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Maths in science

GCSE · AQA

A fixed share of your GCSE science marks is maths — at least 10% in Biology, 20% in Chemistry and 30% in Physics. These are the skills, each drilled inside a real science question. Read the method and the worked example, then try the practice and reveal the answers.

Arithmetic

Standard form

Writing very large or very small numbers as A x 10n, where A is between 1 and 10.

How to do it

  1. 1Move the decimal point so there is one non-zero digit in front of it — that gives A.
  2. 2Count how many places the point moved; that number is the power n.
  3. 3Moving the point left gives a positive power (large numbers); moving it right gives a negative power (small numbers).

Worked example

A red light has a wavelength of 0.00000070 m. Write it in standard form.

  1. Put one non-zero digit before the point: 7.0
  2. The point moved 7 places to the right, so n = -7.
  3. Answer: 7.0 x 10−7 m

Your turn

  • One mole contains 602,000,000,000,000,000,000,000 particles. Write this in standard form.

  • Write the wavelength 0.0000045 m in standard form.

  • Write 3.2 x 104 as an ordinary number.

Comes up in: Atom and wavelength sizes, the Avogadro constant, tiny masses.

Significant figures

The digits in a number that carry meaning. Answers are usually given to 2 or 3 significant figures, and never to more figures than the data.

How to do it

  1. 1Count significant figures from the first non-zero digit.
  2. 2To round to a number of significant figures, look at the next digit: 5 or more rounds up, less than 5 rounds down.
  3. 3Keep place-holding zeros so the number stays the right size.

Worked example

A current of 0.04763 A is measured. Give it to 2 significant figures.

  1. The first significant figure is 4 (leading zeros do not count).
  2. The second is 7; the next digit is 6, so round the 7 up.
  3. Answer: 0.048 A

Your turn

  • Give 5.6741 to 3 significant figures.

  • Give 128600 to 2 significant figures.

  • How many significant figures does 0.0230 have?

Comes up in: Every calculated answer — matching the precision of your data.

Percentages

Percentage yield, percentage by mass and percentage error all use: percentage = (part / whole) x 100.

How to do it

  1. 1Decide which value is the 'part' and which is the 'whole'.
  2. 2Divide the part by the whole.
  3. 3Multiply by 100 to get a percentage.

Worked example

A reaction should make 8.0 g of product but only 6.0 g is collected. Calculate the percentage yield.

  1. Part = actual yield = 6.0 g; whole = theoretical yield = 8.0 g.
  2. 6.0 / 8.0 = 0.75
  3. 0.75 x 100 = 75%

Your turn

  • A 50 g sample of a compound contains 12 g of carbon. Calculate the percentage of carbon by mass.

  • A thermometer reads 21.0 C but the true value is 20.0 C. Calculate the percentage error (error / true value x 100).

  • The mass of water in a dish rose from 40 g to 50 g. Calculate the percentage increase.

Comes up in: Percentage yield, atom economy, percentage by mass, percentage error.

Ratios and proportion

Scaling reacting masses, dilutions and quantities. If two amounts are in proportion, multiply both by the same factor.

How to do it

  1. 1Write the ratio from the numbers you are given.
  2. 2Find the scale factor between the two known quantities.
  3. 3Multiply (or divide) the other quantity by that same factor.

Worked example

2 g of hydrogen reacts with 16 g of oxygen to make water. What mass of oxygen reacts with 5 g of hydrogen?

  1. Ratio hydrogen : oxygen = 2 : 16 = 1 : 8.
  2. So every 1 g of hydrogen needs 8 g of oxygen.
  3. 5 g x 8 = 40 g of oxygen.

Your turn

  • 5 g of salt dissolves in 100 cm3 of water. How much salt is needed for 250 cm3 at the same concentration?

  • In a compound the mass ratio of magnesium to oxygen is 3 : 2. What mass of oxygen combines with 12 g of magnesium?

Comes up in: Reacting masses, dilutions, concentration and scaling recipes.

Converting units

Getting quantities into the right units before you calculate — cm3 and dm3, J and kJ, minutes and seconds, g and kg.

How to do it

  1. 1Find the conversion factor, e.g. 1 dm3 = 1000 cm3.
  2. 2Multiply when moving to a smaller unit; divide when moving to a larger unit.
  3. 3Always convert before you substitute into a formula.

Worked example

Convert 250 cm3 into dm3.

  1. 1 dm3 = 1000 cm3.
  2. dm3 is the larger unit, so divide by 1000: 250 / 1000.
  3. = 0.25 dm3

Your turn

  • Convert 3 minutes into seconds.

  • Convert 2.5 kg into grams.

  • Convert 4000 J into kJ.

Comes up in: Concentration (mol/dm^3), energy calculations, rates and speeds.

Handling data

Mean of repeat readings

Averaging repeats to reduce random error. Mean = sum of values / number of values, after leaving out any anomaly.

How to do it

  1. 1Spot any anomalous result (one that does not fit the pattern) and leave it out.
  2. 2Add up the remaining readings.
  3. 3Divide by how many readings you added.

Worked example

A reaction is timed three times: 42 s, 44 s and 43 s. Calculate the mean time.

  1. No anomalies. Add: 42 + 44 + 43 = 129 s.
  2. Divide by 3: 129 / 3 = 43 s.
  3. Mean = 43 s.

Your turn

  • Titre readings are 12.1, 12.3, 15.0 and 12.2 cm3. Ignore the anomaly and find the mean of the rest.

  • Find the mean of 3.0, 3.5, 4.0 and 3.5 g.

Comes up in: Required practicals — averaging concordant repeats.

Algebra

Rearranging equations

Changing the subject of a formula so the quantity you want is on its own.

How to do it

  1. 1Decide which symbol you want to make the subject.
  2. 2Do the same operation to both sides to undo whatever is attached to it.
  3. 3Divide, multiply or take a root until that symbol stands alone.

Worked example

The wave equation is v = f x wavelength. Make wavelength the subject.

  1. Wavelength is multiplied by f, so divide both sides by f.
  2. v / f = f x wavelength / f
  3. wavelength = v / f

Your turn

  • Density is given by density = m / V. Make V the subject.

  • For P = I2 x R, make R the subject.

  • Rearrange v = u + a x t to make a the subject.

Comes up in: Any physics or chemistry equation where the unknown is not the subject.

Substituting into a formula

Putting numbers, in the correct units, into a formula and calculating the answer.

How to do it

  1. 1Rearrange first if the unknown is not already the subject.
  2. 2Convert every quantity to the correct unit.
  3. 3Substitute the numbers, calculate, and write the unit on the answer.

Worked example

Calculate the kinetic energy of a 2.0 kg ball moving at 3.0 m/s. Use KE = 0.5 x m x v2.

  1. Substitute: KE = 0.5 x 2.0 x 3.02
  2. 3.02 = 9.0, so KE = 0.5 x 2.0 x 9.0
  3. KE = 9.0 J

Your turn

  • Calculate the weight of a 6.0 kg mass. Use W = m x g with g = 10 N/kg.

  • Calculate the number of moles in 36 g of water. Use moles = mass / Mr, with Mr = 18.

  • Calculate the charge when a current of 3 A flows for 20 s. Use Q = I x t.

Comes up in: Every calculation question — energy, moles, forces, charge.

Graphs

Gradient of a graph

The slope of a straight-line graph — e.g. speed from a distance-time graph. Gradient = change in y / change in x.

How to do it

  1. 1Choose two points on the line that are far apart.
  2. 2Read off the change in y (up) and the change in x (across).
  3. 3Divide: gradient = change in y / change in x, and give it the right unit.

Worked example

A distance-time graph passes through (0 s, 0 m) and (4 s, 20 m). Calculate the speed.

  1. Change in distance = 20 - 0 = 20 m.
  2. Change in time = 4 - 0 = 4 s.
  3. Gradient = 20 / 4 = 5 m/s.

Your turn

  • A distance-time line runs from (2 s, 10 m) to (6 s, 30 m). Calculate the speed.

  • A velocity-time graph rises from 0 to 12 m/s in 3 s. Calculate the acceleration.

Comes up in: Speed from distance-time graphs, acceleration from velocity-time graphs, rates.

Area under a graph

The area under a velocity-time graph gives the distance travelled. Split it into rectangles and triangles.

How to do it

  1. 1Split the area under the line into rectangles and triangles.
  2. 2Area of a rectangle = base x height; area of a triangle = 0.5 x base x height.
  3. 3Add the areas together — the answer is a distance.

Worked example

A car travels at a steady 8 m/s for 5 s. Use the velocity-time graph to find the distance.

  1. The area is a rectangle: base = 5 s, height = 8 m/s.
  2. Area = 8 x 5 = 40.
  3. Distance = 40 m.

Your turn

  • On a velocity-time graph the velocity rises steadily from 0 to 10 m/s over 4 s (a triangle). Find the distance.

  • A speed-time graph shows a steady 6 m/s held for 10 s. Find the distance.

Comes up in: Distance from velocity-time graphs, energy from force-extension graphs.

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