A taxi leaves Belby station and drives 4.2 km due east along Mill Road, then turns and drives 5.6 km due north along Fen Lane to a hotel. The taxi firm char...
Question 1 Report
A taxi leaves Belby station and drives 4.2 km due east along Mill Road, then turns and drives 5.6 km due north along Fen Lane to a hotel. The taxi firm charges a fixed fee of £2.60 for every journey, plus £1.15 for each kilometre driven. A new road has been built that runs in a straight line from the station to the hotel.
- Calculate the straight line distance from the station to the hotel. (2)
- Work out the fare for the journey along Mill Road and Fen Lane. (2)
- A second taxi drives the same customer along the new road. Show that this taxi saves the customer more than £3. (2)
Answer Details
The two roads are at right angles, since one runs due east and the other due north, so the new straight road is the hypotenuse of a right-angled triangle. The fares then follow from the distance driven.
(a) Straight line distance. [2]
\[d^2 = 4.2^2 + 5.6^2 = 17.64 + 31.36 = 49\] \[d = \sqrt{49} = 7\ \text{km}\]The result is exact: 4.2, 5.6, 7 is 1.4 times the triple 3, 4, 5.
(b) Fare along Mill Road and Fen Lane. [2]
That route covers
The fixed fee is charged once, whatever the distance.
(c) Show the saving is more than £3. [2]
Along the new road the distance is 7 km, so
Since £3.22 is more than £3, the second taxi does save the customer more than £3.
The saving comes entirely from the shorter distance: the new road is \(9.8 - 7 = 2.8\) km shorter, and \(2.8 \times 1.15 = 3.22\) pounds, which matches. The £2.60 fee is paid either way and so cancels out of the comparison.
Examination point: a "show that" comparison needs both fares and the subtraction written out. Quoting only the saving, however correct, leaves the examiner without the evidence that both fares were worked out properly.
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