2009-I-15

1. The taxi fare

   $\begin{array}{cl}
   = & 30 + \dfrac{x – 2}{0.2} \times 2.4 \\
   = & 30 + 12 x – 24 \\
   = & \$ (6 + 12x) \\
   \end{array}$

2. No, it is because the taxi fare calculated by $\$(6+12x)$ is base on the distance travelled rounding up to the nearest $0.2\text{ km}$, not the actual distance travelled.

$\begin{array}{cl}
= & 6 + 12(3.2) \\
= & \$44.4
\end{array}$

The distance covered by the 99th journey

$\begin{array}{cl}
= & 3.1 + (99 – 1)(0.5) \\
= & 52.1\text{ km}
\end{array}$

Hence, the taxi fare for the 99th journey

$\begin{array}{cl}
= & 6 + 12 (52.2) \\
= & \$632.4
\end{array}$

Therefore, the total taxi fare for the 1st, the 3rd, …, the 99th journey

$\begin{array}{cl}
= & \dfrac{(44.4 + 632.4)(50)}{2} \\
= & \$16~920
\end{array}$

Note that the distance covered by the 2nd, the 4th, …, the 98th journey are multiples of $0.2\text{ km}$, then the distances need not be rounded up. Note also that the taxi fare for the 2nd, the 4th, …, the 98th journey form an arithmetic sequence.

The taxi fare for the 2nd journey

$\begin{array}{cl}
= & 6 + 12 (3.6) \\
= & \$49.2
\end{array}$

The distance cover by the 98th journey

$\begin{array}{cl}
= & 3.1 + (98 – 1)(0.5) \\
= & 51.6\text{ km}
\end{array}$

Hence, the taxi fare for the 98th journey

$\begin{array}{cl}
= & 6 + 12 ( 51.6) \\
= & \$625.2
\end{array}$

Therefore, the total taxi fare for the 2nd, the 4th, …, the 98th journey

$\begin{array}{cl}
= & \dfrac{(49.2 + 625.2)49}{2} \\
= & \$ 16~522.8
\end{array}$

Hence, the total taxi fare

$\begin{array}{cl}
= & 16~920 + 16~522.8 \\
= & \$ 33~442.8 \\
> & \$ 33~000
\end{array}$

Therefore, the claim is not correct.