A football league's finance office displays, in standard form, the total cost of running its stadiums for one season and the number of match days that seaso...
Question 1 Report
A football league's finance office displays, in standard form, the total cost of running its stadiums for one season and the number of match days that season, shown below.
- Work out the mean cost per match day, giving your answer in standard form. (2)
- Next season, both values are multiplied by the same scale factor \(k = 5 \times 10^{-1}\). Show that the mean cost per match day is unchanged. (3)
Answer Details
Mean cost per match day is total cost divided by the number of match days; scaling both quantities by the same factor \(k\) leaves this ratio unchanged, since the scale factor appears in both the numerator and the denominator and cancels out algebraically.
(a) Dividing the total cost by the number of match days: \(\dfrac{3.6\times10^{7}}{1.5\times10^{3}}=\dfrac{3.6}{1.5}\times10^{7-3}=2.4\times10^{4}\) pounds per match day. [2 marks]
(b) Let the original cost be \(C\) and the original number of match days be \(N\), so the original mean is \(\dfrac{C}{N}\). Scaling both by \(k\) gives a new cost of \(Ck\) and new match days of \(Nk\), so the new mean is \(\dfrac{Ck}{Nk}\). The factor \(k\) cancels from the numerator and denominator, leaving \(\dfrac{Ck}{Nk}=\dfrac{C}{N}\), which is the same mean, \(2.4\times10^{4}\) pounds, as found in part (a): the mean is unchanged. [3 marks]
Scaling both the numerator and denominator of a fraction by the same non-zero factor always leaves the fraction's value unchanged; this is exactly why the mean cost per match day is unaffected here, regardless of what the actual value of \(k\) is.
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