李先生和他的妻子准备去泰国进行一次家庭旅行。表格显示了个人和家庭保障方案所需支付的旅游保险保费。

1. Compare the costs if the trip is within the first 3 days.

For $n\leqslant 3$,

Individual policy for Mr Lee and his wife:
$2(\$34)=\$68$

Family policy has a $20\%$ discount:
$\$75\times 80\%=\$60$

Since $\char36{}68>\char36{}60$, the individual policy is not cheaper for $n\leqslant 3$.

2. Form expressions for the costs when the trip is more than 3 days.

For $n>3$,

Individual policy for two adults:
$2\left[34+8(n-3)\right]$
$=2(8n+10)=16n+20$

Family policy:
$75+15(n-3)$
$=15n+30$

3. Set up and solve the inequality for individual policy to be cheaper.

Individual policy is cheaper than family policy:
$16n+20<15n+30$
$n<10$

4. Combine with the condition for this case and state integer possible values of $n$.

This case is for $n>3$, and we found $n<10$.

Since $n$ is the number of days, $n$ is a positive integer.

Therefore,
$3<n<10$

So the possible values are
$n=4,5,6,7,8,9.$

1. Calculate the amount of Thai baht received for SGD \$1000 using the sell rate.

Since the money changer sells Thai baht at $4.2194 for 100 Thai baht,
$\text{Thai baht received}=\frac{1000}{4.2194}\times 100=23700.052\ldots$

2. Find the remaining Thai baht after spending 19836 Thai baht.

$\text{Remaining Thai baht}=23700.052\ldots-19836=3864.052\ldots$

3. Convert the remaining Thai baht back to SGD using the buy rate.

Since the money changer buys Thai baht at $3.5002 for 100 Thai baht,
$\text{SGD received}=\frac{3864.052\ldots}{100}\times 3.5002$
$=135.249\ldots$
$\approx 135.25$