Under 66 visits, pay-as-you-go is cheaper.

  13. Compare many visits

    at 10,pay=50\text{at } 10, \text{pay} = 50

    Over 66 visits, membership is cheaper.

  14. Advise a frequent visitor

    membership\text{membership}

    Someone visiting often should take the membership.

  15. State the answer

    6 visits6 \text{ visits}

    The two options cost the same at 66 visits.

Answer
66 visits
Question 5
6 markschallenging
A car travels a steady 150150 km in 2.52.5 hours. Work out its speed in metres per minute.
Show worked solution

Worked solution

  1. Speed is the gradient

    speed=distancetime\text{speed} = \frac{\text{distance}}{\text{time}}

    On a distance–time graph the gradient is the speed.

  2. Find the speed in km/h

    1502.5\frac{150}{2.5}

    Distance in km over time in hours.

  3. Work it out

    =60= 60

    150150 divided by 2.52.5 is 6060.

  4. State the speed in km/h

    60 km/h60 \,\mathrm{km/h}

    The car travels at 6060 km/h.

  5. Change kilometres to metres

    60 km=60000 m60 \,\mathrm{km} = 60000 \,\mathrm{m}

    One kilometre is 10001000 metres.

  6. So per hour

    60000 m per hour60000 \text{ m per hour}

    The car covers 60000 m each hour.

  7. Change hours to minutes

    1 hour=60 minutes1 \text{ hour} = 60 \text{ minutes}

    There are 6060 minutes in an hour.

  8. Divide by 6060

    6000060\frac{60000}{60}

    Metres per hour divided by 6060 gives metres per minute.

  9. Work it out

    =1000= 1000

    Sixty thousand divided by sixty is one thousand.

  10. State the speed in m/min

    1000 m per minute1000 \text{ m per minute}

    The car travels 10001000 metres each minute.

  11. Check the reasoning

    60 km/h=1 km/min60 \,\mathrm{km/h} = 1 \,\mathrm{km/min}

    6060 km in 6060 minutes is 11 km each minute.

  12. Relate to metres

    1 km=1000 m1 \,\mathrm{km} = 1000 \,\mathrm{m}

    1 km per minute is 1000 m per minute.

  13. Sense check with distance

    1000×150=1500001000 \times 150 = 150000

    In 150150 minutes it would go 150150 km, matching the trip length in minutes.

  14. Note the gradient meaning

    steeper⇒faster\text{steeper} \Rightarrow \text{faster}

    A faster car would give a steeper distance–time line.

  15. State the answer

    1000 m per minute1000 \text{ m per minute}

    The car's speed is 1000 metres per minute.

Answer
10001000 metres per minute

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