, 2014

Solution 1

Each of the 200 passengers who checked exactly one bag is charged $20 to do so.

Each of the 45 passengers who checked exactly two bags is charged 20 for the first bag plus 7 for the second bag, or $27 in total for the two bags.

Thus, the total charge for all checked bags is $(200\times\$20)+(45\times\$27)$ or $5215.

Solution 2

All 245 passengers checked at least one bag.

They were each charged $20 to check this first bag.

The 45 passengers who checked a second bag were each charged an additional $7 to do so.

Thus, the total charge for all checked bags is $(245\times\$20)+(45\times\$7)$ or $5215.
Solution 1

Since each of the 245 passengers checked at least one bag, then the total baggage fees collected for the first bag is $245\times \$20=\$4900$.

A total of $\$5173-\$4900=\$273$ in baggage fees remains to be collected.

Since all passengers checked exactly one or exactly two bags, then the remaining $273 in baggage fees is collected from the passengers who checked a second bag.

The cost to check a second bag is $7.

Thus, the number of passengers who checked exactly two bags is $\frac{273}{7}=39$.

Solution 2

Let the number of passengers who checked exactly one bag be $n$.

Since there were 245 passengers on board, and each checked exactly one bag or exactly two bags, then the remaining $(245-n)$ passengers checked exactly two bags.

Each of the $n$ passengers who checked exactly one bag is charged $20 to do so.

Each of the $(245-n)$ passengers who checked exactly two bags is charged 20 for the first bag plus 7 for the second bag, or $27 in total for the two bags.

Since the total charge for all checked bags is $5173, then$$(n\times20)+((245-n)\times27)=5173$$. Solving,$20n+6615-27n=5173$or$1442=7n$, and so$n=206$.

That is, $245-n=245-206=39$ passengers checked exactly two bags.

Solution 3

All 245 passengers checked at least one bag.

They were each charged $20 to check this first bag.

Let the number of passengers who checked exactly two bags be $m$.

The $m$ passengers who checked a second bag were each charged an additional $7 to do so.

Thus, the total charge for all checked bags is $(245\times\$20)+(m\times\$7)$, so $4900+7m=5173$ or $7m=273$ and $m=39$. Therefore, 39 passengers checked exactly two bags.
Assume that each of the 245 passengers checked at most two bags.

The charge to check exactly two bags is 27, so in this case the total baggage fees collected could not have exceeded $245\times\$27=\$6615\$.$

Since $6825 (which is greater than$6615) was collected in baggage fees on this third flight, then at least one passenger must have checked at least three bags.

(It is possible to have baggage fees total $6825 if 215 passengers check exactly 2 bags, and 30 passengers check exactly 3 bags.

Here, the total baggage fees collected would be $(215\times\$27)+(30\times\$34)= \$6825$.)
Assume that each passenger (of which there are at most 245), checked at most two bags.

Let the number of passengers who checked exactly one bag be $a$ and the number of passengers who checked exactly two bags be $b$.

While it may be the case that there are passengers who checked no bags, they don’t contribute to the $142 collected and so we may ignore them.

Each of the $a$ passengers who checked exactly one bag is charged $20, while each of the$b passengers who checked exactly two bags is charged 27.

Since the total fees collected was $142, then$20a+27b=142$.

Solving for $a$ we get, $a=\frac{142-27b}{20}$ and since both $a$ and $b$ must be non-negative integers,

we systematically try values for $b$ in the table below to see if any gives a non-negative integer value for $a$. Since $27b$ is at most 142, but $27(5)=135$ and $27(6)=162$, then $b$ is at most 5 ($27b$ is larger than 162 when $b$ is larger than 6).

Value of b
Calculation of a

0
$a=\frac{142-27(0)}{20}=7.1$

1
$a=\frac{142-27(1)}{20}=5.75$

2
$a=\frac{142-27(2)}{20}=4.4$

3
$a=\frac{142-27(3)}{20}=3.05$

4
$a=\frac{142-27(4)}{20}=1.7$

5
$a=\frac{142-27(5)}{20}=0.35$

Each of the values of $a$ calculated above is not a non-negative integer.

Thus, there are no non-negative integers $a$ and $b$ that make $20a+27b=142$.

Therefore, there is no combination of passengers who check at most two bags such that the baggage fees collected total $142.

Therefore, there must be at least one passenger who checked at least 3 bags.

(It is possible to have baggage fees total 142 if 4 passengers check exactly 2 bags, and 1 passenger checks exactly 3 bags. Here, the total baggage fees collected would be $(4\times\$27)+(1\times\$34)=\$142\$.)$