A taxi meter records a journey distance as \(8.4\) km, correct to the nearest \(0.1\) km. The taxi firm charges a fixed \(\pounds 2.50\) plus \(\pounds 1.20...
Question 1 Report
A taxi meter records a journey distance as \(8.4\) km, correct to the nearest \(0.1\) km. The taxi firm charges a fixed \(\pounds 2.50\) plus \(\pounds 1.20\) per km travelled.
- Write down the lower and upper bounds of the actual distance travelled. (1)
- Work out the greatest possible fare for this journey. (2)
Answer Details
Because the distance is only known correct to the nearest \(0.1\) km, it has its own error interval; substituting the upper bound of that interval into the fare formula gives the greatest possible fare for the journey.
(a) Rounding to the nearest \(0.1\) km, the true distance lies within \(0.05\) km of \(8.4\): the error interval is \(8.35\leq d\lt8.45\). [1 mark]
(b) Since the fare increases with distance, the greatest possible fare uses the upper bound of the distance: \(2.50+1.20\times8.45=2.50+10.14=\pounds12.64\). [2 marks]
Because the fare formula adds a rate per km to a fixed charge, and both terms increase (or stay the same) as distance increases, using the distance's upper bound is guaranteed to give the fare's upper bound too.
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