Mathematics practice questions.

A cyclist's mean power output over a 20 min time trial rises from 250 W to 280 W after an eight-week training block. Calculate the percentage increase in her mean power output.

• 10.7 %
• 12.0 %
• 30.0 %
• 1.12 %

Answer: 12.0 %

Percentage change divides the change by the ORIGINAL value: (280 - 250) / 250 x 100 = 12.0 %. Dividing by the final value gives 10.7 %, which is the most frequent error in data-analysis questions, and 30.0 is the absolute change in watts, not a percentage.

Two hockey squads record the same mean multi-stage fitness test score, but squad A has a standard deviation of 0.4 levels and squad B a standard deviation of 1.6 levels. Which statement about the two squads is correct?

• The scores in squad B are more widely spread
• Squad B has the higher mean aerobic fitness of the two squads
• Squad A's individual scores must have been measured more accurately
• Squad A must contain more players than squad B

Answer: The scores in squad B are more widely spread

Standard deviation reports how far individual values sit from the mean, so a larger value means a more heterogeneous squad, not a different mean and not a less accurate measurement. Spread and accuracy are separate ideas: a very precise test can still produce a widely spread set of genuine scores.

Lifter P has a body mass of 62 kg and a best back squat of 155 kg. Lifter Q has a body mass of 94 kg and a best back squat of 217 kg. Which comparison of their relative strength is correct?

• Lifter Q is relatively stronger, at 2.5 times body mass against 2.3
• The two are equally strong relative to body mass, at 2.5 times each
• Lifter P is relatively stronger, at 2.5 times body mass against 2.3
• Relative strength cannot be compared while the body masses differ

Answer: Lifter P is relatively stronger, at 2.5 times body mass against 2.3

Normalising divides the load by body mass: 155 / 62 = 2.5 and 217 / 94 = 2.3, so the lighter lifter moves the greater multiple of her own mass even though the heavier lifter moves more kilograms. Comparing absolute loads alone always flatters the heavier athlete.

An electronic timing-gate system reading to 0.01 s records 60 m sprint times of 7.42 s, 7.38 s and 7.49 s for one athlete. Which is the correctly reported mean of these three trials?

• 7.4 s
• 7.4300 s
• 7 s
• 7.43 s

Answer: 7.43 s

The mean is 22.29 / 3 = 7.43 s and is quoted to the same precision as the raw measurements, which the gates resolve to 0.01 s. Adding further decimal places claims a precision the system does not have, while rounding to 7.4 s throws away information that was actually measured.

A student converts a 70 kg runner's absolute oxygen uptake of 4.20 L min-1 into a relative value and writes: 4.20 / 70 = 0.06, so the relative uptake is 0.06 ml kg-1 min-1. Which single error has been made?

• Body mass should have been divided by the oxygen uptake instead, since a relative value is expressed per litre of uptake.
• The value should have been divided by 60 first, because the uptake was recorded per minute rather than per second.
• The value in litres was not converted to millilitres before it was divided by body mass.
• The result should have been rounded to two significant figures rather than to two decimal places.

Answer: The value in litres was not converted to millilitres before it was divided by body mass.

4.20 L min-1 is 4200 ml min-1, and 4200 / 70 = 60 ml kg-1 min-1. A relative value of 0.06 should have been rejected on sight, because figures in ml kg-1 min-1 for trained adults lie in the tens.

A swimmer's mean 200 m time falls from 130.0 s to 124.8 s. A student calculates 5.2 / 124.8 x 100 = 4.2 and reports a 4.2 % improvement. Which single error has been made?

• The two times should have been subtracted the other way round, so that the change came out as a positive number.
• A time cannot be expressed as a percentage change at all, because seconds are not a ratio quantity.
• The improvement should have been divided by 200 m, the distance over which it was recorded.
• The change was divided by the final time instead of the starting time.

Answer: The change was divided by the final time instead of the starting time.

Percentage change is always taken relative to the starting value: 5.2 / 130.0 x 100 = 4.0 %. Using the final value as the base inflates an improvement and understates a decline, so the two errors do not cancel across a season.