National 5 Mathematics: applications

Applications is the part of National 5 Maths that looks least like algebra: money growing or shrinking year on year, fractions, vectors, and the handful of statistics a candidate is expected to calculate and interpret. This set covers appreciation and depreciation over several years, reversing a percentage increase, dividing mixed numbers, adding vectors and finding a magnitude, the interquartile range, what a larger standard deviation actually tells you, reading a prediction off a line of best fit, and working out the saving in a three-for-two offer.

The percentage questions are built around the mistake that costs most marks: treating repeated percentage change as a single multiplication, so that three years of 15 per cent depreciation becomes 45 per cent off. Both of those wrong answers are offered, along with the value after stopping a year early. The statistics questions separate the three numbers that get confused with each other — the range, the median and the interquartile range are all among the options, so only the right calculation picks the right one.

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  • Calculate appreciation and depreciation over several years by compounding, not by multiplying the rate
  • Reverse a percentage increase by dividing by the multiplier
  • Divide mixed numbers by converting to improper fractions and multiplying by the reciprocal
  • Add vectors by components and find a magnitude using Pythagoras
  • Find quartiles from an ordered list and calculate the interquartile range
  • Interpret what a larger standard deviation means when two means are equal
  • Predict a value from the equation of a line of best fit
  • Work out the percentage saved on a multi-buy offer

A car is worth 12 000 pounds and depreciates by 15 per cent each year. What is its value after 3 years? — $12\,000 \times 0.85^3 = 12\,000 \times 0.614125 = 7369.50$ pounds.

Sample question

A house is valued at 180 000 pounds and appreciates by 5 per cent each year. What is its value after 2 years?

See the answer

198 450 pounds

The value is calculated using compound appreciation: $180 000 \times 1.05^2 = 180 000 \times 1.1025 = 198 450$. The option 198 000 pounds is a common error resulting from simple interest ($180 000 + 180 000 \times 0.05 \times 2$).

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