Two taxi ranks: flagfall versus rate per kilometre
Two taxi ranks serve a school. Rank A charges a flagfall of P8.00 plus P1.20 for every kilometre travelled. Rank B charges a flagfall of P5.00 plus P1.50 per kilometre. Both charge for the whole trip, and up to four passengers may share one fare. Teachers usually make a 6 km trip to town, but the sports team's trip to the district stadium is 15 km. After 20:00, Rank A adds a flat P6.00 to the fare, while Rank B doubles the whole fare. A parent tells the sports team, 'Use Rank B - the flagfall is P3 cheaper, so it is always cheaper.'
1. Write each rank's daytime fare as an equation in the distance d, find where the fares are equal, and price the 6 km and 15 km trips.
Read the worked answer
Rank A: F = 8 + 1.2d Rank B: F = 5 + 1.5d
Equal fares: 8 + 1.2d = 5 + 1.5d → 3 = 0.3d → d = 10 km, and both charge 8 + 12 = 5 + 15 = P20.
- 6 km: A = 8 + 7.2 = P15.20; B = 5 + 9 = P14.00 → B is cheaper by P1.20.
- 15 km: A = 8 + 18 = P26.00; B = 5 + 22.5 = P27.50 → A is cheaper by P1.50.
So the answer depends on distance: B wins for trips under 10 km, the fares are identical at exactly 10 km, and A wins beyond 10 km. Rank B's lower flagfall buys a head start of P3, but it loses P0.30 of that advantage on every kilometre, and after 10 km the head start is gone.
2. Describe and compare the two graphs, and refute the parent's claim that Rank B is always cheaper.
Read the worked answer
Both graphs are straight lines with fare on the vertical axis and distance on the horizontal axis. Rank A starts at 8 on the vertical axis and climbs P1.20 per km; Rank B starts at 5 and climbs P1.50 per km. B's line begins lower but is steeper; A's line begins higher but is flatter. They cross at the point (10, 20).
- For 0 < d < 10, B's line is below A's: B is cheaper.
- At d = 10 the fares are equal.
- For d > 10, A's line is below B's: A is cheaper.
The parent is wrong. The flagfall is only the fare at zero distance; it is the intercept, not the price of a journey. What a passenger actually pays depends on the intercept and the gradient multiplied by the distance travelled. The claim survives only for trips shorter than 10 km. Numbers to give: at 15 km A saves P1.50; at 30 km A = 8 + 36 = P44 while B = 5 + 45 = P50, a saving of P6.00. The longer the trip, the more the steeper rate costs, so 'cheaper flagfall' and 'cheaper trip' are different statements.
3. The sports team's 12 km return trip is made after 20:00. Write the night fare of each rank as a function of d, price the trip, and advise the team. Then say what happens if four learners each make a 6 km daytime trip separately.
Read the worked answer
Night functions:
- Rank A adds a flat P6: F = 8 + 1.2d + 6 = 14 + 1.2d
- Rank B doubles everything: F = 2(5 + 1.5d) = 10 + 3d
At d = 12: A = 14 + 14.4 = P28.40; B = 10 + 36 = P46.00. Rank A is cheaper by P17.60 - the opposite of the parent's advice. The reason is that B doubles the per-kilometre part too, while A adds a fixed amount, so doubling hurts B much more on a long trip.
Night break-even: 14 + 1.2d = 10 + 3d → 4 = 1.8d → d ≈ 2.22 km. Beyond about 2.2 km, A is cheaper at night.
Sharing: one fare covers up to four passengers, so with Rank A the team pays 28.40 ÷ 4 = P7.10 each; with Rank B it would be 46.00 ÷ 4 = P11.50 each. Take Rank A and share.
But: four separate daytime 6 km trips give B = 4 × 14.00 = P56.00 against A = 4 × 15.20 = P60.80, so B wins by P4.80 when nobody shares. Sharing the fare, not the flagfall, is what decides these questions.